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• 1. Introduction of Quantitative Technique 1 INTRODUCTION OF QUANTITATIVE TECHNIQUE INTRODUCTION People have been using mathematical tools to help solve problems for thousands of year; however, the formal study and application of quantitative techniques to practical decision making is largely a product of the 20th century. The technique we study in quantitative analysis have been applied successfully to an increasingly wide variety of complex problem in business, government, healthcare, education, and many other areas. To get aware of the mathematics of how a particular quantitative technique works; one must also be familiar with the limitations, assumption and specific applicability of the technique. The successful use of quantitative techniques usually results in a solution that is timely, accurate, flexible, economical, reliable, an easy to understand and use. WHAT IS QUANTITATIVE TECHNIQUE? Quantitative technique is the scientific approach to managerial decision making. The approach starts with data. Like raw material for a factory, these data are manipulated or processed into information that is valuable to people making decision. This processing and manipulating of raw data into meaningful information is the heart of quantitative technique. e.g., we can use quantitative technique to determine how much our investment will be worth in the future when deposited at a bank at a given interest rate for a certain number of years. Quantitative technique can also be used in computing financial ratio for the balance sheets for several companies whose stock we are considering. DEFINITION Different mathematician have given various definition of operation research. “Operation research is a scientific method of producing executive departments with a quantitative basis for decision regarding the operation under their control” -P.M. MORSE & G.E.KIMBALL “Operation research is a act of giving bad answers to problems which otherwise have worse answers” -T.S. SAATY “Operation research is a scientific research an approach to problem solving for executive management” -H.M.WAGNER
• 2. Introduction of Quantitative Technique 2 “Operation research has been described as a method, an approach, set of techniques a team activity, a combination of many disciplines, an extension of particular disciplines (mathematics engineering economics), a new discipline, vocation, even a religion. It is perhaps some of all this things.” -S.L.COOK THE QUANTITATIVE TECHNIQUE APPROACH CHARACTERISTICS OF QUANTITATIVE TECHNIQUE 1. Flexibility:-Models should be capable of adjustment with new formulas without having any significant changes in its frame. 2. Less no of variable: - The number of variable in a model should not be very large but at the time no variable selected to any important factors should be left ones. 3. Less time: - A model should not take much time in construction. ADVANTAGES OF QUANTITATIVE TECHNIQUES 1. It describes the problem much more concisely. 2. It provides systematic and logical approach to the problem. Acquiring Input Data Testing the Solution Analyzing the Results Implementing the Results Defining the problem Developing a Model Developing a Solution
• 3. Introduction of Quantitative Technique 3 3. It indicates the scope and limitation of the problem. 4. It enables the use of high power mathematical tools to analyze the problems. 5. It helps in finding avenues for new research and improvement in a system. TECHNIQUES A brief description of some of the commonly used techniques is given below. Details are covered in relevant chapters of this book. Some of the techniques which is covered in this book are:-  LINEAR PROGRAMMING: It is an allocation technique where the objective function is linear and the constraints are modeled as linear inequalities. Example- Graphical Representation, Simplex Method etc.  TRANSPORTATION MODEL: A special case of linear programming which matches sources of supply to destinations on cost or distance considerations. For example; movement of raw materials from different sources to manufacturing plants at different locations based on availability of raw materials at various sources, the requirements at different plants and the cost of transportation involved.  ASSIGNMENT MODEL: A special case of transportation model where the aim is to assign a number of ‘origins’ to the same number of destinations at a minimum total cost. For example; assigning of man/machines to same number of jobs /tasks.  REPLACEMENT MODEL: These models deal with the formulation of appropriate replacement policies when some items have to be replaced due to obsolescence or wear and tear. Replacement models also deal with equipments and items which fail completely and instantaneously.  QUEUING THEORY: It studies the random arrivals at servicing or processing facility of limited capacity. These models attempt to predict the behavior of waiting lines, i.e.; the time spend waiting for a service. The technique is descriptive and describes behavior that can be expected given certain parameters. It is not prescriptive in nature and does not offer an optimal solution. The models deal with the tradeoffs between cost of providing service and the value of time spent in waiting for a service.  DECISION THEORY: Decision situations can be classified into deterministic or certainty, probabilistic or risk and uncertainty. Decision making under certainty can be dealt with by various optimization techniques. Decision theory deals largely with decision making under risk where the probabilities of certain conditions occurring (such as demand for an item) are predicted and various options assessed based on these probabilistic values. In situations of uncertainty there can be no specific approach. A set of decision rules can be applied and
• 4. Introduction of Quantitative Technique 4 insight gained into the decision maker’s style of functioning. This is particularly applicable to studying a competitor’s style of decision making and then predicting how he would react to a certain condition so as to gain advantage for oneself.  GAME THEORY: This deals with decision making under conditions of competition. Its assumptions currently restrict its usage. APPLICATION & SCOPE OF QUANTITATIVE TECHNIQUES Some of the industrial, government, business problems which can be analyzed by or approach have been arranged by functional areas as follows: ----- Finance and accounting:  Dividend policies, investment and portfolio management, auditing, balance sheet and cash flow analysis  Break-even analysis, capital budgeting, cost allocation and control, and financial planning  Claim and complaint procedure, and public accounting  Establishing costs for by-products and developing standards costs Marketing:  Selection of product-mix, marketing and export planning  Best time to launch a new product  Sales effort allocation and assignment  Predicting customer loyalty  Advertising, media planning, selection and effective packing alternatives Purchasing, procurement and exploration:  Optimal buying and recording under price quantity discount  Bidding policies  Transportation planning  Vendor analysis  Replacement policies Production management:  Facilities planning  Location and size of warehouse or new plant, distribution centers and retail outlets  Logistics, layout and engineering design  Transportation, planning and scheduling
• 5. Introduction of Quantitative Technique 5  Manufacturing  Aggregate production planning, assembly line, blending, purchasing and inventory control  Employment, training, layoffs and quality control  Allocating R&D budgets most effectively  Maintenance and project scheduling  Maintenance policies and preventive maintenance  Maintenance crew size and scheduling  Project scheduling and allocation of resources Personal management:  Manpower planning, wage/salary administration  Negotiation in a bargaining situation  Designing organization structures more effectively  Skills and wages balancing  Scheduling of training programmes to maximize skill development and retention Techniques and general management:  Decision support systems and MIS; forecasting  Making quality control more effective  Project management and strategic planning Government:  Economic planning, natural resources, social planning and energy  Urban and housing problems  Militar, police, pollution, control, etc. Limitation of Quantitative Techniques 1. All the problems cannot be converted into numerical values. So it is not solving by Quantitative technique. 2. Not understandable to everyone who are not aware of the technique of Quantitative technique. 3. The some of the technique of the Quantitative technique are very complex in solving 4. Mathematical model: as most of the operation research techniques are mathematical in nature. It is therefore necessary to put the problem in the term
• 6. Introduction of Quantitative Technique 6 of mathematical model. In many situations it is not possible to represent the problem in mathematical form and hence operation research techniques cannot be applied. 5. Expensive: As application of operation research for the problem solving requires service of specialist and the use of computer. Therefore it is expensive to use operation research technique for solving day to day problems. 6. Imperfection of solution: By operation research technique we cannot obtain the perfect answer to our problems. But only the quality of the solution is improved from worse to bad answer. 7. The techniques used must be applicable to the problem and must reflect the purpose and scope of the problem.
• 7. Assignment Problems 7 ASSIGNMENT PROBLEMS INTRODUCTION In the world of trade Business Organizations are confronting the conflicting need for optimal utilization of their limited resources among competing activities. The course of action chosen will invariably lead to optimal or nearly optimal results. The assignment problem is a special case of transportation problem in which the objective is to assign a number of origins to the equal number of destinations at the minimum cost (or maximum profit). Assignment problem is one of the special cases of the transportation problem. It involves assignment of people to projects, jobs to machines, workers to jobs and teachers to classes etc., while minimizing the total assignment costs. One of the important characteristics of assignment problem is that only one job (or worker) is assigned to one machine (or project). Hence the number of sources are equal the number of destinations and each requirement and capacity value is exactly one unit. Although assignment problem can be solved using either the techniques of Linear Programming or the transportation method, the assignment method is much faster and efficient. This method was developed by D. Konig, a Hungarian mathematician and is therefore known as the Hungarian method of assignment problem. In order to use this method, one needs to know only the cost of making all the possible assignments. Each assignment problem has a matrix (table) associated with it. Normally, the objects (or people) one wishes to assign are expressed in rows, whereas the columns represent the tasks (or things) assigned to them. The number in the table would then be the costs associated with each particular assignment. It may be noted that the assignment problem is a variation of transportation problem with two characteristics.(i)the cost matrix is a square matrix, and (ii)the optimum solution for the problem would be such that there would be only one assignment in a row or column of the cost matrix . Application Areas of Assignment Problem Though assignment problem finds applicability in various diverse business situations, we discuss some of its main application areas: (i) In assigning machines to factory orders. (ii) In assigning sales/marketing people to sales territories. (iii) In assigning contracts to bidders by systematic bid-evaluation. (iv) In assigning teachers to classes. (v) In assigning accountants to accounts of the clients.
• 8. Assignment Problems 8 Format of Assignment Problems In assigning police vehicles to patrolling area. Job Persons j1 j2 ------ jn I1 X11 X12 ----- X1n I2 X21 X22 ----- X2n --- ----- ---- ----- ----- In ----- ---- ----- Xnn Cij is the cost of performing jth job by ith worker. Xij is the number ith individual assigned to jth job. Total cost = X11 * C11 + X12 * C12 + ----- + Xnn * Cnn. Solution Methods There are various ways to solve assignment problems. Certainly it can be formulated as a linear program, and the simplex method can be used to solve it. In addition, since it can be formulated as a network problem, the network simplex method may solve it quickly. However, sometimes the simplex method is inefficient for assignment problems (particularly problems with a high degree of degeneracy). The Hungarian Method used with a good deal of success on these problems and is summarized as follows. Step 1. Determine the cost table from the given problem. (i) If the no. of sources is equal to no. of destinations, go to step 3. (ii) If the no. of sources is not equal to the no. of destination, go to step 2. Step 2. Add a dummy source or dummy destination, so that the cost table becomes a square matrix. The cost entries of the dummy source/destinations are always zero. Step 3. Locate the smallest element in each row of the given cost matrix and then subtract the same from each element of the row. Step 4. In the reduced matrix obtained in the step 3, locate the smallest element of each column and then subtract the same from each element of that column. Each column and row now have at least one zero. Step 5. In the modified matrix obtained in the step 4, search for the optimal assignment as follows:
• 9. Assignment Problems 9 (a) Examine the rows successively until a row with a single zero is found. In rectangle this row (�) and cross off (X) all other zeros in its column. Continue in this manner until all the rows have been taken care of. (b) Repeat the procedure for each column of the reduced matrix. (c) If a row and/or column has two or more zeros and one cannot be chosen by inspection then assign arbitrary any one of these zeros and cross off all other zeros. Step 6. If the number of assignment (�) is equal to n (the order of the cost matrix), an optimum solution is reached. If the number of assignment is less than n(the order of the matrix), go to the next step. Step7. Draw the minimum number of horizontal and/or vertical lines to cover all the zeros of the reduced matrix. Step 8. Develop the new revised cost matrix as follows: (a) Find the smallest element of the reduced matrix not covered by any of the lines. (b) Subtract this element from all uncovered elements and add the same to all the elements laying at the intersection of any two lines. Step 9. Go to step 6 and repeat the procedure until an optimum solution is attained.
• 10. Assignment Problems 10 See diagrammatic Representation of Hungarian Approach
• 11. Assignment Problems 11 MINIMIZATION PROBLEM (BALANCED) Example 1: A Company has 5 machines and 5 jobs. The relevant cost matrix is given below: Find the assignment that minimizes the total cost: Machines Jobs J1 J2 J3 J4 J5 M1 10 4 5 3 11 M2 13 11 9 12 10 M3 12 3 10 1 9 M4 9 1 11 4 8 M5 8 6 7 3 10 , Solutions: Step 1: First we have to check that the given matrix is square matrix or not. Here the given matrix is square matrix. Step 2: Subtract least entry of each row from all the entries of that row. The first reduced cost matrix will be as given below: J1 J2 J3 J4 J5 M1 7 1 2 0 8 M2 4 2 0 3 1 M3 11 2 9 0 8 M4 8 0 10 3 7 M5 5 3 4 0 7 Then in the above matrix subtract least entry of each column from all entries of that column. The second reduced cost matrix or the total opportunity cost matrix will be as follows: J1 J2 J3 J4 J5 M1 3 1 2 0 7 M2 0 2 0 3 0 M3 7 2 9 0 7 M4 4 0 10 3 6 M5 1 3 4 0 6
• 12. Assignment Problems 12 Step3: For testing the optimality we draw minimum number of straight lines to cover all the zeros. Since three lines cover all the zeros, which is not equal to the matrix size (number of row= number of column≠ number of draw line). Step4: Select the smallest entry from all the entries which are not covered by a straight line (here it is 1). Subtract this smallest entry from all the uncovered entries and add it to all those entries which are at the intersection of two lines and other covered entries remain unchanged. The revised reduced cost matrix is given below: J1 J2 J3 J4 J5 M1 2 0 1 0 6 M2 0 2 0 4 0 M3 6 1 8 0 6 M4 4 0 10 4 6 M5 0 2 3 0 5 Again we see that only four straight lines are required to cover all the zeros of the revised cost matrix, therefore optimal assignment cannot be made at this stage. Repeating the step 4 we get the following revised cost matrix: J1 J2 J3 J4 J5 M1 1 0 0 0 5 M2 0 3 0 5 0 M3 5 1 7 0 5 M4 3 0 9 4 5 M5 0 3 3 1 5 Since 5 straight lines (equal to the number of rows and columns) are required to cover all the zeros, optimal assignment can be made at this stage. It is given below:
• 13. Assignment Problems 13 J1 J2 J3 J4 J5 M1 1 0 0 5 M2 0 3 0 5 M3 5 1 7 5 M4 3 9 4 5 M5 3 3 1 5 Step5: Select one row containing exactly one zero and surrounded it by Here we select row no. 3. We can also select 4 or 5. Cut all the zeros of that column (If has). Machine1 – Job3 Cost Rs 5 Machine2 – Job5 Cost Rs 10 Machine3 – Job4 Cost Rs 1 Machine4 – Job2 Cost Rs 1 Machine5 – Job1 Cost Rs 8 Total 25 Q.1) Unbalanced Minimization: A B C D E F I 17 25 11 08 16 31 II 23 13 44 16 19 17 III 32 19 31 28 12 25 IV 26 24 27 21 29 07 V 28 21 19 45 23 43 Ans. (STEP-I) The number of Column are less than number of Row, therefore we add one Dummy in Column with relative cost zero. The assignment problem is given below: 0 0 0 0 0
• 14. Assignment Problems 14 Balancing: A B C D E F I 17 25 11 08 16 31 II 23 13 44 16 19 17 III 32 19 31 28 12 25 IV 26 24 27 21 29 07 V 28 21 19 45 23 43 Dummy 0 0 0 0 0 0 (STEP-II) Subtract least entry of each Row from all the entries of that Row. The matrix will be as given below: Row minimization: A B C D E F I 09 17 03 00 08 23 II 10 00 31 03 07 04 III 20 07 19 16 00 13 IV 19 17 20 14 22 00 V 09 02 0 26 04 24 Dummy 00 00 00 00 00 00 (STEP-III) Subtract least entry of each Column from all the entries of that Column. The matrix will be as given below: Column minimization: A B C D E F I 09 17 03 00 08 23 II 10 00 31 03 07 04 III 20 07 19 16 00 13 IV 19 17 20 14 22 00 V 09 02 0 26 04 24 Dummy 00 00 00 00 00 00
• 15. Assignment Problems 15 (STEP-IV)Draw minimum line & covers maximum zero: A B C D E F I 09 17 03 00 08 23 II 10 00 31 03 07 04 III 20 07 19 16 00 13 IV 19 17 20 14 22 00 V 09 02 00 26 04 24 Dummy 00 00 00 00 00 00 (STEP-V) Make box to assign the particular jobs: A B C D E F I 09 17 03 08 23 II 10 31 03 07 04 III 20 07 19 16 13 IV 19 17 20 14 22 V 09 02 26 04 24 Dummy 00 00 00 00 00 00 I=D=08 II=B=13 III=E=12 IV=F=07 V=C=19 Dummy=A=00 Total= 59(Ans.) 00 00 00 00 00
• 16. Assignment Problems 16 Example 2: Company XYZ has 5 jobs. 5 people applied for those jobs. The company needs to find the minimum salary taken for those jobs. Expected salary of those jobs of person A, B, C, D, E is given below as matrix form: Person Jobs (in thousand) I II III IV V A 4 3 1 5 2 B 7 4 2 10 5 C 8 7 4 6 4 D 3 5 8 7 9 E 5 6 3 8 10 Condition a) A cannot get job 5. b) Job 3 get only C. c) E cannot get any job. Step 2: Since A cannot get job no. 5, therefore we have to put ∞ at job 5 of A. C get job no. 3, therefore we have to remove row of C and job no. 3 column from the matrix. E cannot get any job; therefore we have to remove all the row of E. It is shown in table below: Person Jobs I II IV V A 4 3 5 ∞ B 7 4 10 5 D 3 5 7 9 Step2: Here the matrix is unbalanced so we have to take dummy person as a row. New matrix as follows: Person Jobs I II IV V A 4 3 5 ∞ B 7 4 10 5 D 3 5 7 9 Dm 0 0 0 0
• 17. Assignment Problems 17 Row Minimization: Person Jobs I II IV V A 1 0 2 ∞ B 3 0 6 1 D 0 2 4 6 Dm 0 0 0 0 Column Minimization: Person Jobs I II IV V A 1 0 2 ∞ B 3 0 6 1 D 0 2 4 6 Dm 0 0 0 0 Step3: Select the smallest entry from all the entries which are not covered by a straight line (here it is 1). Subtract this smallest entry from all the uncovered entries and add it to all those entries which are at the intersection of two lines and other covered entries remain unchanged. The matrix is given below: Person Jobs I II IV V A 0 0 1 ∞ B 2 0 5 0 D 0 3 4 6 Dm 0 0 0 0 Here no. of line = matrix size, therefore optimal assignment can be made at this stage. Person Jobs I II IV V A 0 0 1 ∞ B 2 0 5 0 D 0 3 4 6 Dm 0 0 0 0
• 18. Assignment Problems 18 The solution is: A – Job2 = 3 B – Job5 = 5 C – Job3 = 4 D – Job1 = 3 Dm – Job2 =0 Total = 15 MAXIMIZATION Maximization Case – Question: Four sales men are to be assigning to four sales territories (1 to each). Estimates of the sales revenues in thousand of rupees for each sales man are as under: Sales territories Salesman T1 T2 T3 T4 S1 25 38 43 20 S2 45 12 19 4 S3 43 16 29 24 S4 9 40 45 44 You are required: To obtain the optimal assignment pattern that maximizes our sales revenue. Solution: Step 1- As given matrix gives revenues which are to be maximized. In order to use minimization technique obtain relative loss matrix by subtracting all the revenues from maximum revenue i.e. 45 Relative loss matrix 20 7 2 25 0 33 26 41 2 29 16 21 36 5 0 1
• 19. Assignment Problems 19 Step 2- We have to check that the given matrix is square or not. Here the matrix is square. Step 3- Subtract least entry of each row from all entries of that row to obtain row reduce matrix. It is given below: 18 5 0 23 0 33 26 41 0 27 14 19 36 5 0 1 Step 4- In the above matrix subtract least entry of each column from all the entries of that column to obtain total reduce matrix. Try to cover all the zeros of this matrix by using minimum number of lines. It is shown below- 18 0 0 22 0 28 26 40 0 22 14 18 36 0 0 0 Since the number of covering lines is 3 which is less than size of the matrix optimal solution cannot be obtained. Subtract minimum uncovered element (14 in this case) from all the uncovered entries and add it to all those entries which are at the intersection of two lines. Draw minimum number of lines to cover all the zeros of this new matrix. 32 0 0 22 0 14 12 26 0 8 0 4 50 0 0 0 Since number of covering lines is 4, which is equal to size of the matrix. Optimal assignment can be made. It is given in the following table-
• 20. Assignment Problems 20 32 0 0 22 0 14 12 26 0 8 0 4 50 0 0 0 Thus optimal assignment is as follows: Salesman Sales Territory Sales revenue S1 T2 38000 S2 T1 45000 S3 T3 29000 S4 T4 44000 1,56,000 MAXIMIZATION UNBALANCED Ques.1. There is 5 jobs for product selling and only 4 executives applied for it. You have to find maximum selling out of them and assign the job. A B C D E I 26 33 14 53 27 II 37 17 22 08 11 III 55 13 24 41 12 IV 42 38 32 27 49 Sol. Here we have to find maximum profit so we have to reduce all the values from the maximum value. For maximization to minimization. So the new table is A B C D E I 29 22 41 02 28 II 18 38 33 47 44 III 00 42 31 14 43
• 21. Assignment Problems 21 IV 13 17 23 28 06 This is not a square matrix so we have to balance it with the help of dummy row. A B C D E I 29 22 41 02 28 II 18 38 33 47 44 III 00 42 31 14 43 IV 13 17 23 28 06 Dm 00 00 00 00 00 Row minimization A B C D E I 27 20 39 00 26 II 00 20 15 29 26 III 00 42 31 14 43 IV 07 11 17 22 00 Dm 00 00 00 00 00 Colum minimization A B C D E I 27 20 39 00 26 II 00 20 15 29 26 III 00 42 31 14 43 IV 07 11 17 22 00 Dm 00 00 00 00 00
• 22. Assignment Problems 22 Draw minimum line for covers maximum zeros A B C D E I 27 20 39 00 26 II 00 20 15 29 26 III 00 42 31 14 43 IV 07 11 17 22 00 Dm 00 00 00 00 00 Numbers of lines is not equal to numbers of row or Colum. Then we have to find least number(11) of free numbers and subtract it from free numbers and add with there when lines are intersects and don’t do any things where lines are pass. And again draw the minimum lines which covers maximum zeros. Free numbers are those which are not in lines. A B C D E I 27 09 28 00 26 II 00 09 04 29 26 III 00 31 20 14 43 IV 07 00 06 22 00 Dm 11 00 00 11 11 Number of lines is not equals to row or column then we have to apply the same step. A B C D E I 27 05 24 00 22 II 00 05 00 29 22 III 00 27 16 14 39 IV 11 00 06 26 00 Dm 15 00 00 15 11
• 23. Assignment Problems 23 Candidate I assign the job D, II assign the job C, III Assign the job A, IV assign the job E and Dm will be assign the job B. So the minimum salaries which is given to employees is First option is I*D= 53 II*C= 37 III*A= 55 IV*E= 49 Dm*B= 0 194 So maximum sell is 194 Ans.
• 24. Assignment Problems 24 MULTIPLE CONDITIONED QUESTIONS If there is any unbalanced, multiple conditioned question, first we will minimize it (if question is maximization) then we will balance the question. After that fulfill the conditions and if needed balance the question again. Then after follow the general procedure. Q.1) Profit earned by different sales man in different territories are as follows:- TERRITORY SALES MAN T1 T2 T3 T4 T5 T6 A 31 16 14 13 15 30 B 25 19 18 17 19 26 C 38 17 22 21 23 22 D 15 22 26 25 27 18 E 14 23 30 29 31 14 CONDITIONS 1. C has to be assigned job T5 2. E should not get T2 & T6 3. D should not get any job SOLUTION Minimization and applying conditions. STEP-I TERRITORY SALES MAN T1 T2 T3 T4 T5 T6 A 31 16 14 13 15 30 B 25 19 18 17 19 26 C 38 17 22 21 23 22 D 15 22 26 25 27 18 E 14 ∞ 30 29 31 ∞
• 25. Assignment Problems 25 STEP-II Balancing the problem TERRITORY SALES MAN T1 T2 T3 T4 T6 A 7 22 24 25 8 B 13 19 20 21 12 E 24 ∞ 8 9 ∞ D1 0 0 0 0 0 D2 0 0 0 0 0 STEP-III Row minimization TERRITORY SALES MAN T1 T2 T3 T4 T6 A 0 15 17 18 1 B 1 7 8 9 0 E 16 ∞ 0 1 ∞ D1 0 0 0 0 0 D2 0 0 0 0 0 STEP-IV Column minimization TERRITORY SALES MAN T1 T2 T3 T4 T6 A 0 15 17 18 1 B 1 7 8 9 0 E 16 ∞ 0 1 ∞ D1 0 0 0 0 0 D2 0 0 0 0 0
• 26. Assignment Problems 26 STEP-V Assigning the job. SALES MAN TERRITORIES SALES PROFIT A T1 31 B T6 26 C T5 23 D NO JOB 00 E T3 30 D1 T2/T4 00 EXERCISES MINIMIZATION PROBLEMS 1. A company has 5 machines and 5 jobs. The revelant cost is given below Find the assignment that minimizes total cost. Machines JOBS J1 J2 J3 J4 J5 M1 10 4 5 3 11 M2 13 11 9 12 10 M3 12 3 10 1 9 M4 9 1 11 4 8 M5 8 6 7 3 10 Ans. Machine 1 JOB 3 COST RS.3 Machine 2 JOB 5 COST RS.1O Machine 3 JOB 4 COST RS.1 Machine 4 JOB 2 COST RS. 1 Machine 5 JOB 1 COST RS. 8 TOTAL. 25
• 27. Assignment Problems 27 2. A car hire company has one car at each of the five depots a,b,c,d,e. A customer requires car at each town vit. A,B,C,D and E. Distance (in kms) between deposits (origins) and towns (destination) are given in the following distance matrix. How should the cars be assigned to customers so as to minimize the total distance travelled. A B C D E A 160 130 175 190 200 B 135 120 130 160 175 C 140 110 155 170 185 D 50 50 80 80 110 E 55 35 70 80 105 Ans. A * e =200 km B * c =130 km C * b =110 km D * a =50 km E * d =80 km Total = 570 km MAXIMIZATION PROBLEMS 1. A marketing manager has 5 sales man & 5 sales area, considering the capabilities of the salesmen nature of areas, the maraketing manager estimates that sales per month (in thousands of rupees) for salesman in each area would be as follow: Salesman AREAS A1 A2 A3 A4 A5 S1 42 48 50 38 50 S2 50 34 38 31 36 S3 51 37 43 40 47 S4 32 48 51 46 46 S5 39 43 50 45 49
• 28. Assignment Problems 28 Final Optimal Assignment Ans. TOTAL COST 241000 S1=A2 S1=A2 S2=A1 S2=A5 S3=A5 OR S3=A1 S4=A3 S4=A4 S5=A4 S5=A5 < 2. Quantity of clothes of different brands sold in different cities per mnth are as follows Wrangler Pantaloon Pvogue Lee Levis Delhi 25 22 55 42 48 Gurgaon 78 41 40 46 41 Noida 18 33 52 50 37 Jaipur 32 18 30 37 20 Chandigarh 40 20 42 48 51 Find the showroom for different brands which should be opened in different cities. Ans = 254 Unbalance Assignment problem (Minimization) Q. 1) A company is face with the problem of assigning six different machines to Five different job. The cost are estimated as follows (in hundred of Rs.) Solve the problem assuming that the objective is to minimize the total cost. A1 A2 A3 A4 A5 M1 05 10 02 12 02 M2 04 10 03 14 06 M3 06 13 04 16 06 M4 07 14 04 18 09 M5 08 14 06 18 12 M6 12 18 10 20 12
• 29. Assignment Problems 29 Ans. Machine Cost Machine Cost M1 A5 02 M1 A5 02 M2 A1 04 M2 A4 14 M3 A4 16 M3 A1 06 M4 A3 04 M4 A3 04 M5 A2 14 M5 A2 14 M6 A6 00 M6 A6 00 Total=40 Total=40 Q. 2) Solve the following unbalance Assignment problem of the minimizing total time for doing all jobs. Jobs J1 J2 J3 J4 J5 O1 16 12 15 12 16 O2 12 15 18 17 17 O3 17 18 16 19 18 O4 16 12 13 14 15 O5 19 13 18 19 17 O6 14 17 14 16 18 Ans: Operation Job Time O1 J4 12 O2 J1 12 O3 J6 00 O4 J5 15 O5 J2 13 O6 J3 14 Total=66
• 30. Assignment Problems 30 Unbalance Assignment problem(Maximization) Q.1) Four different Airplanes are to be assigned to handle three cargo consignment with a view to maximize profit . The profit matrix is a follow in thousand of Rupees Cargo Consignment Airplane I II III W 08 11 12 X 09 10 10 Y 10 10 10 Z 12 08 09 Ans: W to III profit 12,000 W to III profit 12,000 X to II profit 10,000 X to Dummy profit 0 Y to Dummy profit 00 Y to II profit 10,000 Z to I profit 12,000 Z to I profit 12,000 Total=34,000 Total=34,000 Q.2) Solve the following assignment problem. The data give him the table refer to production in central unit. Machine A B C D I 10 05 07 08 II 11 04 09 10 III 08 04 09 07 IV 07 05 06 04 V 08 09 07 05
• 31. Assignment Problems 31 Ans. Operation Machine No. of Units 1 A 10 2 D 10 3 C 9 4 Dummy 0 5 B 9 Unit=38 CONDITION EXERCISE- Solve the following assignment problem in which 5 jobs and 7 peoples are given and some conditions are considered when jobs assign, the conditions are-  Job 3 has to be assigned to B.  Job 1 and 4 can not be assigned to G.  D has to get job 5 1 2 3 4 5 A 36 33 25 30 26 B 42 25 35 26 45 C 54 50 52 35 28 D 28 35 40 28 36 E 20 28 32 40 52 F 26 30 48 25 27 G 24 28 22 26 30 ANSWER- 128
• 32. Transportation Problem 32 TRANSPORTATION PROBLEMS “INTRODUCTION” The Transportation model deals with situations where some commodity or product is distributed from multiple sources to multiple destinations. The model may be used to find the minimum transportation cost or the maximum profit, depending upon the amounts shipped from each source to each destination. The problem solution will be the optimum distribution scheme, showing exactly how much of the commodity should be transported via each possible route. The transportation algorithm discussed in this chapter is applied to minimize the total cost of transporting a homogeneous commodity from supply origins to demand destinations. However, it can also be applied to the maximization of some total value or utility, for example, financial resources are distributed in such a way that the profitable return is maximized. ADVANTAGES  Proper utilization of resources take place without much of the losses.  The selections and allocations of resources to their destinations become more accurate.  This process helps in cost minimization and profit maximization, which major objective of organizations.  It helps in planning and decision making. Some Important Terms or Definitions 1) Feasible solution-: Set of Non Negative values xij=1,2,3,4….m,j=1,2,3,4……….n. which satisfy the following condition is called feasible solution. 2) Basic feasible solution-: a feasible solution with an allocation of (m+n-1) number of variables . xij,i=1,2,3…..m, j=1,2,3,4…….n. is called a basic feasible solution. 3) Optimum Solution-: A basic feasible solution of transportation problem which minimizes the total transportation cost or maximizes total revenue. 4) Rim requirement-: The quantity required or available are called rim requirement. 5) Balanced transportation problem-: If in any transportation problem total number of units available is equal to the total number of units required, then it is called balanced transportation problem.
• 33. Transportation Problem 33 STEPS OR PROCEDURE OF SOLVING TRANSPORTATION PROBLEMS Make Empty Cells(4) YES NO NO Calculate Value ofRi and Cj (5) Xij =TCij – (Ri + Cj) YES Is Xij <0 ? Select Minimum Value ofXij Construct Loop Revise solution and check for rim value condition Find aBasic feasible Solution (3) f YES Balance the Problem (2) Is R + C -1= Filled Cell? ? NO Is Problem Balanced? ? NO START YES Is Problem of maximization? Convert to minimization (1) OptimalSolution
• 34. Transportation Problem 34 Notes: 1) Convert to minimization-: If the problem is maximization then convert it into minimization by subtracting the whole value of table from the largest. 2) Balance the problem-: If the problem is unbalanced then balance the problem by using the dummy and give the extra capacity to the dummy. Now the problem will be balanced. 3) Find a Basic Feasible Solution-: After balancing the problem find a basic feasible solution that which one is right for problem among following methods .north west corner method, least cost method, Vogel’s. 4) Make empty cell-: If R+C-1= filled cell does not satisfy the table then make a empty cell on any unfilled cell. 5) Calculate value of R and C-: By using formula x= TC-(R+C).We will calculate the value of R and C. THE TRANSPORTATION METHOD There are several methods available to obtain an initial basic feasible solution. But the general steps are discussed below: Step 1. The solution algorithm to a transportation problem may be summarized into the following steps: The formulation of the transportation problem is similar to the LP problem formulation. Here the objective function is the total transportation cost and the constrains are the supply and demand available at each source and destination, respectively. Step 2. Obtain an initial basic feasible solution. North-West Corner Method (NWCM) It is a simple and efficient method to obtain an initial solution. This method does not take into account the cost of transportation on any route of transportation. The method can be summarized as follows; Step1 Start with the check at the upper left (north-west) corner of the transportation matrix and allocate as much as possible equal to the minimum of the rim value for the first row and first column.
• 35. Transportation Problem 35 Step 2 (a) If allocation made in Step 1 is equal to the supply available at first source (a1, in first row), then move vertically down to the cell (1, 2) in the second row and first column and apply Step 1 again, for next allocation. (b) If allocation made in Step 1 is equal to the demand of the first destination (b1 is first column), then move horizontally to the cell (1, 2) in the first row and second column and apply Step 1 again for next allocation. (c) If a1 = b1, allocate w11 = a1 or b1 and move diagonally to the cell (2, 2). Step 3. Continue the procedure step by step till an allocation is made in the south-east corner cell of the transportation table. Example-: A construction company wants cement at three of its project sites P1, P2 and P3. It procures cement from four plants C1, C2, C3 and C4. Transportation costs per ton, capacities and requirements are as follows: P1 P2 P3 Capacity(tons) C1 5 8 12 300 C2 7 6 10 600 C3 13 4 9 700 C4 10 13 11 400 Requirement 700 400 800 Determine optimal allocation of requirements. Solution: P1 P2 P3 Capacity(tons) C1 5 (300) 8 12 300 C2 7 (400) 6 (200) 10 600 C3 13 4 (300) 9 (400) 700 C4 10 13 11 (400) 400 Requirement 700 500 800
• 36. Transportation Problem 36 The cell (C1, P1) is the North-West corner cell in the given transportation table, the rim values for row C1 and column P1 are compared. The smaller of two, i.e. 300 is assign as the first allocation. This means that 300 units of commodity are to be transported from plant C1 to project site P1. However, this allocation leaves a supply of 700 – 300 = 400 unit of commodity at C1. Move vertically and allocated as much as possible to cell (C2, P1). The rim value for column P1 is 400 and for row C2 is 600. The smaller of the two, i.e. 400, is placed in the cell. Proceeding to column P2, the rim value for column P2 is 500 and for row C2 is 200. The smaller of the two, i.e. 200, is placed in the cell. Proceeding to column P3, the rim value for column P3 is 800 and for row C3 is 400. The smaller of the two, i.e. 400, is placed in the cell. Now, the rim value for column P3 is 400 which is balanced by the rim value of row C4 that is 400. The total transportation cost of the initial solution derived by the North-West corner method is obtain by multiplying the quantity in the occupied sales with the corresponding unit cost and adding. Thus the total transportation cost of the solution is-: Total cost = 300 x 5 + 400 x 7 + 200 x 6 + 300 x 4 + 400 x 9 + 400 x 11 = Rs. 14,700 Least Cost Method Since the objective is to minimize the total transportation cost, we must try to transport as much as possible through those routes where the unit transportation cost is lowest. Step 1 – Select the cell with the lowest unit cost in the entire transportation table and allocate as much as possible to this cell and eliminate that row and column in which either supply or demand is exhausted. Step 2 – Repeat the procedure until the entire available supply at various sources and demand at various destinations is satisfied Example-: A construction company wants cement at three of its project sites P1, P2 and P3. It procures cement from four plants C1, C2, C3 and C4. Transportation costs per ton, capacities and requirements are as follows:
• 37. Transportation Problem 37 P1 P2 P3 Capacity(tons) C1 5 8 12 300 C2 7 6 10 600 C3 13 4 9 700 C4 10 13 11 400 Requirement 700 500 800 Determine optimal allocation of requirements. Solution: P1 P2 P3 Capacity(tons) C1 5 (300) 8 12 300 C2 7 (400) 6 10 (200) 600 C3 13 4 (500) 9 (200) 700 C4 10 13 11 (400) 400 Requirement 700 500 800 In the above Solution we have, The cell (C3, P2) contain the lowest transportation cost in the given transportation table, the rim values for row C3 and column P2 are compared. The smaller of two, i.e. 500 is assign as the first allocation. This means that 500 units of commodity are to be transported from plant C3 to project site P2. However, this allocation leaves a supply of 700 – 500 = 200 unit of commodity at C3. Now, search for the least cost in table without considering P2 because the rim value of P2 is zero. The lowest transportation cost is at cell (C1, P1) the rim values for row C1 and column P1 is compared. The smaller of two, i.e. 300 is assigned. However, this allocation leaves a supply of 700 – 300 = 400 unit of commodity at P1. Now, search for the least cost in table without considering column P2 and row C1 because the rim value of P2 and C1 is zero. The lowest transportation cost is at cell (C2, P1) the rim values for row C2 and column P1 is compared. The smaller of two, i.e. 400 is assigned. However, this allocation leaves a supply of 600 – 400 = 200 unit of commodity at c2.
• 38. Transportation Problem 38 Now, the rim value for row C2, C3, and C4 are respectively 200, 200 and400 which is balanced by the rim value of column P3 that is 800. The total transportation cost of the initial solution derived by the “Least Cost Method” is obtained by multiplying the quantity in the occupied sales with the corresponding unit cost and adding. Thus the total transportation cost of the solution is-: Total Cost = (C1, P1) + (C2, P1) + (C2, P3) + (C3, P2) + (C4, P3) = 300 X 5 + 400 X 7 + 200 X 10 + 500 X 4 + 400 X 11 = 1500 + 2800 + 2000 + 2000 + 4400 = Rs. 12,700 VOGELS APPROXIMATION METHOD Step 1: Construct the transportation tableau as described earlier. Step 2: For each row and column, the difference between the two lowest cost entries .If the lowest cost entries are tied, the difference is Zero. Step 3: Select the row or column that has the largest difference .In the event of a tie selection is arbitrary. Step 4: In the row or column, identified in step 3, select the cell that has lowest cost in tree. Step 5: Assign maximum possible number of unit to the cell selected in step 4(the smaller of two between demand and the availability).This will completely exhaust a row or a column. Omit the exhausted row and column. Step 6: Reapply step 2 to step 5. Iteratively using the remaining row and columns until the total demand is met and supply exhausted. Example-: A construction company wants cement at three of its project sites P1, P2 and P3. It procures cement from four plants C1, C2, C3 and C4. Transportation costs per ton, capacities and requirements are as follows: P1 P2 P3 Capacity(tons) C1 5 8 12 300 C2 7 6 10 600 C3 13 4 9 700 C4 10 13 11 400 Requirement 700 500 800 Determine optimal allocation of requirements.
• 39. Transportation Problem 39 Solution: P1 P2 P3 CAPACITY ROW DIFFERENCE C1 5 300 8 12 300 3 7 - - C2 7 400 6 10 200 600 200 1 3 3 3 C3 13 4 500 9 200 700 200 5 4 4 - C4 10 13 11 400 400 1 1 1 1 REQUIREMENT 700 400 500 800 600 2000COLUMN DIFFERENCE 2 2 1 2 - 1 3 - 1 3 - 1 Total cost = C1×P1+C2×P1+C2×P3+C3×P2+C3×P3+ C4×P3 = 300×5+400×7+200×10+500×4+200×9+400×11 = 1500+2800+2000+2000+1800+4400 = Rs. 14,500 Firstly we will find, for each row (C1, C2, C3, and C4) and each column (P1, P2, and P3), the difference for two lowest entries. After that we select the row C3, because it has largest difference. In C3, find the least value is 4 after that assign the maximum possible number 500 to the cell so that it exhausted the column P2 completely. After that we will find remaining rows and columns the difference for two lowest entries. After that we select the row C1, because it has largest difference. In C1, find the least value is 5 after that assign the maximum possible number 300 to the cell so that it exhausted the row C1 completely. After that we will find remaining rows and columns the difference for two lowest entries. After that we select the row C3, because it has largest difference. In C3, find the least value is 9 after that assign the maximum possible number 200 to the cell so that it exhausted the Row C3 completely.
• 40. Transportation Problem 40 After that we will find remaining rows and columns, the difference for two lowest entries. After that we select the row C2, because it has largest difference. In C7, find the least value is 4 after that assign the maximum possible number 400 to the cell so that it exhausted the column P1 completely. By continuous doing this process we will assign the all unit to the cell and will find the minimum transportation cost. Optimality Test Once initial solution has been found the next step is to test that solution for optimality. Modified Distribution Method (MODI method) is generally used for testing the optimality of the existing solution. MODIFIED DISTRIBUTION METHOD (MODI Method) The following steps involved in the MODI method:- Step 1- R+C−1= Filled cell (In this step we will subtract 1 from the sum of rows and columns, after that we will compare with filled cells. which will be equal to the filled cell.) Step 2- Assume any row or column equal to zero. Step 3- Filled cell – R+C= TC (In this step we find the value of row and column) Step 4 – Unfilled cell – Find the sign of unfilled cell with the help of TC−(R+C) Example-: A construction company wants cement at three of its project sites P1, P2 and P3. It procures cement from four plants C1, C2, C3 and C4. Transportation costs per ton, capacities and requirements are as follows: P1 P2 P3 Capacity(tons) C1 5 8 12 300 C2 7 6 10 600 C3 13 4 9 700 C4 10 13 11 400 Requirement 700 500 800 2000 Determine optimal allocation of requirements.
• 41. Transportation Problem 41 Solution: C1=5 C2=3 C3=8 P1 P2 P3 CAPACITY ROW DIFFERENCE R1=0 C1 5 300 8 12 300 3 7 - - R2=2 C2 7 400 6 10 200 600 200 1 3 3 3 R3=1 C3 13 4 500 9 200 700 200 5 4 4 - R4=3 C4 10 13 11 400 400 1 1 1 1 Requirement 700 400 500 800 600 2000 Column Difference 2 2 1 2 - 1 3 - 1 3 - 1 Total cost = C1×P1+C2×P1+C2×P3+C3×P2+C3×P3+C4×P3 = 300×5+400×7+200×10+500×4+200×9+400×11 = 1500+2800+2000+2000+1800+4400 = Rs. 14,500 Step 1- R+C–1 = Filled cell 4+3–1 = 6 (BALANCED MATRIX) Step 2- For filled cell Let R1= 0 R1+C1 = TC 0+C1 = 5 C1 = 5 R2+C1 = TC R2+5 = 7 R2 = 2 R2+C3 = TC 2+C3 = 10 C3 =8 R3+C3 = TC R3+8 = 9 R3 = 1 R3+C2 =TC 1+C2 = 4 C2 = 3
• 42. Transportation Problem 42 R4+C3 = TC R4+8 = 11 R4 =3 Step 3- For unfilled cell (Find sign) TC–(R1+C2) = 8–(0+3) = 5 TC–(R1+C3) = 12-(0+8) = 4 TC–(R2+C2) = 6–(2+3) = 1 TC–(R3+C1) = 13–(1+5) = 7 TC–(R4+C1) = 10–(3+5) = 2 TC–(R4+C2) = 13–(3+3) = 7 There is no negative sign so that the total cost will be same. Question - 2 C1 =13 C2 = 12 C3 = 10 C4 W1 W2 W3 W4 PRODUCTION CAPACITY ROW DIFFERENCE R1 = 0 F1 13 (2) 11 15 40 2 2 - - - R2 = 2 F2 17 14 (3) 12 (3) 13 6 3 1 1 - 4 R3 = 5 F3 18 (1) 18 15 (1) 12 (5) 7 2 3 3 3 6 W.H. CAPACITY 3 1 3 4 1 5 COLUMN DIFFERENCE 4 3 3 1 1 4 3 1 1 - 3 1 1 - 3 - Total Cost = F1W1+F2W2+F2W3+F3W1+F3W3+F3W4 = 26+42+36+18+15+60 = Rs. 197 Step 1- First we will find filled cell TC-(R1+C4) = 40-(0+7) = 33 TC-(R2+C1) = 17-(2+13) = 2 TC-(R2+C4) = 13-(2+7) = 4
• 43. Transportation Problem 43 TC-(R3+C2) = 18-(5+12) = 1 If there will be negative sign than we will make house in L shape or Square shape. In this process, we will assign the sign alternatively. Then we will subtract the whole corner vales from the least positive value which will also at any corner of house. R+C−1 = 3+4−1 = 6 Step 2 -After that we will find the value of row and column by assuming any row is equals to zero. For filled cell (let R1 = 0) R1+C1 = TC = 0+C1 = 13 = C1 = 13 R3+C1 = TC = R3+13 = 18 = R3 = 5 R3+C3 = TC = 5+C3 = 15 = C3 = 10 R3+C4 = TC = 5+C4 = 12 = C4 = 7 R2+C3 = TC = R2+10 = 12 = R2 = 2 R2+C2 =TC = 2+C2 =14 = C2 = 12 Step 3- After that we will find the sign. For Unfilled Cell TC-(R1+C2) = 11-(0+12) = -1 TC-(R1+C3) = 15-(0+10) = 5 C1 = 15 C2 =14 C3 =12 C4=9 W1 W2 W3 W4 PRO. CAPACITY R1=-2 F1 13 + (2) 11 - (H) 15 40 2 R2=0 F2 17 14 + (3) 12 - (3) 13 6 R3=3 F3 18 - (1) 18 15 + (1) 12 (5) 7
• 44. Transportation Problem 44 W.H. CAPACITY 3 3 4 5 15 W1 W2 W3 W4 PRO. CAPACITY F1 13 (1) 11 (1) 15 40 2 F2 17 14 (2) 12 (4) 13 6 F3 18 (2) 18 15 12 (5) 7 W.H. CAPACITY 3 3 4 5 15 Total Cost = F1W1+F1W2+F2W2+F2W3+F3W1+F3W4 = 13+11+28+48+60+36 = Rs. 196 Special Situations There is certain special situation or case which generally happened in practical life, some of them are discussed below: Maximizing Transportation Problem: The main objective of an organization is to cost minimization and profit maximization. The problem of maximizing can be solved by following process: A maximization transportation problem can be converted into the usual minimizing problem by subtracting all the contributions from the highest contribution involved in the problem. Example: Priyanka Steel Co. has three factories and five customers, profit (` per unit) are given. Find a schedule where co. gets maximum profit. Factory Customer CapacitiesC1 C2 C3 C4 C5 F1 4 2 3 2 6 8 F2 5 4 5 2 1 12
• 45. Transportation Problem 45 Solutions: Factory Customer CapacitiesC1 C2 C3 C4 C5 F1 3 5 4 5 1 8 F2 2 3 2 5 6 12 F3 1 2 3 0 0 14 Required 4 4 6 8 12 In above problem the highest contribution or per unit profit is 7 and now we subtract all the contribution no from 7, Now applying the VAM method in the above table. Total Cost according to VAM and after applying MODI method is as follows: = 8 X 1 + 2X 2 +3X 4 + 2 X 6 + 1 X 2 + 0 X 8 + 0 X 4 =8 + 4 +12 +12 +2 + 0 + 0 = Rs. 38 F3 6 5 4 7 7 14 Required 4 4 6 8 12 Factory Customer Capacities P1 P2 P3 P4 C1 C2 C3 C4 C5 C1=2 C2=3 C3=2 C4=1 C5=1 F1 R1=0 3 5 4 5 1 (8) 8 2 2 X X F2 R2=0 2 (2) 3 (4) 2 (6) 5 6 12 0 O 1 1 F3 R3=- 1 1 (2) 2 3 0 (8) 0 (4) 14 0 1 1 1 Required 4 4 6 8 12 P1 1 1 1 5 1 P2 1 1 1 x 1 P3 1 1 1 x 6 P4 1 1 1 x x
• 46. Transportation Problem 46 Unbalanced Transportation problem Transportation problems that have the supply and demand equal is a balanced transportation problem. In other words requirements for the rows must equal the requirements for the columns. An Unbalanced transportation problem is that in which the supply and demand are unequal. There are 2 possibilities that make the problem unbalanced which are (i) Aggregate supply exceeds the aggregate demand or (ii) Aggregate demand exceeds the aggregate supply. Such problems are called unbalanced problems. It is necessary to balance them before they are solved. The following table gives the cost of transportation, the availabilities and requirements of an organization: Ware House Demand Points Available Capacity A B C D W1 10 20 20 15 50 W2 15 40 15 35 100 W3 25 30 40 50 150 Requirements 25 115 60 30 Solution: In the above problem total demand is less than total supply by (300-230) 70 units. Step 1: So, we will create a new demand point and name this as “DUMMY”. Step 2: Assume Transportation cost as to be “Zero” and write in DUMMY column in place of Transportation cost. Step 3: Write the difference of total demand and supply in the required column cell. Aggregate supply exceeds the aggregate demand Aggregate demand exceeds the aggregate supply 2 Possibilities That Make the Problem Unbalanced
• 47. Transportation Problem 47 A B C D DUMMY CAPACITY W1 10 20 20 15 0 50 W2 15 40 15 35 0 100 W3 25 30 40 50 0 150 25 115 60 30 70 300 Step4: Apply method of basic feasible solution. (By Vogel’s Approximation Method). A B C D DUMMY CAPACITY P1 P2 P3 P4 P5 W1 10 (20) 20 20 15 (30) 0 50 10 5 10 - - W2 15 (5) 40 (35) 15 (60) 35 0 100 5 0 0 0 25 W3 25 30 (80) 40 50 0 (70) 150 30 5 5 5 5 25 115 60 30 70 300 P1 5 10 5 20 0 P2 5 10 5 20 - P3 5 10 5 - - P4 10 10 25 - - P5 10 10 - - - Total Cost =10×20+40×35+15×30+15×5+40×35+15×60+30×80 =200+1400+450+75+1400+900+2400 = Rs. 6825 Step 5: Now, we will test this solution for optimality. C1= 15 C2= 30 C3= 15 C4= 25 C5= 0 A B C D DUMMY CAPACITY R1= - 10 W1 10 20 (20) 20 15 (30) 0 50 R2= 0 W2 15 (25) 40 15 (60) 35 0 (15) 100 R3= 0 W3 25 30 (95) 40 50 0 (55) 150 25 115 60 30 70 300 The Optimal Cost for above sum = 20×20+15×30+15×25+15×60+30×95 = 400+450+375+900+2850 = Rs. 4975
• 48. Transportation Problem 48 Prohibited Routes At times there are transportation problems in which one of the sources is unable to ship to one or more of the destinations. When this occurs, the problem is said to have an unacceptable or prohibited route. In a minimization problem, such a prohibited route is assigned a very high cost to prevent this route from ever being used in the optimal solution. After this high cost is placed in the transportation table, the problem is solved using the techniques previously discussed. In a maximization problem, the very high cost used in minimization problems is given a negative sign, turning it into a very bad profit. Question: Solve the following transportation problem with the restriction that nothing can be transported from factory F2 to warehouse WE3 : FACTORY W1 W2 W3 W4 NO OF UNITS AVAILABLE F1 40 60 60 10 15 F2 70 30 50 20 4 F3 30 90 20 40 11 REQUIREMENT 10 5 10 5 30 Transportation root is restricted in the cell F2-W3. So we have to replace the particular cell with any maximum value. Here we are taking maximum value as M and solve the problem as usual method. Now, apply VAM W1 W2 W3 W4 Availability P1 P2 P3 P4 F1 40 (9) 60 (1) 60 10 (5) 15/10 30 30 20 20 F2 70 30 (4) M 20 4/0 810 10 40 40 F3 30 (1) 90 20 (10) 40 11/1/0 10 10 60 Requirement 10/9/0 5/1/0 10/0 5/0 P1 10 30 40 10 P2 10 30 10 P3 10 30 P4 30 30
• 49. Transportation Problem 49 Now test for optimality. C1 = 40 C2 =60 C3= 30 C4 = 10 W1 W2 W3 W4 Availability R1 = 0 F1 40 (9) 60 (1) 60 10 (5) 15 R2 = -30 F2 70 30 (4) M 20 4 R3 = -10 F3 30 (1) 90 20 (10) 40 11 Requirement 10 5 10 5 Total Optimal Cost = 9 x 40 + 1 x 60 + 5 X 10 + 4 X 30 + 1 X 30 + 10 X 20 = 360 + 60 + 50 + 120 + 30 + 200 = Rs. 820 Some Necessary Allocation In our today’s business it is often happens that company has to send their product to pre determined or fixed customers or warehouse or distribution centre, in spite of their new or any other customer’s .so, in this type of problem, transportation cost is pre – allocated between two particular parties. The following examples illustrate the details of the problem and put more light on the concept. For example: A firm producing a single product has three plants and four distributors. The three plants will produce 60, 80 and 100 units respectively during the next period. The firm has made a commitment to sell 90 units to distributor a,40 units to distributor B, 60 units to distributor C and 50 units to D. if the management wants to transport at least 40 units from plant P1 to distributor A .find the optimal transportation schedule. The net distribution cost to a distributor is given in the following table: Plant Distributor A B C D P1 8 7 5 2 P2 5 2 1 3 P3 6 4 3 5
• 50. Transportation Problem 50 Solution: Since the firm wants to apply at least 40 units from plant P1 to distributor C, we subtract 40 units from the number of units available at plant P1and from the requirement of distributor A i.e. now the number of units available at plant P1 will be only 20 (60-40) and demand of distributor A will be only50 (90-40) units. Plant Distributor A B C D No. of units available P1 8 7 5 2 20 P2 5 2 1 3 80 P3 6 4 3 5 100 Demand 50 40 60 50 200 Now, apply Vogel’s approximation method, the following table comes: C1= 6 C2= 4 C3= 3 C4= 5 A B C D R1= -3 P1 8 7 5 2 (20) 20 3 - - R2= -2 P2 5 2 (20) 1 (60) 3 80 1 1 1 R3= 0 P3 6 (50) 4 (20) 3 5 (30) 100 1 1 1 Demand 50 40 60 50 200 1 2 2 1 1 2 2 2 1 2 - 2 Total Cost = 20 X 2 + 20 X 2 + 60 X 1 + 50 X 6 + 20 X 4 + 30 X 5 = 40 + 40 +60 + 350 + 80 + 150 = Rs. 720 Unsolved Problems Q.1 A manufacturing firm must produce a product in sufficient quantity to meet contractual sales in next four months. The production capacity and unit cost of production vary from month to month. The product produced in any month may be held for sale in later months but at an estimated storage cost of Re.1 per unit per month. No storage cost is incurred for goods sold in the same
• 51. Transportation Problem 51 month in which they are produced. There is no opening inventory and none is desired at the end of fourth month. The necessary details are given in the following table: Month Contractual Sales Max. Production Unit cost of Production 1 20 40 14 2 30 50 16 3 50 30 15 4 40 50 17 How much should the firm produce each month to minimize total cost? Ans. Cost Rs. 2210 Q.2 A company has four manufacturing plants and five warehouses. Each plant manufactures the same product which is sold to different warehouse at different prices. The details are given below. Plants 1 2 3 4 Manufacturing cost 12 10 8 7 Raw material cost 8 7 7 5 Capacity 100 200 120 80 Warehouses Transportation Cost Demand Sales Price1 2 3 4 A 4 7 4 3 80 30 B 8 9 7 8 120 32 C 2 7 6 10 150 28 D 10 7 5 8 70 34 E 2 5 8 9 90 30 Formulate the above as a transportation problem to maximize profit and obtain the optimal transportation schedule. Ans: Rs. 4580
• 52. Transportation Problem 52 Q.3 The following table gives all the necessary information on the available supply to each warehouse, the requirement of each customer and unit transportation cost from warehouse to each customer: Warehouse Customer Available 8 9 6 3 18 20 18 6 11 5 10 3 8 7 9 Required 15 16 12 13 56 Find optimal transportation schedule. Ans. Rs. 301
• 53. Transportation Problem 53 Case Study:
• 54. Game Theory 54 GAME THEORY INTRODUCTION In practical life, it is required to take decisions in a competing situation when there are two or more opposite parties with conflicting interest and the outcome is controlled by the decisions of all the parties are concerned. In all the above problems where the competitive situations are involved one act in a rational manner and tries to resolve the conflict of interest in his favor. In these situations GAME THEORY was developed in the twentieth century. However, the mathematical treatment of the game theory was given in 1944 by John- Von-Newmann through “THEORY OF THE GAMES AND ECONOMIC BEHAVIOR”. DEFINITION It may be defined as “a body of knowledge that deals with making decisions when 2 or more intelligent and rational opponents are involved under conditions of conflict and competition”. The approach is to seek to determine a rival’s most profitable counter –strategy to one’s own best moves and to formulate the appropriate defensive measures. ASSUMPTIONS OF GAME THEORY  The number of competitors in the competition is known.  The participants simultaneously choose their respective course of action.  The profit and loss in the competition is fixed and determined in advance.  It is assumed that each competitor behaves rationally and intelligently.  It is assumed that players attempts to maximize gains and minimize losses.  The decision of the game can be positive, negative or zero to each player.  It is assumed that players have the knowledge of all the information relating to the game they play. KEY CONCEPTS AND TERMS  Game – A game represents a conflict between two parties / countries / persons / and is played with certain predetermined rules.
• 55. Game Theory 55  Dominance – One of the strategies of either player may be inferior to atleast one of the remaining ones. The superior strategies are said to dominate the inferior ones. A procedure that is used to reduce the size of the game.  Mixed strategy – In a game without saddle point, the optimal policy is to use mixed strategies i.e. to use some optimal combination of available strategies.  Saddle point – A saddle point of a pay off matrix is that position in the matrix where the maximum of the row minima is equal to the minimum of the column maxima. The pay off at the saddle point is called the value of the game and the corresponding strategies are the pure strategies.  Strategy – The number of competitive actions that are available for a player are called strategies to that player.  Value of the game – It is the expected pay off of the play when all players of the game follow their optimal strategies. The game is called fair if the value of game is zero and unfair if it is non zero.
• 56. Game Theory 56 NO YES NO YESYES Law of dominance START Find the saddle point If saddle is there If Matrix is 2*2 If matrix is m*2 or 2*n NO YES
• 57. Game Theory 57 FLOW CHART REPRESENTING FLOW OF GAME THEORY FINDING SADDLE POINT  See the highest number in column draw  on that number  And See the lowest number in row draw  on that number.  If  and  on a particular number. That is Saddle point.  The saddle point is the value of game. PRINCIPLE OF DOMINANCE If no saddle point:  When the elements of a row are less than or equal to the elements of another row then that lesser row is cut.
• 58. Game Theory 58  When the elements of a row are less than or equal to the average elements of another two rows, then that lesser row is cut.  When the elements of a column are greater than or equal to the elements of another row then that greater column is cut.  When the elements of a Column are Greater than or equal to the average elements of another Column, then that Greater column is cut. After the law of dominance we find the following situations After the law of dominance if m*n matrix become 2*2 matrix then we find the solution as follows; 2*2  Find the difference of second row and put it in front of first row  Find the difference of first row and put it in front of second row  Find the difference of second column and put it in below of first column  Find the difference of first column and put it in below of Second column  Calculate the total of the difference of the columns or rows that should be equal  For calculating the value of game we multiply difference of columns with corresponding any of the column, and divided by total of the difference m*n 2*2 Arithmetic Method m*n LPP (Linear Programming Problem) m*2 Arithmetic method/graphical method 2*n Arithmetic method/Graphical method (lower envelop & upper point)
• 59. Game Theory 59  For calculating the strategies of players we divided the value putted in front of the player strategies GRAPHICAL METHOD 2*n Steps to solve:  For 2*n game, first draw to vertical parallel lines.  Let line first be the first row and line second be the second row.  Now plot scale on these lines.  Then plot the element of first row and first column on first line and the element of second row and first column on second line. Similarly second column third column so on.  We find the lower envelop and upper point and intersections lines of upper point.  Then with the help of two parallel lines strategies and upper point intersections lines we make 2*2 matrix we find the value of game and strategies. m*2 Steps to solve:  For m*2 game, first draw to vertical parallel lines.  Let line first be the first column and line second be the second column.  Now plot scale on these lines.  Then plot the element of first row and first column on first line and the element of first row and second column on second line. Similarly second column third column so on.  We find the upper envelop and lower point and intersections lines of lower point.  Then with the help of two parallel lines strategies and lower point intersections lines we make 2*2 matrix we find the value of game and strategies. ARITHMETIC METHOD Steps to solve  In the game matrix of m*2 and 2*n, make various combinations of 2*2 matrix.
• 60. Game Theory 60  Now out of these combinations we calculate value of game, the maximum value of game in case of m*2 and minimum value of game in case of 2*n, we take and their corresponding strategies. APPLICATIONS OF GAME THEORY  Used by poker and chess player to win their games.  Army generals make use of this technique to plan war strategies.  Used to analyze a board range of activities, including dating and mating strategies, legal and political negotiations, and economic behavior.  The game theory technique is designed to evaluate situations where individuals and organizations can have conflicting objectives.  In wage negotiations between unions and firms SIGNIFICANCE IT IS BASED ON TWO BASIC ASSUMPTIONS-  A least in two person zero sum game this theory outlines a scientific quantitative technique which can be fruitfully used by players to arrive at an optimum strategy, given firms objective  Game theory gives insight into several less known aspects which arise in situation of conflicting interest.
• 61. Game Theory 61 LIMITATIONS  The assumptions that the player makes is unrealistic as he can only make a guess of his own and his rival’s strategy.  It becomes complex and difficult as the number of players increases in the game  The assumption made does not seem practical.  Mixed strategies are also not very helpful. Example: Suppose there are two players A and B with different strategies. A a1 a2 a3 a4 b1 20 15 12 35 b2 25 14 08 10 B b3 40 02 10 05 b4 - 5 04 11 00 Now make a square on the largest number column wise and encircle the lowest number row wise. A a1 a2 a3 a4 b1 20 15 12 35 b2 25 14 08 10 B b3 40 02 10 05 b4 - 5 04 11 00 Thus, Saddle point is at no. 12 Value of game= 12 B(1,0,0,0) and A(0,0,1,0) Ans- Now let's take another example: Suppose there are two competitors Reliance and Airtel with different strategies against each other.
• 62. Game Theory 62 Reliance R1 R2 R3 A1 10 09 10 Airtel A2 15 10 16 A3 20 12 13 Now make square around largest number column wise and encircle the smallest number rpw wise. Solution : Reliance R1 R2 R3 A1 10 09 10 Airtel A2 15 10 16 A3 20 12 13 So, Value of game= 12 Reliance(0,1,0) Airtel (0,0,1) Ans- 1. Problem without saddle point: When there is no saddle point then we require to apply the law of dominance. Steps to solve: 1. Apply law of dominance- In this we remove the highest column and remove the lowest row compared to other columns and rows. 2. If we get a 2/2 matrix then difference of the column will be multiplied with the columns and difference of the rows will be multiplied by rows. Case of Henry and Dave Suppose there are two brokers accused of fraudulent trading activities: Dave and Henry. Both Dave and Henry are being interrogated separately and do not know what the other is saying. Both brokers want to minimize the amount of time spent in jail and here lies the dilemma. The sentences vary as follows:
• 63. Game Theory 63 Henry A1 A3 B1 02 05 2/5 B2 05 03 3/5 2/5 3/5 5 V= 2x2+5x3 = 19 5 5 Strategies Henry (2/5,3/5) Dave (2/5,3/5) Ans- 1) If Dave pleads not guilty and Henry confesses, Henry will receive the minimum sentence of one year, and Dave will have to stay in jail for the maximum sentence of five years. 2) If nobody makes any implications they will both receive a sentence of two years. 3) If both decide to plead guilty and implicate their partner, they will both receive a sentence of three years. Dave
• 64. Game Theory 64 4) If Henry pleads not guilty and Dave confesses, Dave will receive the minimum sentence of one year, and Henry will have to stay in jail for the maximum five years. Obviously, pleading guilty is the most attractive should the other plead not guilty since the sentence is only one year. However, if the other party also chooses to plead guilty, both will have to serve three years. On the other hand, if both parties plead not guilty, they'd have to serve two years in jail. Consequently, the risk of pleading not guilty is a five-year sentence, should the other choose to confess. Example: There are two player A and B with their a different strategies: A A1 A2 A3 B1 30 20 15 B B2 25 18 10 B3 19 15 25 SOLUTION :- A A1 A2 A3 B1 30 20 15 B B2 25 18 10 B3 19 15 25 B1 B2 B3 A1 30 20 15 A2 25 20 15 A3 19 15 25
• 65. Game Theory 65 B2 B3 A1 20 15 10/15 A3 15 25 5/15 10/15 5/15 15 V = 20X10 + 15X5 = 275 15 15 V = 15X10 + 25X5 = 275 15 15 B ( 0, 10/15, 5/15) A ( 10/15, 0, 5/15) Ans- EXAMPLE 2: There are two competitors Ramesh and Suresh having different strategies. Suresh S1 S2 S3 R1 10 5 -2 Ramesh R2 13 12 15 R3 16 14 10 Suresh S1 S2 S3 R1 10 5 -2 Ramesh R2 13 12 15 R3 16 14 10
• 66. Game Theory 66 S2 S3 R1 12 15 4/7 R3 14 10 3/7 5/7 2/7 7 V = 12X4+ 14x3 = 90 = 12.85 7 7 V = 12X5+ 15X2 = 90 = 12.85 7 7 Strategies: S(0,5/7,2/7) R(4/7,0,3/7) ANS- Example 2: The two car companies Maruti and Hyundai are competing with each other having different strategies. Maruti M1 M2 M3 M4 Hundai H1 18 13 5 10 H2 7 14 15 11 H3 6 3 14 16 Maruti M1 M2 M3 M4 Hundai H1 18 13 5 10 H2 7 14 15 11 H3 6 3 14 16
• 67. Game Theory 67 Maruti M1 M2 M3 Hundai H1 18 13 5 H2 7 14 15 Maruti M1 M2 Hundai H1 18 13 7 H2 7 14 5 1 11 12 V =18x7 +7x5 = 161 = 13.4 12 12 Maruti M1 M3 Hundai H1 18 5 8/21 H2 7 15 13/21 10/21 11/21 21 V =18x8 +7x13 = 235 = 11.19 21 21 In this case we will take the minimum value of the game that is 11.19 Strategies: Maruti ( 10/21,0,11/21,0) Hyundai ( 8/21,13/21,0) Ans- Graphical Representation of the above problem: Maruti M1 M2 M3 Hundai H1 18 13 5 H2 7 14 15H1 H2
• 68. Game Theory 68 18 18 17 17 M3 16 16 15 15 M2 14 14 13 13 12 12 11 11 10 10 9 9 8 8V=11.19 7 7 6 6 5 5 4 4 M1 3 3 2 2 1 1
• 69. Game Theory 69 Graphic Method at a glance: Row × Column Value of game Envelope Intersecting point 2 × m minimum Lower Highest M × 2 Maximum Upper Lowest Practice Problems: 1). There are two players Rafael Nadal and Roger Federer. They both have different strategies. Solve the game problem and find the Value of the game. Rafa R1 R2 R3 F1 20 28 14 Fedex F2 16 15 11 F3 24 19 12 Ans. Value of game=14 2). There are two players Lisa and Mona with different strategies. Find the value of game. Lisa L1 L2 L3 L4 M1 12 20 15 28 Mona M2 15 16 11 15 M3 16 25 17 26 Ans: Value of game=16 3). Two competitors X and Y are competing with each other.Both have different strategies against each other. Find value of game. Player x X1 X2 X3 Y1 16 25 13 Player y Y2 12 20 7 Y3 8 10 40 Ans: Value of the Game= 15.31
• 70. Game Theory 70 4). There are two players A and B having different strategies against each other. Find the Value of game. Player B B1 B2 B3 A1 1 7 2 Player A A2 6 2 7 A3 5 1 6 Ans: Value of game= 4 < 5). There are two players Gillette and Vi John. Both have different strategies to acquire better market share. Find the Value of Game. Show graphical representation. Gillette G1 G2 G3 V1 07 16 12 Vi John V2 09 17 19 V3 14 16 08 V4 12 15 11 Ans: Value of Game=12.125 6). Times of India and The Hindu are competing with each other having different strategies to win the market. Find the value of game and give the graphical representation. TOI T1 T2 T3 T4 T5 H1 03 06 03 07 09 Hindu H2 07 09 02 02 06 H3 09 10 07 04 15 Ans: Value of game= 5.28
• 71. Replacement Theory 71 REPLACEMENT THEORY The aim of the replacement theory was to show that quantitative analysis could help decision making within an organisation. INTRODUCTION Replacement theory is concerned with the problem of machinery, men or equipment is one which arises in every organisation. In any organisation, sooner or later all equipment needs to be replaced. Suppose, an organisation purchase an equipment and after few years or within a month it become less effective or useless due to either sudden or gradual deterioration in their efficiency, failure or breakdown and needs substantial expenditure on its maintenance. The problem, in such situation is, to determine the best policy to be adopted with respect to replacement of the equipment. The replacement theory provides answer to the following three types of situation in terms of optimal replacement period. a. Items such as machines, vehicles, tyres, etc. whose efficiency deteriorates with age due to constant use and which need increased operating and maintenance cost. In such cases deterioration level is predicted and is represented by (a) increased maintenance/operation cost, (b) its waste or scrap value and damage to them and safety risk. b. It such as light bulbs and tubes, radio and television parts, etc. which do not give any indication of deterioration with time but fail all of sudden and become completely useless. Such cases requires an anticipation of failure involving probabilities if failure. The optimum replacement policy is formulated to balance the wasted life of items replaced before against the costs incurred when items fail in service. c. In respect of replacement of employee in an organisation gradually reduces due to retirement, death, retrenchment and other reasons. DEFINITION: “Replacement theory deals with the analysis of assets/equipment which deteriorates with time and fix the optimal time of their replacement so that the total cost is the minimum.” Advantage of Replacement Theory: 1. It helps in decision making within an organisation. 2. It deals with the analysis of men, machines, equipment, etc which deteriorates with time and fix the optimal time to solve the issue. 3. By fixing the optimal time of the organisation with the help of replacement, it reduces the cost of maintenance and increased the production.
• 72. Replacement Theory 72 4. It helps management to taking decision when the old item has deteriorate and requires expensive maintenance. 5. It helps the management to taking the decision whether to replace now the expensive item which has deteriorate, or if not, when to reconsider replacement of the item in question. Limitation of Replacement Theory: 1. Future costs and resale prices of an equipment are predictable is quite Unrealistic. 2. Replacement theories in general are likely to be random variables calling for probabilistic approach to the analysis. 3. The historical costs in respect of an equipment requiring replacement are appropriate estimates of the costs for the equipment by which it is sought to be replaced may not hold. 4. Maintenance costs do not always follow a smooth pattern over a time. Very often, they are incurred in discrete lumps caused by such things as, for example, installing a new gearbox in van or reconditioning the body of the truck. In such cases, the best policy might be to be decide whether to scrap the vehicles (or other equipment) or undertake the major repairs to work and use it further. 5. In respect of the group replacement decision in case of items that fail suddenly, a practical objection is that all the items are replaced when group replacement take place whether they are working or not even if they are the ones that were installed only recently the ones that were. 6. The replacement could be carried out only at certain fixed time at the year ends or weekends. 7. In reality, the replacement can be, and is, done at any point during a given time. There are two types of items that lend themselves to replacement application. a. Items that deteriorate with age. b. Items that fail completely. When consideration items that deteriorates with age, chief elements of concern are the factors of: a. Increased operating cost, b. Idle time, and c. Increased repair cost.
• 73. Replacement Theory 73 Replacement models for the above situation should determine: Timing – When the equipment should be replaced. Selection – The type of replacement. Timing is complicated by decreasing salvage value, increasing Operating and maintenance cost and technological innovation. In essence, the problem also becomes that of estimating future cost, value and developments. There are given below graph of two possible approaches to the problem. First, equipment may be replaced when its performance declines to the point where it is no longer acceptable, the out put may be too low, quality too poor, breakdowns too frequent, etc. A drawback of this approach is that its response is too late as the equipment is already unsatisfactory. A better alternative would be to analyse costs and keep the equipment operating for the specific time which minimize to total costs. Performance of new equipment Minimal acceptable performance Replacement Replacement Time Performance (a) Operatingcostof newequipment Cost of operating Replacement Replacement Time (b)
• 74. Replacement Theory 74 Performance varying with age: (a) Performance deteriorating with time until replacement when a minimal acceptable is reached; (b) Increasing operating cost over time minimized by replacement policy. Types of Failure There are two types of failure are discussed below; Gradual Failure: It is a progressive in nature. That is, as the life of an item increases, its operational efficiency also deteriorates resulting in a. Increased the running cost. b. Decreased in its productivity c. Decreased in the resale or salvage value Mechanical items like pistons, rings, bearing tyres etc. fall under this category. Sudden Failure: This type of failure occurs in terms after some period of giving desired service rather than deterioration while in service. The period of giving desired services is not constant but follows some frequency distribution which may be progressive, retrogressive or random in nature. Types of replacement 1. Replacement of equipment which deteriorates with time 2. Replacements of items that fail completely 3. Staffing problems 1. Replacement of equipment which deteriorates with time Replacement decisions of such items are based on the economic life cycle cost concept or average monthly/ annual, in economic life cycle of, maintenance and operating cost in general increase with time and a stage come when these costs become so large that it is better to replace the item with new one. Some times in a Failure Gradual failure Sudden failure
• 75. Replacement Theory 75 system there are a number of alternatives in which we have to make comparison between the choices on the basis of cost involvement. 2. Replacement of items that fail completely. The second type of replacement problem is concerned with items that either work or fail completely. It is sometimes more economical to replace the group as a whole even if some of the items are functioning satisfactorily then to replace each item as it fails. Failures items have to be replace as they occur but it may be profitable at some stage. It is of two types:- a) Individual Replacement b) Group Replacement 3. Staff replacement The staff of an organization calls for replacement because people leave the organization due to inefficiency, resignation, retirement or death of employee from time to time. Therefore to maintain suitable strength of employee in system there is a need to formulate some recruitment policy. For this we assume that life distribution of the staff in a system is known to us. Steps for solving the problem; Step 1: Find the present value of the maintenance cost for each of the years, by multiplying the cost value by an appropriate present value factor (based on the given rate of discount and the time). Step 2: Accumulate present values obtained in step 1 up to each of the years 1, 2, 3... Add the cost of the equipment to each of these values. Step 3: Accumulate present value factors up to each of the year 1, 2, 3.… Step 4: Divide the cost plus cumulative maintenance costs for each year, obtained instep 2, by the corresponding cumulative present value factors in step 3. This gives the annualized costs for the various years. It may be noted that the annualized cost is also the weighted average of costs- the price value (in short PV) factors serve as weight here and the average is calculated as in the usual way, by dividing the summation of the products of the costs and their respective weights by the summation of the weights. KEY CONCEPTS: Year of Service- It is the year from purchasing year of the equipment till the year we are taking the service from it.
• 76. Replacement Theory 76 Running cost- It is the day to day cost incurred in operating equipment or performing unit. Cumulative running cost- It is the cost which become greater by stages or increasing by successive addition of running cost. Depreciation cost- It is the cost when there is a decrease in price or value of equipment due to its obsolescence or use. Total cost- It is the sum of cumulative cost and depreciation cost. Average cost- It is the total cost for all units produced divided by the year of service. Illustration: 1 A machine owner finds from his past records that costs pre year of maintaining a machine whose purchase price in ` 6,000 in as follows: Year 1 2 3 4 5 6 7 8 Maintenance Cost 1000 1200 1400 1800 2300 2800 3400 4000 Resale Price 3000 1500 1500 375 200 200 200 2000 Determine at what age machine should be replaced. Solution: Purchase Price (C) = ` 6,000 Determination of optimal replacement period Year Maintenance cost Cumulative Maintenance cost Resale Price (S) Capital Loss C –S Total Cost (T) Average Cost T/Year 1 2 3 4 5 6 7 8 1,000 1,200 1,400 1,800 2,300 2,800 3,400 4,000 1,000 2,200 3,600 5,400 7,700 10,500 13,900 17,900 3,000 1,500 1,500 375 200 200 200 2,000 3,000 4,500 5,250 5,625 5,800 5,800 5,800 4,000 4,000 6,700 8,050 11,025 13,500 16,300 19,700 23,700 4,000 3,350 2,950 2,756.25 2,700 2,716.6 2,814.29 2,962.50 The above table shows that the lowest average cost per year is at the end of 5th year, which is Rs. 2700.
• 77. Replacement Theory 77 Decision: Machine should be replaced after 5 yrs. Illustration: 2 The data of the operating cost per year and resale prices of equipment A whose purchase price is RS. 10000 are given here. Year 1 2 3 4 5 6 7 Operating Cost 1500 1900 2300 2900 3600 45000 55000 Resale Value 5000 2500 1250 600 400 400 400 A: What is the optimum period for replacement? B: When equipment A is 2 year old, equipment B , which is a new model for the same usage, is available the optimum period for replacement is 4 year with an AVG cost of RS.3600 should we change equipment A with that of B? If so when? Solution: Year Maintenance cost Cumulative Maintenance cost Capital Loss C –S Total Cost (T) Average Cost T/Year 1 1500 1500 5000 6500 6500.0 2 1900 3400 7500 10900 5450.0 3 2300 5700 8750 14450 4816.7 4 2900 8600 9400 18000 4500.0 5 3600 12200 9600 21800 4360.0* 6 4500 16700 9600 26300 4383.3 7 5500 22200 9600 31800 4542.9 Since the AVG. cost corresponding to the 5 yearly is the least, the optimal period of replacement = 5 years. (B) As the minimum average cost for equipment B is smaller than that for equipment. A, it is product to change the equipment. To decide the time of change, we would determine the cost of keeping the equipments in its 3rd, 4th, 5th year of life and compare each of these values with RS.3600 (the AVG. cost of equipment B). The equipment A shall be held as long as the marginal cost of holding it would be smaller than the minimum AVG. Cost for equipment B. The calculations are given below.
• 78. Replacement Theory 78 YEAR OPERATING COST DEPRECIATION TOTAL COST 3 2300 1250(2500-1250) 3550 4 2900 650(1250-600) 3550 5 3600 200(600-400) 3800 Illustration: 2 An airline requires 200 assistant hostesses, 300 hostesses, and 50 supervisors, Women are recruited at the age of 21, and if still in service retire at 60. Given the following life table, determine: a) How many women should be recruited in each year? b) At what age should promotion take place? Airline Hostesses Life Record Age No in services 21 1,000 22 600 23 480 24 384 25 307 26 261 27 228 28 206 Age No in services 29 190 30 181 31 173 32 167 33 161 34 155 35 150 36 146 Age No in services 37 141 38 136 39 131 40 125 41 119 42 113 43 106 44 99 Age No in services 45 93 46 87 47 80 48 73 49 66 50 59 51 53 52 46 Age No in services 53 39 54 33 55 27 56 22 57 18 58 14 59 11 - - Solution: If 1,000 women had been recruited each year for the past 39 years, then the total number of them recruited at the age of 21 and those serving up to the age of 59 is 6,480. Total numbers of women recruited in the airline are: 200+300+50=550. (a) Number of women to be recruited every year in order to maintain a strength of 550 hostesses 550 × (1,000/6,480) = 85 approx. (b) If the assistant hostesses are promoted at the age of X, then up to age (x - 1), 200 assistant hostesses will be required. Among 550 women, 200 are assistant hostesses. Therefore, out of strength of 1,000 there will be:
• 79. Replacement Theory 79 200 × (1,000/550) = 364 assistant hostesses. But from the life table given in the question, this number is available up to the age of 24 years. Thus, the promotion of assistant hostesses is due in the 25th year. Since out of 550 recruitments only 300 hostesses are needed, if 1,000 girls are recruited, then only 1000 × (300/500) = 545 (approx). will be hostesses. Hence, total number of hostesses and assistant hostesses in recruitment of 100 will be: 545 + 364 = 909. This means, only 1,000 – 909 = 91 supervisors are required. But from life table this number is available up to the age of 46 years. Thus promotion of hostesses to supervisors will be due in 47th year. ILLUSTRATION 3: A truck owner finds from his past records that maintained cost per year of a truck whose purchase price is Rs. 80000 are given below: Year 1 2 3 4 5 6 7 8 Maintenance cost 10000 13000 17000 22000 29000 38000 48000 60000 Resale price 40000 20000 12000 6000 5000 4000 4000 4000 Find the optimal replacement period. SOLUTION: Using the given data the minimum average annual cost of the equipment is computed in the table: Year Running cost M Total running cost ∑Mi Depreciation C-S Total cost C-S+∑Mi Average annual cost 1 10000 10000 40000 50000 50000 2 13000 23000 60000 83000 41500 3 17000 40000 68000 108000 36000 4 22000 62000 74000 166000 34000 5 29000 91000 75000 205000 33200 6 31000 129000 76000 136000 34167 7 48000 177000 76000 253000 36143 8 60000 237000 76000 313000 39125
• 80. Replacement Theory 80 Clearly the minimum average cost is in fifth year and is Rs. 33200. Hence the equipments must be replaced after 5 years ILLUSTRATION 4: A plant manager is considering replacement policy for a new machine. He estimates the following: Year Cost in Rupees Replacement Cost at the beginning of the year Salvage Value at the end of the Year (S) Operating Cost 1 30,000 18,000 7,500 2 33,000 15,000 9,000 3 37,500 12,000 12,000 4 42,000 7,500 15,000 5 48,000 3,000 19,500 6 57,000 0 24,000 SOLUTION: Table for different replacement costs will be as under: Year Maintenance Cost Cumulative Maintenance Cost ∑Mi Replacement Cost at the beginning of the year R Capital Loss R-S (Salvage Value) Total Cost ∑Mi + R-S Average Cost ∑Mi-R- S/i 1 7,500 7,500 30,000 12,000 19,500 19,500 2 9,000 16,500 33,000 18,000 34,500 17,250 3 12,000 28,500 37,500 25,500 54,000 18,000 4 15,000 43,500 42,000 34,500 78,000 19,500 5 19,500 63,000 48,000 45,000 1,08,000 21,600 6 24,000 87,000 57,000 57,000 1,44,000 24,000 Since the average cost for 2nd year (17250) is minimum the optimal replacement period is two year.
• 81. Replacement Theory 81 Numerical Question: 1. The following table gives the running costs per year and resale values of certain equipment whose purchase price is Rs6500. At what year is the replacement are optimally. Year 1 2 3 4 5 6 7 8 Running Cost 1400 1500 1700 2000 2400 2800 3300 3800 Resale Price 4000 3000 2200 1700 1300 1000 1000 1000 (Ans: After 5 years) 2. The cost of a machine is ` 10,000 and the other relevant information is given below: Period Running Cost Resale Price 1 1000 7000 2 1200 5000 3 1400 4750 4 1800 4375 5 2300 4200 6 2800 4200 7 3400 4200 8 4000 4200 Find the optimal replacement period for the machine. (Ans: 5yrs optimal replacement cost per year ` 2700.) 3. The Cost of a Scooter is ` 30000 and other details are given below: Year 1 2 3 4 5 6 Maintenance Cost 300 1000 1500 3000 4800 6000 Resale Value 20000 15300 12550 11000 6100 3400 Find the Optimal replacement period for the scooter. (Ans: 4years)
• 82. Replacement Theory 82 4. A machine owner finds from his past records that costs pre year of maintaining a machine whose purchase price in Rs. 6000 in as follows: Year 1 2 3 4 5 6 7 8 Maintenance cost 1000 1200 1400 1600 2300 2800 3400 4000 Resale price 3000 1500 1500 315 200 200 200 2000 Determine at what age machine should be replace? 5. A Fleet owner finds from his past records that the cost per year of running a vehicle, whose purchase price is RS 50,000 is: YEAR 1 2 3 4 5 6 7 RUNNING COST (RS) 5000 6000 7000 9000 11500 16000 18000 RESALE VALUE(RS) 30000 15000 7500 3750 2000 2000 2000 Thereafter, running cost increases by Rs 2000, but resale value remains constant at Rs 2000. At what age is a replacement due? 6. The cost of a machine is Rs 6100 and it’s scrap value is only Rs 100. The maintenance cost are found from experience to be: Year 1 2 3 4 5 6 7 8 Maintenance cost(Rs) 100 250 400 600 900 1250 1600 2000 When should the machine be replaced?
• 83. Queuing Theory 83 QUEUING THEORY INTRODUCTION Queuing theory deals with problems that involve waiting (or queuing). It is quite common that instances of queue occurs everyday in our daily life. Examples of queues or long waiting lines might be  Waiting for service in bank and at reservation counter.  Waiting for a train or bus.  Waiting at barber saloon.  Waiting at doctors’ clinic. Whenever a customer arrives at a service facility, some of them usually have to wait before they receive the desired service. This form a queue or waiting line and customer feel discomfort either mentally or physically because of long waiting queue. We infer that queues from because the service facilities are inadequate. If service facilities are increased, then the question arise how much to increase? For example, how many buses would be needed to avoid queues? How many reservation counters would be needed to reduce the queue? Increase in number of buses and reservation counters requires additional resources. At the same time, cost due to customer dissatisfaction must also be considered. Symbols and notations: n = total number of customers in the system, both waiting and in service µ = average number of customers being serviced per unit of time. λ = average number of customers arriving per unit of time. C = number of parallel service channels Ls or E(n) = average number of customers in the system, both waiting in the service. Lq or E(m) = average number of customers waiting in the queue Ws or E(w) = average waiting time of a customer in the system both waiting and in service Wq or E(w) = average waiting time of a customer in the queue Pn (t = probability that there are n customer in the queue
• 84. Queuing Theory 84 Queuing system The customers arrive at service counter (single or in a group) and attended by one or more servers. A customer served leaves the system after getting the service. In general, a queuing system comprise with two components, the queue and the service facility. The queue is where the customers are waiting to be served. The service facilivy is customers being served and the individual service stations. SERVICE SYSTEM The service is provided by a service facility (or facilities). This may be a person (a bank teller, a barber, a machine (elevator, gasoline pump), or a space (airport runway, parking lot, hospital bed), to mention just a few. A service facility may include one person or several people operating as a team. There are two aspects of a service system—(a) the configuration of the service system and (b) the speed of the service. Configuration of the service system The customers’ entry into the service system depends upon the queue conditions. If at the time of customers’ arrival, the server is idle, then the customer is served immediately. Otherwise the customer is asked to join the queue, which can have several configurations. By configuration of the service system we mean how the cost of waiting cost of service Level of Service FastLow High Total cost of the service service Cost Low optical service level
• 85. Queuing Theory 85 service facilities exist. Service systems are usually classified in terms of their number of channels, or numbers of servers. Single Server – Single Queue The models that involve one queue – one service station facility are called single server models where customer waits till the service point is ready to take him for servicing. Students’ arriving at a library counter is an example of a single server facility. Several (Parallel) Servers – Single Queue In this type of model there is more than one server and each server provides the same type of facility. The customers wait in a single queue until one of the service channels is ready to take them in for servicing Several Servers – Several Queues This type of model consists of several servers where each of the servers has a different queue. Different cash counters in an electricity office where the customers can make payment in respect of their electricity bills provide an example of this type of model.
• 86. Queuing Theory 86 Service facilities in a series In this, a customer enters the first station and gets a portion of service and then moves on to the next station, gets some service and then again moves on to the next station. …. and so on, and finally leaves the system, having received the complete service. For example, machining of a certain steel item may consist of cutting, turning, knurling, drilling, grinding, and packaging operations, each of which is performed by a single server in a series. Service Facility. Characteristics of Queuing System In designing a good queuing system, it is necessary to have a good. Information about the model. The characteristic listed below would Provide sufficient information. 1. The Arrival pattern. 2. The service mechanism. 3. The queue discipline. 4. The number of service channels. 5. Number of Service Stages 1. The Arrival pattern. Arrivals can be measured as the arrival rate or the interarrival time (time between arrivals).
• 87. Queuing Theory 87 Interarrival time =1/ arrival rate These quantities may be deterministic or stochastic (given by a propbability distribution). Arrivals may also come in batches of multiple customers, which is called batch or bulk arrivals. The batch size may be either deterministic or stochastic. (i) Balking: The customer may decide not to enter the queue upon Arrival, perhaps because it is too long. (ii) Reneging: The customer may decide to leave the queue after waiting a certain time in it. (ii i) Jockeying: If there are multiple queues in parallel the customers may switch between them. (iv) Drop-o®s: Customers may be dropped from the queue for reasons outside of their control. (This can be viewed as a generalisation of reneging.) 2. Service Pattern As with arrival patterns, service patterns may be deterministic or stochastic. There may also be batched services. The service rate may be state-dependent. (This is the analoge of impatience with arrivals.) Note that there is an important di®erence between arrivals and services. Services do not occur when the queue is empty (i.e. in this case it is a no-op). 3. Queue Discipline This is the manner by which customers are selected for service. (i) First in First Out (FIFO). (ii) Last in First Out (LIFO), also called (iii) Service in Random Order (SIRO). (iv) Priority Schemes. Priority schemes are either: Preemptive: A customer of higher priority immediately displaces any customers of lower priority already in service. The displaced customer's service may be either resumed from where it was left o®, or started a new. Non-Preemptive: Customers with higher priority wait current service completes, before being served.
• 88. Queuing Theory 88 4. The number of service channels 5. Number of Service Stages Customers are served by multiple servers in series. In general, a multistage queue may be a complex network with feedback Application of queuing theory: Queing theory has been applied to a great variety of business situations. Here we shall discuss a few problem s where the theory may be applied- 1) Waiting line theory can be applied to be determine the number of checkout counters needed to secure smooth and economic operations of its stored at various time during the day of a super market or a departmental store . 2) Waiting line theory can be used to analyze the delays at the toll booths of bridges and tunnels.
• 89. Queuing Theory 89 3) Waiting line theory can be used to improve the customers service at restaurants, cafeteria, gasoline service station, airline counters, hospitals etc, 4) Waiting line theory can be used to determine the proper determine the proper number docks to be constructed in the building of terminal facilities for trucks &ships. 5) Several manufacturing firms have attacked the problems of machine break down & repairs by utilizing this theory. Waiting line theory can be used to determine the number of personnal to be employed so that the cost of the production loss from down time & the cost f repairman is minimized. 6) Queuing theory has been extended to study a wage incentive plan Queuing theory (Limitations) 1) Most of the queuing models are quite complex & cannot be easily understood. 2) Many times form of theoretical distribution applicable to given queuing situations is not known. 3) If the queuing discipline is not in” first in, first out”, the study of queuing problems become more difficult. BASIC POINTS Customer: (Arrival) The arrival unit that requires some services to performed. Queue: The number of Customer waiting to be served. Arrival Rate (λ): The rate which customer arrive to the service station. Service rate (µ): The rate at which the service unit can provide services to the customer If Utilization Ratio Or Traffic intensity i.e λ /µ λ / µ > 1 Queue is growing without end. λ / µ < 1 Length of Queue is go on diminishing. λ /µ = 1 Queue length remain constant. When Arrival Rate (λ) is less than Service rate (µ) the system is working. i.e λ< µ (system work)
• 90. Queuing Theory 90 Formulas µ=Service Rate λ= Arrival Rate 1. Traffic Intensity (P)= λ /µ 2. Probability Of System Is Ideal (P0) =1-P P0 = 1- λ /µ 3. Expected Waiting Time In The System (Ws) = 1/ (µ- λ) 4. Expected Waiting Time In Quie (Wq) = λ / µ(µ- λ) 5. Expected Number Of Customer In The System (Ls)= λ / (µ-λ) Ls=Length Of System 6. Expected Number Of Customers In The Quie (Lq)= λ 2/ µ(µ- λ) 7. Expected Length Of Non-Empty Quie (Lneq)= µ/ (µ- λ) 8. What is the Probability Track That K Or More Than K Customers In The System. P >=K (P Is Greater Than Equal To K) = (λ /µ)K 9. What Is The Probability That More Than K Customers Are In The System (P>K)= (λ /µ)K+1 10. What Is The Probability That Atleast One Customer Is Standing In Quie. P=K=(λ /µ)2 11. What Is The Probability That Atleast Two Customer In The System P=K=(λ /µ)2 Solved Example Question 1. People arrive at a cinema ticket booth in a poison distributed arrival rate of 25per hour. Service rate is exponentially distributed with an average time of 2 per min. Calculate the mean number in the waiting line, the mean waiting time, the mean number in the system , the mean time in the system and the utilization factor? Solution: Arrival rate λ=25/hr Service rate µ= 2/min=30/hr
• 91. Queuing Theory 91 Length of Queue (Lq) = λ 2/ µ(µ- λ) = 252/(30(30-25)) =4.17 person Expected Waiting Time In Quie (Wq) = λ / µ(µ- λ) =25/(30(30-25)) =1/6 hr= 10 min Expected Waiting Time In The System (Ws) = 1/ (µ- λ) =1/(30-25) =1/5hr= 12 min Utilization Ratio = λ /µ =25/30 =0.8334 = 83.34% Question 2. Assume that at a bank teller window the customer arrives at a average rate of 20 per hour according to Poisson distribution .Assume also that the bank teller spends an distributed customers who arrive from an infinite population are served on a first come first services basis and there is no limit to possibl e queue length. 1. What is the value of utilization factor? 2. What is the expected waiting time in the system per customer? 3. What is the probability of zero customer in the system? Solution: Arrival rate λ=20 customer per hour Service rate µ= 30 customer per hour 1. Utilization Ratio = λ /µ = 20/30 = 2/3 2. Expected Waiting Time In The System (Ws) = 1/ (µ- λ) =1/(30-20) =1/10 hour = 6 min 3. Probability of zero customers in the system P0 = 1 – P =1- 2/3 = 1/3
• 92. Queuing Theory 92 Question 3 : Abc company has one hob regrinding machine. The hobs needing grinding are sent from company’s tool crib to this machine which is operated one shift per day of 8 hours duration. It takes on the average half an hour to regrind a hob. The arrival of hobs is random with an average of 8 hobs per shift. 1. Calculate the present utilization of hob regrinding machine. 2. What is average time for the hob to be in the regrinding section? 3. The management is prepared to recruit another grinding operator when the utilization of the machine increases to 80%. What should the arrival rate of hobs then be? Solution: Let us calculate arrival rate and service rate per shift of 8 hours. Arrival rate λ=8 shift Service rate µ=8x60/30=16 /shift 1. Percentage of the time the machine is busy Pb =arrival rate/service rate=8/16=0.50=50% 2. Average time for the hob to be in the grinding section. i.e., average time in the queue system=ws ws = 1/( µ- λ)=1/16-8=1/8 shift=1/8x8=1 hour 3. Let λ’ =arrival rate for which utilization of the machine will be 80%, Therefore, Pb ’ = λ’ / µ i.e., λ’ = Pb ’ . µ=0.80x16=12.8 per shift. Question: 4 (a) calculate expected number of persons in the system if average waiting time pf a customer is 45 or more than 45 minutes . b) if service rate is same. c) if arrival rate is same. Solution:-(a)expected no. of persons in a system(Ls ) =λ/μ- λ =45/65-45 =9/4 =3/4=1/65- λ λ =191/3 (b)Ws = 1/ μ –λ=1/65-45
• 93. Queuing Theory 93 =1/20 x60/1=3 mins. (c)ws =1/ μ- λ=1/6-4 = 3/4=1/ μ-45 =3 μ-135=4 =3 μ=139 μ =46.33 Question 5: In a factory, the machines break down and require service according to a poission Distribuation at the average of per day. What is the probability that exactly six Machines. Solution : Given λ = 4, n = 6, t = 2 p P(n,t) = (6,4) when λ = 4 We know, p (n,t) = (λt)n e-λt/ n! p(6,2) = (4×2)6 e-4×2/ 6! =86 e-8/720 =0.1221 Question 6: On an average , 6 customer arrive in a coffee shop per hour. Determine the probability that Exactly 3 customers will reach in a 30 minute period, assuming that the arrivals follow poison Distribution. Solution: Given, λ = 6 customers / hour t = 30 minutes = 0.5 hour n = 2 we know, p(n,t) = (λt)n e-λt/n! p(6,2) = (6×0.5)2 e-6×0.5/2! = 0.22404 Question 7: In a bank with a single sever, there are two chairs for waiting customers. On an average one customer arrives 12 minutes and each customer takes 6 minutes for getting served. Make suitable assumption, find (i) The probability that an arrival will get a chair to sit on, (ii) The probability that an arrival will have to stand, and (iii) Expected waiting time of a customer.
• 94. Queuing Theory 94 Solution following assumption are made for solving the given queuing problem: 1. The arrival rate is randomly distributed according to poission distribution. 2. The mean value of the arrival rate is λ . 3. The services time distribution approximated by an exponiential distribution and a mean rate of services is μ. 4. The rate of services is greather than the rate of arrival (μ>λ) 5. The queue discipline id FIFO. Arrival rate λ= 12min or 5 customer / hr Services rate μ = 6 min or 10 customer/ hr λ/μ = 5/10 = ½ there are two chairs including services one. (i) The probality that an arrival get a chair to seat on is given by: Pn (n<=2)= 1- Pn(n>2) 1-(λ/μ)3 1-(1/2)3 = 7/8 (II) The probability that an arrival will have to stand is given by 1-(P0+p1+P2) = 1-(7/8)= 1/8 (III) Expected waiting time of a customer in the queue is given by Wq =λ/μ(μ-λ) =5/10(10-5) = 1/(2*5) hr = 6 min Question 8: A television repairman finds that the time spent on his jobs has an expontial distribution with a mean of 30 minutes. If he repairs sets in the order in which they came in, and if the arrival of sets follow a passion distribution approximately with an average rate of 10 per 8- hour day, what is the repairman’s expected idle time each day? How many jobs are ahead of the average set just brought in? Solution from data of problem, we have Λ= 10/8=5/4 set per hour; and μ=(1/30)60= 2set per hour
• 95. Queuing Theory 95
• 96. Queuing Theory 96 (i) Expected idle time of repairmen each day Number of hour for a repairman remains busy in 8 hour day( traffic intensity) is given by (8) (λ/μ)=(8) (5/8)= 5 hour Hence, the idle time for a repairman in an 8 hour day will be : (8-5) =3 hour (ii) Expected (or average) number of TV set in the system LS = λ/μ-λ = 5/4/2-(5/4) =5/3 =2 (APPROX) T.V sets Unsolved question Question 1: Calculate expected number of person in the system. If average waiting time of customer is 30 min or more than 30 min , then services provider starts another windows. Calculate Arrival rate if service rate is same. Calculate service rate if arrival rate is same. (answer: Ws=1/5 hr, λ =13 µ = 2) Question 2: At a certain petrol pump, Customer arrive according to a passion process with a average time at 5 min between the arrivals. The service time is exponential distribution with mean 2 mins on the basic of this information. Find out:- a. Traffic intensity b. What would be the average quieting length? c. What is the expected number of customer at petrol pump? d. What is the expected number time one spend at petrol pump? e. What would we expected waiting time? f. What would be the proportion time the petrol pump is idle?
• 97. Queuing Theory 97 Answer a. 0.4 b. 0.26 c. 0.66 d. 0.02 e. 0.05 f. 0.6 < Question 3. The machines in production shop breakdown at an average of 2 per hour. The non productive time of any machine costs rs.30 per hour. If the cost of repairman is Rs.50 per hour. Calculate: a. Number of machines not working at any point of time. b. Average time that a machine is waiting for the repairman. c. Cost of non-productive time of the machine operator. d. Expected cost of system per hour. Answer. a:: 2 machines b: 2/3 hours c: Rs. 60 d: Rs.110 Question 4. In a bank cheques are cashed at a single ‘teller’ counter. Customers arrived at the counter in a Poisson manner at an average rate of 30 customers /hour. The teller takes on an average, a minute and a half to cash cheque. The service time has been shown to be exponentially distributed a) Calculate the percentage of time the teller is busy. b) Calculate the average time a person is expected to wait. Answer a) 3/4 b) 6 minutes
• 98. Queuing Theory 98 Question 5 : Telephone users arrive at a booth following a Poisson distribution with an average time of 5 minutes between one arrival and the next. The time taken for a telephone call is on a average 3 minutes and it follows an exponential distribution. What is the probability that the booth is busy? How many more booths should be established to reduce the waiting time less than or equal to half of the present waiting time. Answer a) 0.6 b) wq=3/40hrs. Question 6: Assume that goods trains are coming in a yard @ 30 trains per day and suppose that the inter arrival times follow an exponential distribution . the service time for each train is assumed to be exponential with an average of 36 minutes if the yard can admit 9 trains at a time(there being 10 lines one of which is reserved for shunting purpose).calculate the probability that the yard is empty and find the average queue length. Answer λ=1/48 μ=λ/16 p=0.75 Po=o.28 Lq=1.55 Question 7. At what average rate must a clerk at a supermarket in order to ensure a probability of 0.90 so that the customer will not wait longer than 12 minutes ? It is assumed that there is only one counter at which customers arrive in a Poisson fashion at an average rate of 15/hour. The length of service by the clerk has an exponential distribution. Answer: 2.48 minutes /service Question 8. The beta company ‘s quality control deptt. Is managed by a single clerk, who takes an average 5 minutes in checking part of each of the machine coming for inspection. The machine arrive once in every 10 min. on the average one hour of the machine is valued at Rs 25 and cost for the clerk is at rs 5 per hour. What are the average hourly queueing system cost associated with the quality control department. Answer Rs 30 per hour
• 99. Decision Theory 99 DECISION THEORY The success or failure of an individual or organization experiences, depends upon the ability to make appropriate decision. For making appropriate decision it requires certain course of action or strategies which should be feasible (possible) and viable (exists) in nature. Decision theory provides an analytical and systematic approach to depict the expected result of a situation when alternative managerial actions and outcomes are compared. Decision theory is the combination of descriptive and prescriptive business modeling approach i.e., it is concerned with identifying the best decision to take, assuming an ideal decision maker who is fully informed, able to compute with perfect accuracy, and fully rational. The practical application of this prescriptive approach (how people actually make decisions) is called decision analysis, and aimed at finding tools, methodologies and software to help people make better decisions which can be classified as a degree of knowledge. The knowledge of degree is divided into four categories which are given below:- Characteristics of problem formulation : A. Decision alternatives: In this case, N numbers of alternatives are available with the decision maker whenever the decision is made. These alternatives may depend on the previous decisions made. These alternatives are also called courses of action which are under control and known to decision maker. B. States of nature: These are the future conditions ( also known as consequences, events, or scenarios) which are not under the control of decision maker. A state of nature can be inflation, a weather condition, a political development etc. it usually is not determined by an action of an individual or an organization. But it may identify through some technique such as scenario analysis. Ex- stakeholders, long-time managers.
• 100. Decision Theory 100 C. Pay off: A numerical outcome resulting from each possible combination of alternatives and states of nature is called payoff. The payoff values are always conditional values because of unknown states of nature. The payoff is measured within a specified period (e.g. after one year). This period is sometimes called decision horizon. Payoff can be measured in terms of money market share, or other measures. D. Pay off table: A tabular arrangement of these conditional outcomes (profit or loss values) is known as payoff matrix. To construct a payoff matrix, the decision alternatives (courses of action or strategies) and states of nature are represented in the tabular form as below: States of nature(events) Decision alternative ( courses of action) A1 A2 A3 …… AM E1 A11 A12 A13 …… A1m E2 A21 A22 A23 …… A2m E3 A31 A32 A33 …… A3m ….. ….. …… ….. …… …… EN An1 An2 An3 …… Amn Steps in decision theory approach  Identify and define the problem.  Listing of all possible future events, called states of nature. Such events are not under control of decision maker.  Identification of all the courses of action which are available to the decision- maker.  Evaluating the alternatives such as, cost effectiveness, performance, quality, output, profit.  Expressing the pay-offs resulting from each pair of course of action and state of nature.  Choosing an appropriate course of action from the given list on the basis of some criterion that result in the optimal pay-off.  The next step is to implement the decision.
• 101. Decision Theory 101 Types of decision making environments Decisions are made based upon the data available about the occurrence of events as well as the decision situation. There are four types of decision making environment which are follows: 1. Decision–making under certainty: In this type the decision maker has the perfect information about the consequences of every course of action or alternatives with certainty. Definitely he selects an alternatives that gives the maximum return (pay-off) for the given state of nature. For ex- one have a choices either to purchase national saving certificate, Indira Vikas Patra or deposit in national saving scheme. Obviously he will invest in one the scheme which will give him the assured return. In these decision model only one possible state of nature exists. 2. Decision-making under risk: in this type, the decision maker has less information about the certainty of the consequence of every course of action because he is not sure about the return. In these decision model more than one state of nature exists for which he makes an assumption of the probability with each state of nature which will occur. For ex- probability of getting head in the toss of a coin is 50%. 3. Decision-making under uncertainty: In this type, the decision-maker is unable to predict the probabilities of the various states of nature which will occur. Here the possible states of nature are known but still there is a less information than Define the problem Gather information Search for alternatives Evaluates alternatives Select alternative for action Implement decision
• 102. Decision Theory 102 the decision under risk. For ex- The probability that Mr. Y will be the captain of the Indian cricket team for coming 10 years from now is not known. 4. Decision-making under conflict: In this type, the consequences of each act of the decision maker are influenced by the acts of opponent. An example of this is the situation of conflict involving two or more competitors marketing the same product. The technique used to solve this category is the game theory. Decision Making under risk: Decision under risk is a probabilistic decision situation, in which more than one state of nature exists and the decision maker has sufficient information to assign probability values to the likely occurrence of each of these states. Knowing the probability distribution of the states of nature, the best decision is to select the course of action which has the largest expected payoff value. The expected (average) payoff of an alternative is the sum of all possible payoffs of that alternative weighted by the probabilities of those payoffs occurring. The most widely used criterion for evaluating various courses of action under risk : 1. Expected Monetary Value (EMV) or Expected utility. 2. Expected opportunity Loss (EOL). 3. Expected value of Perfect Information (EVPI) Expected monetary value: The expected value (EMV) for a given course is the weighted sum of possible payoffs for each alternative. It is obtained by summing the payoffs for each course of action multiplied by the probabilities associated with each state of nature. The expected (or mean) value is the long-run average value that result if the decision were repeated a large number of times. Steps for calculating EMV: The various steps involved in the calculation of EMV are as follow: 1. Construct a payoff matrix listing all possible courses of action and states of nature. Enter the conditional payoff values associated with each possible combination of course of action and state of nature along with probabilities of the occurrence of each state of nature. 2. Calculate the EMV for each course of action by multiplying the conditional payoffs by the associated probabilities and add these weighted values for each course of action. 3. Select the course of action that yields the optimal EMV. Expected opportunity loss (EOL) : An alternative approach to maximizing expected monetary value (EMV) is to minimize the expected opportunity loss (EOL) also called expected value of regret. The EOL is defined as the difference between the
• 103. Decision Theory 103 highest profit (highest payoffs) for a state of nature and the actual profit obtained for the particular course of action taken. In other words, EOL is the amount of payoff that is lost by not selecting the course of action that has greatest payoff for the state of nature that actually occur. The course of action due to which EOL is minimum is recommended. Since EOL is an alternative decision criterion for decision making under risk, therefore the results will always be the same as those obtained by EMV criterion. Thus only one of the two methods should be applied to reach to a decision. It is stated as follows: Steps for calculating EOL: The steps which are involved in the calculation of EOL are as follows: 1. Prepare a conditional profit table for each course of action and state of nature combination along with the associated probabilities. 2. For each state of nature calculate the conditional opportunity loss (COL) values by subtracting each payoff from the maximum payoff for that outcome. 3. Calculate EOL for each course of action by multiplying the probability of each state of nature with the COL value and then adding the values. 4. Select a course of action for which the EOL value is minimum. Expected value of perfect information (EVPI): In these decisions making under risk each state of nature as associated with the probability of its occurrence. Perfect information about the future demand would remove uncertainty for the problem. With these perfect information the decision maker would know in advance exactly about the future demand and he will be able to select a course of action that yields the desired payoff for whatever state of nature that actually occurs. EVPI represents the maximum amount the decision maker has to pay to get to this additional information about the occurrence of various events. EVPI = (expected profit with perfect information)- (expected profit without perfect information). Examples: Q) A shopkeeper buys apple for Rs 20/kg and sells them for Rs 30/kg. The past records of the sales are as follows: No of customers: 50 80 100 120 150 No of Days: 20 30 20 10 20
• 104. Decision Theory 104 Ans: Profit =Selling price – Cost price. = 30-20 = 10 Probability Demand Supply 50 80 100 120 150 20/100= 0.20 50 500 -100 -500 -900 -1500 30/100= 0.30 80 500 800 400 0 -600 20/100= 0.20 100 500 800 1000 600 0 10/100= 0.10 120 500 800 1000 1200 600 20/100= 0.20 150 500 800 1000 1200 1500 Total : 2500 3100 2900 2100 0 Maximum value : 500 800 1000 1200 1500 Minimum value : 500 -100 -500 -900 -1500 Maxi max: 1500 out of all maximum values. Maxi min: 500 out of all minimum values. Laplace: 2500/5 3100/5 2900/5 2100/5 0/5 = 500 = 620 = 580 = 420 = 0 620 in 80 units. step 1. Write the demand & probability in column and supply in row. Step 2. Calculate the Expected Pay off table (probability × pay off) Supply Probability Demand 50 80 100 120 150 20/100= 0.20 50 100 -20 -100 -180 -300 30/100= 0.30 80 150 240 120 0 -180 20/100= 0.20 100 100 160 200 120 0 10/100= 0.10 120 50 80 100 120 60 20/100= 0.20 150 100 160 200 240 300620 A
• 105. Decision Theory 105 Total 500 620 520 300 -120 Expected monetary value (EMV): 620 (the highest one among the total). Expected profit on perfect information (EPPI): (100+240+200+120+300) = 960. Expected value of perfect competition (EVPI): 960 – 620 = 340. Expected opportunity loss table: Deduct the highest number from expected payoff table from each row. Probability Demand Supply 50 80 100 120 150 20/100= 0.20 50 0 120 200 280 400 30/100= 0.30 80 90 0 120 240 420 20/100= 0.20 100 100 40 0 80 200 10/100= 0.10 120 70 40 20 0 60 20/100= 0.20 150 200 140 100 60 0 Total 460 340 440 660 1080 Expected opportunity loss table will be 340. The minimum among all total. Expected value of perfect information (EVPI)= Expected opportunity of loss table (EOL). Decision making under uncertainty: In the absence of information about the probability of any state of nature occurring, the decision-maker must arrive at a decision only on the actual conditional pay-offs values, together with a policy. There are several different criteria of decision making in these situation. The criteria are as follows:- i. Optimism (Maximax or Minimin) criterion. ii. Pessimism (Maximin or Minimax) criterion. iii. Equal probabilities (Laplace) criterion. iv. Coefficient of optimism (hurweiz) criterion. v. Regret (salvage) criterion. i. Optimism criterion: In this criterion the decision-maker always looks for the maximum possible profit (Maximax) or lowest possible (Minimin).Therefore he selects the alternatives that maximum of the maxima (or minimum of the minima) pay-offs. The method are as follows: 340
• 106. Decision Theory 106 a) Find the maximum (or minimum) payoff values corresponding to each alternative courses of action. b) Select the alternative with the best anticipated payoff value i.e., maximum profit and minimum profit. Examples: Strategies States of nature S1 S2 S3 P1 2,00,000 5,00,000 3,00,000 P2 4,00,000 1,50,000 9,00,000 P3 0 4,50,000 7,00,000 Strategies States of nature S1 S2 S3 P1 2,00,000 5,00,000 3,00,000 P2 4,00,000 1,50,000 9,00,000 P3 0 4,50,000 7,00,000 Column maximum 4,00,000 5,00,000 9,00,000 . Ans: The maximum of column maxima is 9,00,000. Hence the company should adopt strategy S3. ii. Pessimism criterion: in this criterion the decision-maker ensures that he should not earn no less (or pay no more) than some specified amount. Thus, he selects the alternative that represents the maximum of the minima payoff in case of profits. The methods are as follows : a) Find the minimum (or maximum in case of profits) payoff values in case of loss (or cost) data corresponding to each alternative. b) Select an alternative with the best anticipated payoff value (maximum for profit and minimum for loss or cost). Maximax
• 107. Decision Theory 107 Examples: Strategies States of nature S1 S2 S3 P1 2,00,000 5,00,000 3,00,000 P2 4,00,000 1,50,000 9,00,000 P3 0 4,50,000 7,00,000 Solution:- Strategies States of nature S1 S2 S3 P1 2,00,000 5,00,000 3,00,000 P2 4,00,000 1,50,000 9,00,000 P3 0 4,50,000 7,00,000 Column minimum 0 1,50,000 3,00,000 Ans: The maximum of the row is 3,00,000. iii. Equal probabilities (Laplace) criterion : The probabilities of states of nature are not known, so it is assumed that all states of nature will occur with equal probability. As state of nature are mutually exclusive and collectively exhaustive, so the probability of each of these must be 1/(number of states of nature). The methods are as follows:- a) Assign equal probability value to each state of nature by using the formula: = 1/(number of states of nature). b) Calculate the expected (or average) payoff for each alternative (course of action) by adding all the payoffs and dividing by the number of possible states of nature or by applying the formula: = (probability of state of nature) *(payoff value for combination of alternative, and state of nature) c) Selected the best expected payoff value (maximum profit and minimum cost). Maximin
• 108. Decision Theory 108 Strategy Expected return S1 2,00,000 + 4,00,000 + 0 = 6,00,000/3= 2,00,000 S2 5,00,000 + 1,50,000 + 4,50,000 = 11,00,000/3 = 3,66,666 S3 3,00,000 + 9,00,000 + 7,00,000 = 19,00,000/3 =6,33,333 Since the largest expected return is from strategy S3.. the executive must select strategy S3. iv. Coefficient of optimism (hurwicz) criterion: In this criterion a decision maker should neither be completely optimistic nor of pessimistic. It should be a mixture of both. Hurwicz who suggested this criterion, introduced the idea of coefficient of optimism (denoted by α) to measure the degree of optimism. This coefficient lies between 0 and 1 represents a complete pessimistic attitude about future and 1 a complete optimistic attitude about future. Thus if α is the coefficient of optimistic, then (1-α) will represent the coefficient of pessimism. Hurwicz approach suggests that the decision maker must select an alternative that maximizes H(criterion of realism) =α (maximum in column)+ (1-α) minimum in column. The methods are as follows: a) Decide the coefficient of optimism α and then coefficient of pessimism(1 – α) b) For each alternative select the largest and the lowest payoff value and multiply these with α and (1-α) values, respectively. Then calculate the weighted average, H by using above formula. c) Select an alternative with best anticipated weighted average payoff value. Examples: Let the degree of optimism being 0.7. Strategies States of nature S1 S2 S3 P1 2,00,000 5,00,000 3,00,000 P2 4,00,000 1,50,000 9,00,000 P3 0 4,50,000 7,00,000
• 109. Decision Theory 109 Strategies States of nature Maximum pay off (i) Minimum pay off (ii) H =α(i) +(1-α) (ii) P1 5,00,000 2,00,000 0.7×5,00,000 +0.3×2,00,000=4,10,000 P2 9,00,000 1,50,000 0.7×9,00,000 +0.3×1,50,000=6,75,000 P3 7,00,000 0 0.7×7,00,000+0.3×0=4,90,000 The maximum value of H = 6,75,000 v. Regret (savage) criterion: In this, criterion is also known as opportunity loss decision criterion or minimax regret decision criterion because decision maker feels regret after adopting a wrong course of action resulting in an opportunity loss of payoff. Thus he always intends minimize this regret. The method are as follows: a) Find the best payoff corresponding to each state of nature. b) Subtract all other entries (payoff values) in that row from this value. c) For each course of action identify the worst or maximum regret table . Record this number in a new row. d) Select the course of action with the smallest anticipated opportunity- loss value. Examples: Strategies States of nature S1 S2 S3 P1 2,00,000 5,00,000 3,00,000 P2 4,00,000 1,50,000 9,00,000 P3 0 4,50,000 7,00,000
• 110. Decision Theory 110 Solution:- Strategies States of nature S1 S2 S3 P1 5,00,000- 2,00,000=3,00,000 5,00,000-5,00,000=0 5,00,000- 3,00,000=2,00,000 P2 9,00,000- 4,00,000=5,00,000 9,00,000- 1,50,000=7,50,000 9,00,000-9,00,000=0 P3 7,00,000-0=7,00,000 7,00,000- 4,50,000=2,50,000 7,00,000-7,00,000=0 Column Maximum 7,00,000 7,50,000 2,00,000 Q 1) The following matrix gives the payoff of different strategies (alternatives) S1,S2,S3, against conditions N1,N2,N3, and N4. N1 Rs. N2 Rs. N3 Rs. N4 Rs. S1 4,000 -100 6000 18000 S2 20,000 5000 400 0 S3 20,000 15000 -2000 1000 Calculate the decision taken under the following approach : i. Pessimistic (maximin). Ans: 0 with strategy S2. ii. Optimistic (maximax) Ans: Rs 20,000 with strategy S2 and S3. iii. Regret Ans: Rs 16,000 with strategy S1. iv. Equal probability (Laplace) Ans: Rs 8500 with strategy S3. v. Hurwicz criterion, the degree Of optimism being 0.7 Ans: Rs 14,000 with strategy S2. MiniMax Regret
• 111. Decision Theory 111 Q 2) A producer of boats has estimated the following distribution of demand for a particular kind of a boat. No. of demanded : 0 1 2 3 4 5 6 Probability : 0.14 0.27 0.27 0.18 0.09 0.04 0.01 Each boat costs him Rs 7000 and sells them for Rs 10,000 each. Any boats that are left unsold at the end of the season must be disposed of for Rs 6000 each. How many boats should be in stock so as to maximize his expected profit? Ans: 3 boats to get a profit of Rs.4080
• 112. Linear Programming Problem 112 LINEAR PROGRAMMING INTRODUCTION In every sphere of Human endeavor, we come across with the problem of allocation of limited resources in competitive activities. In business situations also we often face problems of allocating of fixed set of resources among a set of competing demands e.g. Situations where we have to decide a product mix within the manufacturing and marketing constraints, preparation of a schedule for transporting goods from the places where they are available to the places of demand, etc.. In business problems the criteria for selecting among alternatives is minimizing total cost, maximization total contribution, maximizing total labour utilization etc. What is Linear Programming? Linear Programming (LP) is a particular type of technique used for economic allocation of ‘scarce’ or ‘limited’ resources, such as labour, material, machine, time, warehouse space, capital, energy, etc. to several competing activities, such as products, services, jobs, new equipment, projects, etc. on the basis of a given criterion of optimality. The phrase ‘scarce resources’ means resources that are limited in availability during the planning period. We learn to optimum use of these resources through linear planning. The criterion of optimality generally is either performance, return on investment, profit, cost, utility, time, distance, etc. Before discussing in detail the basic concepts and applications of linear programming, let us be clear about the two words, ‘linear and programming’. The term “linear” is used because of the fact that all the relations among the variables are linear. On the other hand, the word “programming” refers to modeling and solving a problem mathematically that involves the economic allocation of limited resources by choosing a particular course of action or strategy among various alternative strategies to achieve the desired objective. Historical Development George Bernard Dantzig (November 8, 1914 – May 13, 2005) was an American mathematician, and the Professor Emeritus of Transportation Sciences and Professor of Operations Research and of Computer Science at Stanford. Dantzig is known for his development of the simplex algorithm, an algorithm for solving linear programming problems, and his work with linear programming, some years after it was initially invented by Soviet economist and mathematician Leonid Kantorovich. Dantzig is the subject of a tale, often thought to be fictional, of a student solving an important unsolved problem after mistaking it for homework.
• 113. Linear Programming Problem 113 In 1946, as mathematical adviser to the U.S. Air Force Comptroller, he was challenged by his Pentagon colleagues to see what he could do to mechanize the planning process, "to more rapidly compute a time-staged deployment, training and logistical supply program." In those pre-electronic computer days, mechanization meant using analog devices or punched-card machines. "Program" at that time was a military term that referred not to the instruction used by a computer to solve problems, which were then called "codes," but rather to plans or proposed schedules for training, logistical supply, or deployment of combat units. The somewhat confusing name "linear programming," Dantzig explained in the book, is based on this military definition of "program." In 1963, Dantzig’s Linear Programming and Extensions was published by Princeton University Press. Rich in insight and coverage of significant topics, the book quickly became “the bible” of linear programming. General Structure of Linear programming Model The general structure of LP model consists of three basic components which are as under: Decision Variables (Activities): First we evaluate various alternatives for arriving at the optimum value of objective function. The evaluation of various alternatives is guided by the nature of objective function and availability of resources. For this we use certain activities usually called decision variables and denoted by X1, X2,………………………….Xn in an LP model all decision variables are continuous, controllable and non negative. That is X1 ≥0, X2≥0……………….. , Xn≥0. Objective function: The objective function of each LP problem is expressed in terms of decision variables to optimize the criterion of optimality (also known as measure of performance ) such as profit, cost, revenue, distance etc. In general form it is represented as Optimize (maximize or minimize) Z = C1X1 + C2X2+…………………………… + CnXn. Constraints: There are certain limitations on the use of resources like labour, machine, raw material, space, money etc. that limit of which an objective can be achieved. Such constraints must be expressed as linear equalities or inequalities in terms of decision variables. The solution of an LP model must satisfy these constraints.
• 114. Linear Programming Problem 114 Assumptions of Linear Programming The following are the four basic assumptions which are necessary for all linear programming models: 1. CERTAINTY: In all LP models, it is assumed, that all model parameters such as availability of resources, profit (or cost) contribution of a unit of decision variable and consumption of resources by a unit of decision variable must be known and may constant. In some cases, these may be either random variables represented by a known distribution (general or may statistical) or may tend to change, then the given problem can be solved by a stochastic LP model or parametric programming. 2. DIVISIBILITY (or CONTINUITY): The solution values of decision variables and resources are assumed to have either whole numbers (integers) or mixed numbers (integer and fractional). However, if only integer variables are desired, e.g. machines, employees, etc. the integer programming method may be applied to get the desired values. 3. ADDITIVELY: The value of the objective function for the given values of decision variables must be equal to the sum of the contributions (profit or cost) earned from each decision variable and the total sum of resources used, must be equal to the sum of the resources used by each decision variable. For example, the total profit earned by the sale of two products A and B must be equal to the sum of the profits earned separately from A and B. Similarly, the amount of a resource consumed by A and B must be equal to the sum of resources used for A and B individually. 4. LINEARITY (or PROPORTIONALITY): All relationships in the LP model (i.e. in both objective function and constraints) must be linear. In other words, for any decision variable, the amount of particular resource used and its contribution to the cost one in objective function must be proportional to its amount. For example, if production of one unit of a product uses 5 hours of a particular resource, then making 4 units of that product uses 4x5=20 hours of that resource. Formulating a problem as an LP model 1) Once the problem has been described, the next step is to transform it into a proper mathematical structure. General steps taken are as under:
• 115. Linear Programming Problem 115 2) Define the decision variables that are relevant to the problem and, ensure that their units of measurement are explicitly stated. 3) Identify the contribution coefficient (the cj’s) associated with each variable. 4) Formulate the objective function quantitatively and express it as a linear function of decision variables. 5) Identify the physical rate of substitution coefficient(the aij’s) 6) Identify the variable resources or requirement ,i.e. , the right-hand-side coefficient(the bj’s) 7) Formulate suitable mathematical constraints related to each respective resource or requirement as linear equalitiesinequalities in terms of decision variables. 8) Mention the non-negativity condition associated with the decision variables. APPLICATION AREAS OF LINEAR PROGRAMMING: Linear programming is the most widely used technique of decision-making in business and industry and in various other fields. Here, we will discuss a few of the broad application areas of linear programming.  MILITARY APPLICATIONS: Military applications include the problem of selecting an air weapon system against enemy so as to keep them pinned down and at the same time minimizing the amount of aviation gasoline used. A variation of the transportation problem that maximizes the total tonnage of bombs dropped on a set of targets and the problem of community defence against disaster, the solution of which yields the number of defence units that should be used in a given attack in order to provide the required level of protection at the lowest possible cost.  AGRICULTURAL APPLICATIONS: The study of farm economics deals with inter-regional competition and optimum allocation of crop production. Efficient production patterns can be specified by a linear programming model under regional land resources and national demand constraints. Linear programming can be applied in agricultural planning, e.g. allocation of limited resources such as acreage, labour, water supply and working capital, etc. in a way so as to maximize net revenue.
• 116. Linear Programming Problem 116  PRODUCTION MANAGEMENT: (i) PRODUCT MIX: A company can produce several different products, each of which requires the use of limited production resources. In such cases, it is essential to determine the quantity of each product to be produced knowing its marginal contribution and amount of available resource used by it. The objective is to maximize the total contribution, subject to all constraints. (ii) PRODUCTION PLANNING: This deals with the determination of minimum cost production plan over planning period of an item with a fluctuating demand, considering the initial number of units in inventory, production capacity, constraints on production, manpower and all relevant cost factors. The objective is to minimize the total operation costs. (iii) ASSEMBLY-LINE BALANCING: This problem is likely to arise when an item can be made by assembling different components. The process of assembling requires some specified sequence(s). The objective is to minimize the total elapse time. (iv) BLENDING PROBLEMS: These problems arise when a product can be made from a variety of available raw materials, each of which has a particular composition and price. The objective is to determine the minimum cost blend, subject to availability of the raw materials, and minimum and maximum constraints on certain product constituents. (v) TRIM LOSS: When an item is made to a standard size (e.g. glass, paper sheet), the problem that arises is to determine which combination of requirements should be produced from standard materials in order to minimize the trim loss.  FINANCIAL MANAGEMENT: 1) PORTFOLIO SELECTION : This deals with the selection of specific investment activity among several other activities. The objective is to find the allocation which minimizes the expected return of minimizes risk under certain limitations.
• 117. Linear Programming Problem 117 2) PROFIT PLANNING: This deals with the maximization of the profit margin from investment in plant facilities and equipment, cash in hand and inventory.  MARKETING MANAGEMENT: 1) MEDIA SELECTION: Linear programming technique helps in determining the advertising media mix so as to maximize the effective exposure, subject to limitation of budget, specified exposure rates to different market segments, specified minimum and maximum number of advertisements in various media. 2) TRAVELLING SALESMAN PROBLEM: The problem of salesman is to find the shortest route from a given city, visiting each of the specified cities and then returning to the original point of departure, provided no city shall be visited twice during the tour. Such type of problems can be solved with the help of the modified assignment technique. 3) PHYSICAL DISTRIBUTION: Linear programming determines the most economic and efficient manner of locating manufacturing plants and distribution centers for physical distribution.  PERSONNEL MANAGEMENT: 1) STAFFING PROBLEM: Linear programming is used to allocate optimum manpower to a particular job so as to minimize the total overtime cost or total manpower. 2) DETERMINATION OF EQUITABLE SALARIES: Linear programming technique has been used in determining equitable salaries and sales incentives. 3) JOB EVALUATION AND SELECTION: Selection of suitable person for a specified job and evaluation of job in organizations has been done with the help of linear programming technique. Other applications of linear programming lie in the area of administration, education, fleet utilization, awarding contracts, hospital administration and capital budgeting, etc.
• 118. Linear Programming Problem 118 Advantages and disadvantages of Linear Programming: Advantages Drawbacks 1) Insights and perspective into problem solutions. It helps in organization and study of the information in the same way that the scientific approach to the problem requires. 2) Consideration of all possible solutions to the problem. Many management problems are so complex that the difficulty is encountered in planning and feasible solution. By using the LP technique, the manager makes sure that he is considering the best (Optimal) solution for solutions) 3) Better and more successful decisions. With linear programming the executive builds into his planning a true reflection of the limitations and restrictions under which he must operate. When it becomes necessary to deviate from the best programme, he can evaluate the cost or penalty involved. 4) Better tools for adjusting to meet changing conditions once a basic plan is arrived at through linear programming, it can be reevaluated for changing conditions. If conditions change when the plan is partly carried out, they can be determined so as to adjust the remainder of the plan for best results. 5) Highlighting of bottlenecks in the production process is the most significant advantage of this technique E.g.When bottleneck occurs, some machines cannot meet demand while others remain idle for some of the time 6) It also helps the mangers to shave better understanding about the phenomenon and various activities of the organization for the organization construction of suitable mathematical model visualizing the relationship between variables if any ad making improvement over them. 1) In LP problem, fractional values are permitted for the decision variables However, many decision Problems require that the solution for decision variables should be obtained in non- fractional values. Rounding-of the values obtained by linear programming techniques may not result into an optimal Solution. 2) In LP problems, coefficients in the objective function and the constraint equations must be completely known and they should not change during the period of study. In practical situation, it may not be possible to state all coefficients with certainty. 3) It may not possible to solve those problems using LP in which non- linearity arises because of joint interactions between some of the activities regarding the total measure of effectiveness or total usage of some resources. 4) Does not take into consideration the effect of time and uncertainty 5) Parameters appearing in the model are assutmed to be constants but in real-life situations they are frequently either known nor constants. 6) Many real-world problems are so complex in terms of the numbers of variables ad relations constrained in them, that they tax the capacity of even the largest computer.
• 119. Linear Programming Problem 119 PROBLEM FORMULATION: Q: manufacturer wants to manufacture tables and chairs. He will make the profit of Rs.50 per table and Rs.35 per chair. The quantity of raw material which he has- Raw Material Quantity Available (maximum) Ply 40 meter Steel 50 meter Plastic 60 meter Quantity used in one Table and one Chair Raw Material Table Chair Ply 2m 2m Steel 5m 2m Plastic 1m 4m How many tables and chairs he should make to maximize his profit? SOL: PROBLEM FORMULATION: Let assume x1 unit of table and x2 unit of chair made. Then, Max.Z= 50x1 + 35x2 Z= profit Max.Z= maximum profit, Subjective Equations: 2x1 + 2x2 ≤ 40 (for ply) 5x1 + 2x2 ≤ 50 (for steel) x1 + 4x2 ≤ 60 (for plastic) x1, x2 ≥ 0. Graphical method: The graphic solution procedure is one method of solving two variable linear programming. The problems involve following steps: 1. Formulate the problem in terms of mathematical constraints and an objective function.
• 120. Linear Programming Problem 120 2. Plot each of the constraints and convert each equality in the constraints equation to equality. 3. Identify the feasible region i.e. the area which satisfy all the constraints simultaneously Some important definition: 1. Solution: Values of decision variables xj(j=1,2,……..,n)which satisfy the constraints of general L.P problem is called the solution to that L.P problem. 2. Feasible solution: Any solutions that also satisfy the non-negative restriction of the general L.P problem is called a feasible solution. 3. Basic solution: For a set of m simultaneous equation in n unknowns (n>m), a solution obtained by setting (n-m) of the variables equal to zero and solving the remaining m equations in m unknowns is called a basic solution. 4. Basic feasible solution: A feasible solution to a general L.P problem which is also basic is called a basic feasible solution. 5. Optimal feasible solution: Any basic feasible solutions which optimize (maximize or minimize) the objective function of a general L.P problem is called an optimal feasible solution to that L.P problem 6. Degenerate solution: A basic solution to the system of the equations is called degenerate if one or more of the basic variables become equal to zero. ILLUSTRATION: Q1) Max Z = 50x1 + 35 x2 Subject to: 2x1 + 2x2 ≤ 40 5x1 + 2x2 ≤ 50 x1 + 4x2 ≤ 60 Where x1 ,x2 ≥ 0
• 121. Linear Programming Problem 121
• 122. Linear Programming Problem 122 Soln: x1 0 20 x2 20 0 2x1 + 2x2 ≤ 40 x1 0 10 x2 25 0 5x1 + 2x2 ≥ 50 x1 0 60 x2 15 0 x1 + 4x2 ≥ 60 So the coordinates for A 5x1 + 2x2 = 50 x1 + 4x2 = 60 …………….. (Multiply the equation By 5) 5x1 + 2x2 = 50 ……………….. (Subtract the equation) 5x1 + 20x2= -300 ------------------ -18x2 = -250 x2 = 125/9  x1 + 4 x2 =60  x1 + 500/9 = 60 ∴ x1 =60 - 500/9 = 40/9 (x1 = 40/9) (x2 = 125/9) A = (40/9, 125/9) B = (0,15) C = (10,0)
• 123. Linear Programming Problem 123 Max Z =50 x1 + 35x2  (50 × 40/9) + (35 ×125/9)  2000/9 + 4375/9  6375/9--------------------------------------(1)  708.34 Max Z = (50 × 0) + (35 ×15)  525-------------------------------------------(2) Max Z = (50 × 10) + (35 ×0)  500-------------------------------------------(3) Max Z = (6375/9)---------------------(ANSWER) Q2) Min Z = -x1 + 2 x2 Subject to : -x1 + 3x2 ≤ 10 x1 + x2 ≤ 6 x1 - x2 ≤ 2 Where x1 ,x2 ≥ 0 Soln: x1 0 -10 x2 10/3 0 -x1 + 3x2 ≤ 10 x1 0 6 x2 6 0 x1 + x2 ≤ 6 x1 0 2 x2 -2 0 x1 - x2 ≤ 2
• 124. Linear Programming Problem 124 So the coordinates for A -x1 + 3x2 = 10 x1 + x2 = 6 -x1 + 3x2 = 10 ………………………………. (Addition of the equation) x1 + x2 = 6 ------------------ 4x2 = 16 (x1 = 2) (x2 = 4) For the coordinates for B x1 + x2 = 6 x1 + x2 = 6 (Addition of the equation) x1 - x2 = 2 x1 - x2 = 2 ------------------ 2x1 = 8 (x1 = 4) (x2 = 2) A = (2,4) B = (4,2) C = (2,0) D = (0, 10/3) Min Z = -x1 +2 x2  -2 + 8 = 6--------------------------------------(A) Min Z = -4 + 4 = 0-----------------------------------(B) Min Z = -2 + 0 = -2-----------------------------------(C) Min Z = 0 + 2 ×10/3 = 20/3------------------------(D) Min Z = -2 (ANSWER)
• 125. Linear Programming Problem 125
• 126. Linear Programming Problem 126 Q3) Max Z = 6x1 - 4x2 Subject to : 2x1 + 4x2 ≤ 4 4x1 + 8x2 ≥ 16 Where x1, x2 ≥ 0 Soln: x1 0 2 x2 1 0 2x1 + 4x2 ≤ 4 x1 0 4 x2 2 0 4x1 + 8x2 ≥ 16 Q4) Max Z = 3x1 + 2 x2 Subject to: x1 - x2 ≥ 1 x1 + x2 ≥ 3 Where x1, x2 ≥ 0 Soln: x1 0 1 x2 -1 0 x1 + x2 ≥ 1 x1 0 3 x2 3 0 x1 + x2 ≥ 3
• 127. Linear Programming Problem 127
• 128. Linear Programming Problem 128 According to the given condition x1 , x2 ≥ 0, Hence 1st coordinate cannot exist. This is an UNBOUNDED SOLUTION. x1 - x2 = 1 x1 + x2 = 3 -------------- 2 x1 = 4 x1 = 2 (x1 = 2) (x2 = 1) Max Z = 3x1 + 2 x2  6 + 2  8-------------------------------------------(1) Max Z = 3x1 + 2 x2  0 + 6  6-------------------------------------------(2) It is an feasible solution. Q5) Max Z = 7x1 + 3x2 Subject to : x1 + 2x2 ≥ 3 x1 + x2 ≤ 4 0 ≤ x1 ≤ 5/2 0 ≤ x2 ≤ 3/2 Where x1 ,x2 ≥ 0 Soln: x1 0 3 x2 3/2 0 x1 + 2x2 ≥ 3---------------------------(1)
• 129. Linear Programming Problem 129 x1 0 4 x2 4 0 x1 + x2 ≤ 4---------------------------(2) x1 5/2 0 x2 0 0 0 ≤ x1 ≤ 5/2---------------------------(3) x1 0 0 x2 0 3/2 0 ≤ x2 ≤ 3/2---------------------------(4) For – A (1 and 4) x1 + 2x2 = 3 0 + x2 = 3/2 A = (0, 3/2) x1 + 3 = 3 (x1 = 0) (x2 = 3/2) For – B (2 and 4) x1 + x2 = 4 0 + x2 = 3/2 B = (5/2, 3/2) (x1 = 5/2) (x2 = 3/2) For – C (1 and 3) x1 + 2x2 = 3 x1 + 0x2 = 5/2 A = (5/2, 1/4) x2 = 3 – 5/2
• 130. Linear Programming Problem 130
• 131. Linear Programming Problem 131 =1/2 (x1 = 1/4) (x2 = 5/2) Max Z = 7x1 + 3x2  6 + 9/2  9/2-------------------------------------------(A) Max Z = 7x1 + 3x2  (7 × 5/2) + (3 × 3/2)  22---------------------------------------------(B) Max Z = 7x1 + 3x2  (7 × 5/2) + (3 × 1/4)  35/2 + 3/2  70+3 4  73 4 -----------------------------------------------(C) MAX Z = 22 (ANSWER) SIMPLEX METHOD The simplex Algorithm is a systematic ad efficient algebraically procedure for finding corner point solutions and taking them for optimality. BASIC TERMS INVOLVED IN SIMPLEX PROCEDURE Certain terms relevant for solving a linear programming problem through simplex procedure are introduced below: 1. Standard Form: A linear programme in which all of the constraints are written as equalities. The optimal solution of the standard form of a linear programme is the same optimal solution of the original formulation of the linear programme. 2. Slack Variable: A Variable added to the left-hand side of a ‘less-than or equal to’ constraint to convert the constraint into an equality is called a slack variable in economic
• 132. Linear Programming Problem 132 terminology; the value of the negative variable can usually be interpreted as the amount of unused resources. 3. Surplus variable: A variable subtracted from the left hand side of the greater than or equal to constraints to convert the constraints into equality is called a surplus variable. 4. Basic variable : For a system of m simultaneous linear equations in n variables(n>m), a solution obtained by setting (n-m)variables equal to zero and solving for the remaining m variables is called a basic variable. 5. Basic feasible solution: A basic feasible solution to a linear programming problem is a basic solution fro which the variables solved for, are all greater than or equal to zero. 6. Optimal solution: Any basic feasible solution which optimizes the objective function of a general LP problem is called an optimal basic feasible solution to the general LP problem. Tableau Form in which a linear programme must be written prior to setting up the initial simplex tableau. When a linear programme is written in this form. 7. Simplex Tableau: A table used to keep track of the calculation made at each iteration when the simplex solution method is employed. 8. Zj Row: The numbers in this row under each variable represents the total contribution of out going profit when one unit of a non basic variable is introduced in to this in place of a basic variable. 9. Cj – Zj (or Net Evaluation of Index) Row: The row containing the net net profit (for loss) that will result from introducing one unit of the variable indicated in that column in the solution numbers in index rows are also known as shadow prices). 10. Pivot (or Key) Column: The column with the largest positive number in the et evaluation row of maximization problem, or the largest negative number in the net evaluation row in a maximization problem.
• 133. Linear Programming Problem 133 11. Pivot (or Key) Row: The row corresponding to the variable that will leave the basis in order to make room for the entering variable (as indicated by the new pivot column) 12. Pivot (Number Key: The element at the intersection of the pivot row and pivot column. 13. Iteration: An iteration of the simplex method consists of the sequence of steps performed in moving form one basic feasible solution to another. NUMERICALS Q1. Max Z = 3X1+ 5X2+ 4x3 Subject to : 2X1 + 3X2 ≤ 8 2X2+ 5X3 ≤10 3X1+ 2X2 + 4X3≤ 15 X1, X2, X3 ≥0 Balance these equation 2X1 + 3X2 + S1= 8 2X2+ 5X3 + S2=10 3X1+ 2X2 + 4X3+ S3= 15 MaxZ = 3X1+ 5X2+ 4x3+S1+S2+ S3+ Initial Solution Cj 3 5 4 0 0 0 Qty X1 X2 X3 S1 S2 S3 Ratio 0 S1 8 2 3 0 1 0 0 8/3→ 0 S2 10 0 2 5 0 1 0 5 0 S3 15 3 2 4 0 0 1 15/2 Zj 0 0 0 0 0 0 0 (Cj-Zj) 3 5 4 0 0 0 0 ↑
• 134. Linear Programming Problem 134 Key Value = 3 Improved table Cj → 3 5 4 0 0 0 Qty ↓ X1 X2 X3 S1 S2 S3 Ratio 5x2 8/3 2/3 1 0 1/3 0 0 - 0S2 14/3 -4/3 0 5 -2/3 1 0 14/15→ 0S3 29/3 5/3 0 4 -2/3 1 9 29/12 Zj 40/3 10/3 5 0 5/3 0 0 (Cj-Zj) -1/3 0 4 -5/3 0 0 0 ↑ R2(NEW): R3(NEW): 10-8/3X2=14/3 15-8/3X2=29/3 0-2/3X2=-4/3 3-2/3X2=5/3 2-1X2=0 2-1X2=0 5-0X2=5 4-0X2=4 0-1/3X2=-2/3 0-1/3X2=-2/3 1-0X2=1 0-0X2=0 0-0X2=0 1-0X2=1 Improved table-1 Cj 3 5 4 0 0 0 Qty X1 X2 X3 S1 S2 S3 Ratio x2 8/3 2/3 1 0 1/3 0 0 4 x3 14/15 -4/15 0 1 -2/15 1/5 0 - S3 89/15 41/15 0 0 -2/15 -4/5 1 89/41 Zj 56/15 34/15 5 4 17/15 4/5 0 (Cj-Zj) 11/15 0 0 -17/15 -4/5 0
• 135. Linear Programming Problem 135 R1 (New): R2 (New): 8/3 – (0X 14/15) + 5/3 29/3 – (4X14/15) = - 89/15 8/3-(0X (-) 4/15) =2/3 5/3 – (4X -4/15) = 41/15 1- (0X0) = 1 0- (4X0) = 0 0- (0X1) = 0 4- (4x1) = 0 1/3- (0X -2/15) = 1/3 -2/3 – (4X 2/5) = -2/15 0-(0X1/5) = 0 0-(4X1/5) = -4/5 0- (0X0) = 0 1 – (4X0) = 1 Optimal Table Cj 3 5 4 0 0 0 Qty X1 X2 X3 S1 S2 S3 Ratio 5 x2 50/41 0 1 0 15/41 8/41 -10/41 - 4 x3 62/41 0 0 1 -6/41 5/41 4/41 - 0 S3 89/41 1 0 0 -2/41 -22/41 -15/41 - Zj 785/41 3 5 4 45/41 24/41 11/41 - (Cj-Zj) 0 0 0 -45/41 -24/41 4/41 R1 (New) R2 (New) 14/15 – (89/41 X – 4/5) = 62/41 8/3X 89/4) = 50/41 -4/15 – (-4/5X0) = 0 2/3 – (2/3 X1) = 0 – (-4/5 X0) = 0 1 – (2/3 X0) = 1 1-(-4/5x0)=1 0-(2/3x0)=0 -2/5-(-4/5x -2/41)=-6/41 1/3-(2/3x -2/41)=15/41 1/5-(-4/5 x -12/41)= 5/41 0-(2/3x -12/41)=8/41 0-(-4/5 x -15/41)=4/41 0-(2/3 x -15/41)=-10//41 MaxZ = 3X1+ 5X2+ 4x3 =3x(89/41)+5x(50/41)+4x(62/41) =765/41(ans.)
• 136. Linear Programming Problem 136 Q 2: Max Z=X1+2X2+3X3-X4 Sub to X1+2X3+3X3=15 2X1+X2+5X3=20 X1+2X2+X3+X4=10 X1, X2, X3, X4≥0 Balancing the equation X1+2X2+3X3+A1=15 2X2+X2+5X3+A2=20 X1+2X2+X3+X4=10 Max Z=X1+2X2+3X3-X4-MA1-MA2 Initial Table Cj → 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio - MA1 15 1 2 3 0 1 0 5 - MA2 20 2 1 5 0 0 1 4 -X4 10 1 2 1 1 0 0 10 Zj (35M-10) -(3M+1) -(3M+2) -(8M+1) -1 -M -M Cj-Zj (3M+2) (3M+4) (8M+42) 0 0 0 Key Value = 5 Improved Table Cj 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio - MA1 3 -1/5 7/5 0 0 1 -3/5 15/7 → - MA2 4 2/5 1/5 1 0 0 1/5 20 -1X4 6 3/5 9/5 0 1 0 -1/5 30/9 Zj (6M- 3M) (3/5+M/ 5) - (7M/5+6/ 5) 3 -1 -M 4/5 +3M/5 Cj-Zj -(M/5 - 2/5) (7M/5 +16/5) 0 0 0 2M/5- 4/5) ↑ Key Value = 5
• 137. Linear Programming Problem 137 R1(New) R3 (New) 15-(3X4) =3 10-(1X4) =6 (3X2/5) = -1/5 1-(1X2/5) = 3/5 (3X1/5) =7/5 2- (1X1/5) = 9/5 (3X1) = 0 1- (1X1) =0 0-(3X0) = 0 1-(1X0) = 1 (3X0) =1 0-(1X0) = 0 (3X1/5) = -3/5 0-(1X1/5) = -1/5 Key Value = 7/5 Improved Table – 1 Cj 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio 2X2 15/7 -1/7 1 0 0 5/7 -3/7 - 3X3 25/7 3/7 0 1 0 -1/7 2/7 25/3 -1X4 15/7 6/7 0 0 1 -9/7 4/7 15//6 Zj 90/7 1/7 2 3 -1 16/7 -4/7 Cj-Zj 6/7 0 0 0 -(M+16/7) 4/7-M KEY VALUE=6/7 R2(NEW: R3(NEW) 4-(1/5x15/7)=25/7 6-(9/5X15/7)=15/7 2/5-(1/5x1/7)=3/7 3/5-(9/5X-1/7)=6/7 1/5-(1/5x1)=0 9/5-(9/5X1)=0 1-(1/5x0)=1 0-(9/5X0)=0 0-(1/5x0)=0 1-(9/5X 0)=1 0-(1/5x5/7) =-1/7 0-(9/5X5/7)=-9/7 1/5-(1/5x -3/7)=2/7 -1/5-(9/5X-3/7)=4/7
• 138. Linear Programming Problem 138 Improved Table - 2 Cj 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio 2X2 15/6 0 1 0 1/6 1/2 1/3 3X3 15/6 0 0 1 -1/2 1/2 0 1X1 15/6 1 0 0 7/6 -3/2 2/3 Zj 15 1 2 3 0 1 4/3 Cj-Zj 0 0 0 -1 -(M+1) -(M+4/3) R1(NEW): R2(NEW): 15/7-(-1/7X15/6)=15/6 25/7-(3/7X15/6)=15/6 -1/7-(-1/7X1)=0 3/7-(3/7X1)=0 1-(-1/7X0)=1 0-(3/7X0)=0 0-(-1/7X0)=0 1-(3/7X0)=1 0-(-1/7X7/6)=1/6 0-(3/7X7/6)=-1/2 5/7-(-1/7X-3/2)=1/2 -1/7-(3/7X-1/3)=1/2 -3/7-(-1/7X2/3)=-1/3 2/7-(3/7X2/3)=0 MAX Z=X1+2X2+3X3-X4 15/6+15/3+15/2-0 =15(ANS.) Q.3 Max Z = X1 + 2X2+ 3X3 – X4 Subject to: X1 + 2X2 + 3X3=15 2X1+ X2+ 5X3 = 20 X1+ 2X2+ X3 + X4 = 10 X1, X2, X3, X4 ≥0 Balance the equation X1+ 2X2 + 3X3 + A1 = 15 ………………….. (1) 2X1+ X+ 5X3 + A2 = 20 ……………………. (2)
• 139. Linear Programming Problem 139 X1+ 2X2+ X3 + X4 = 10…………………….. (3) Maxz = X1 + 2X2 + 3X3 – X4 – MA1 – MA2 INITIAL TABLE Cj → 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio -MA1 15 1 2 3 0 1 0 5 -MA2 20 2 1 5 0 0 1 4→ -1X4 10 1 2 1 1 0 0 10 Zj = (35M-10), -(3M+1), -(3M+2), -(8M+1), -1, -M, -M (Cj-Zj) = (3M+2), (3M+4), (8M+4), 0 0 0 0 Key Value = 5 Cj → 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio -MA1 3 -1/5 7/5 0 0 1 -3/5 15/7 → 3X3 4 2/5 1/5 1 0 0 1/5 20 -1X4 6 3/5 9/5 0 1 0 -1/5 30/9 Zj = (6-3M), (3/5 +M/5), -(7M/5 + 6/5), 3, -1, -M (4/5 + 3M/5) (Cj-Zj) = (-M/5- 2/5, (7m/5 + 16/5, 0, 0,0, (2M/5 – 4/5) R1 (New): R3 (New): 15-(3X4) = 3 10 – (1X4)=6 1-(3X2/55) = -1/5 1- (1X2/5) = 3/5 2-(3X1/5) = 7/5 2-(1X1/5) = 9/5 3-(3X1)= 0 1-(1X1) = 0 0-(3X0) = 0 1- (1X0) =1 1-(3X0) = 1 0-(1X0)=0 0-(3X1/5) = -3/5 0-(1X1/5) = -1/5 Key Value = 7/5
• 140. Linear Programming Problem 140 Cj → 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio 2X2 15/7 -1/7 1 0 0 5/7 -3/7 - 3X3 25/7 3/7 0 1 0 -1/7 2/7 25/3 -1X4 15/7 6/7 0 0 1 -9/7 4/7 15/6 Zj = 90/7 1/7 2 3 -1 16/7 -4/7 (Cj-Zj) = 6/7 0 0 0, -(M+ 116/7), (4/7 – M) R2 (New): R3 (New) : 4-(1/5 X 15/7) = 25/7 6- (9/5 X 15/7) = 15/7 2/5 – (1/5 X -1/7) = 3/7 3/5 – (9/5 Xx – 1/7) = 6/7 1/5 – (1/5 X1) = 0 9/5 – (9/5X0) = 0 1-(1/5X0) =0 1- (9/5X0) = 1 0-(1/5X0) = 0 1- (9/5 X0) = 1 0-(1/5 X 5/7) = -1/7 0-(9/5 X 5/7) = -9/7 1/5 – (1/5 X -3/7) = 2/7 -1/5 – (9/55X – 3/7) = 4/7 Key Value = 6/7 Cj → 1 2 3 -1 -M -M Qty ↓ X1 X2 X3 X4 A1 A2 Ratio 2X2 15/6 0 1 0 1/6 ½ 1/3 3X3 15/6 0 0 1 -1/2 ½ 0 1X1 15/6 1 0 0 7/6 -3/2 2/3 Zj = 15 1 2 3 0 1 4/3 (Cj-Zj)= 0 0 0 0 -1 (M+1) -(M+4/3) R1 (New): R2 (New): 15/7 – (-1/7 X 15/6) = 15/6 25/7 – (3/7 X 15/6) = 15/6 -1/7 – (-1/7 X1) = 0 3/7 – (3/7 X1) = 0 1- (- 1/7 X 0) = 0 0- (3/7 X0 ) = 0 0 – (-1/7 X 0) = 0 1- (3/7 X0) =1
• 141. Linear Programming Problem 141 0-(-1/7 X – 3/2) = 1/6 0- (3/7 X 7/6) = -1/2 5/7 – (-1/7 X-3/2) = ½ -1/7 – (3/7 X -3/2) = ½ -3/7- (1/7 X 2/3) = -1/3 2/7 – (3/7 X 22/3)=0 Max Z = X1 + 2X2 + 3X3 – X4 15/6 + 15/3 + 15/2 – 0 15 Ans Q.4 MinZ = 5X1 + 3X2 Subject to 2X1 + 4X2 ≤ 12 2x1 + 2X2 = 10 5x1 + 2X2 ≥ 10 X1, X2 ≥0 Balancing the equation 2X1 + 4X2 + s1 = 12 …………………(1) 2x1 + 2X2 + A1 = 10………………..(2) 5x1 + 2X2 - s1 + A2 =10---------(3) Z* = -Z MaxZ* = -5X1 - 3x2 – 0s1 – 0s2 – MA1 – MA2 Cj → -5 -3 0 0 -M -M Qty ↓ X1 X2 s1 s2 A1 A2 Ratio 0s1 12 2 4 1 0 0 0 6 -MA1 10 2 2 0 0 1 0 5 -MA2 10 5 2 0 -1 0 1 2 → Zj = -20M, -7M -4M 0 M -M -M (Cj-Zj) (7M-5)(4M-3)0 -M 0 0 ↑ Key Value = 5
• 142. Linear Programming Problem 142 Cj → -5 -3 0 0 -M -M Qty ↓ X1 X2 s1 s2 A1 A2 Ratio 0s1 8 0 16/5 1 2/5 0 -2/5 5/2 → -MA1 6 0 6/5 0 2/5 1 -2/5 5 -5x1 2 1 2/5 0 -1/5 0 1/5 5 Zj = -(6M + 10), -5 -(6M/2 + 2) 0 (1 – 2M/5) -M (2M/5 – 1) (Cj-Zj) 0 (6M/5 - 1) 0 (2M/5-1) 0 (1 – 7M/5) ↑ Key Value = 16/ 5 R1 (New): R2 (New): 12 – (2 X 2) = 8 10 – (2 X 2) = 6 2 – (2 X 1) = 0 2 – (2X1) = 0 4 – (2 X 2/5) =16/5 2 – (2 X 2/5) =6/5 1 – (2 X 0) = 1 0 – (2 X 0) = 0 0 – (2 X-1/5) = 2/5 0 – (2 X-1/5) = 2/5 0 – (2 X 0) = 0 1 – (2 X 0) = 1 0 – (2 X-1/5) = -2/5 0 – (2 X-1/5) = 2/5 Cj → -5 -3 0 0 -M -M Qty ↓ X1 X2 s1 s2 A1 A2 Ratio -3x2 5/2 0 1 5/16 1/8 0 -1/8 8 -MA1 3 0 0 3/8 -7/20 1 -1/4 8 → -5x1 1 1 0 -1/8 -1/4 0 1/4 - Zj = -(25//2 + 3M), -5,-3, -(3M/8 +5/16),(3M/8 +5/16),(7/8 – 7M/20),-M,(7/8 – M/4) (Cj-Zj) 0 0 (3M/8+5/16),(7/8 – 7M/20, 0,(7//8-5M/4) ↑ Key Value = 13/8 R1 (New): R2 (New): 6 – (6/5 X 5/2) = 3 2 – (2/5 X 5/2) = 1 0 – (6/5 X 0) = 0 1 – (2/5 X 0) = 1
• 143. Linear Programming Problem 143 6/5 –(6/5 X1 )=0 2/5 – (2/5 X1 ) =0 0 – (6/5 X 5/16) = 3/8 0 – (2/5 X 5/16) = -1/8 2/5– (6/5 X 1/8) = -7/20 -1/5 – (2/5 X 1/8) = -7/4 1 – (6/5 X 0) = 1 0 – (2/5 X 0) = 0 -2/5– (6/5 X -1/18) = -1/4 1/5 – (2/5 X-1/8) = 1/4 Cj → -5 -3 0 0 -M -M Qty ↓ X1 X2 s1 s2 A1 A2 Ratio -3x2 0 0 1 0 5/12 -5/6 1/12 0 → 0s1 8 0 0 1 -14/15 8/3 -2/3 - -5x1 2 1 0 0 -11/30 1/3 1/6 12 Zj = -10, -5 -3 0 7/12 5/6 -13/12 (Cj-Zj) 0 0 0 -7/12 -(M + 5/6) (13/12 –M) Key Value = 1/12 R1 (New): R2 (New): 5/2 – (5/16 X 8) = 0 1 – (-1/8 X 8) = 2 0 – (5/16 X 0) = 0 1 – (-1/8 X 0) = 1 1 – (5/16 X 0) =1 0 – (-1/8 X 0) =0 5/6 – (5/16 X 1) = 0 -1/8 – (-1/8 X1 ) = 0 1/8 – (5/16 X -14/15) = 5/12 -1/4 – (-1/8 X -14/15) = -11/30 0 – (5/16 X 8/3) = -5/6 0 – (-1/8 X 8/3) = 1/3 -1/8 – (5/16 X -2/3) = 1/12 1/4 – (-1/8 X -2/3) = 1/6 Cj → -5 -3 0 0 -M -M Qty ↓ X1 X2 s1 s2 A1 A2 Ratio -MA2 0 0 12 0 5 -10 1 0s2 8 0 8 1 36/15 -4 0 -5x1 2 1 -2 0 -6/5 2 0 Zj = -10, -5 (10 -12M) 0 (6- 5M),(10M -10),-M (Cj-Zj) 0 (12M-13) 0 (5M -6) (10-11M),0
• 144. Linear Programming Problem 144 R1 (New): R2 (New): 8 – (-2/3 X 0) = 8 2 – (1/6 X 0) = 2 0 – (-2/3 X 0) = 0 1 – (1/6 X 0) = 1 0 – (-2/3 X 12) =8 0 – (1/6 X 12) = -2 1 – (-2/3 X 0) = 1 0 – (1/6 X 0) = 0 -14/15 – (-2/3 X 5) = 36/15 -11/30 – (1/6 X 5) = -6/5 8/3 – (-2/3 X -10) = -4 1/3 – (1/6 X -10) = 2 -2/3 – (-2/3 X 1) = 0 1/6 – (1/6 X 1) = 0 Since this is a continuous so its solution is UNBOUND. Q.5 MinZ = X1 – 2X2 – 3X3 Subject to -2X1 + X2 + 3X3 = 2 2x1 + 3X2 + 4 X3 =1 X1, X2, X3 ≥0 Balancing the equation -2X1+ X2 + 3X3+ A1 = 2 …………………(1) 2x1 + 3X2 + 4 X3 +A2 =1………………..(2) Z* = -Z MaxZ* = -X1 + 2x2 + 3X3 – MA1 – MA2 Cj → -1 +2 3 -M -M Qty ↓ X1 X2 X3 A1 A2 Ratio -MA1 2 -2 1 3 1 0 2/3 -MA2 1 2 3 4 0 1 1/4 → Zj = -3M, 0 -4M -7M -M -M (Cj-Zj) -1 (2+4M) (3+7M) 0 0 ↑ Key Value = 4
• 145. Linear Programming Problem 145 Cj → -1 +2 3 -M -M Qty ↓ X1 X2 X3 A1 A2 Ratio -MA1 5/4 -7/2 -5/4 0 0 -3/4 2/3 -MA2 1/4 1/2 3/4 1 1 1/4 1/4 → Zj = ¾ - 5M/4), 3/2 +7M/2 0 -4M -7M -M -M (Cj-Zj) = -(5/2+ 7M/2), -(1/4 + 5M/4), 0, 0, -(7M/4 + ¾) R1(New) : 2-(3X1/4) = 5/4 -2- (3X1/2) = -7/2 1- (3X3/4) = -5/4 3- (3X1) = 0 (3X0) = 1 (3X ¼) = -3/4 Max Z* = 2X2 + 3X3 – X1 = 0+ ¾ -0 = ¾ (Optimal But not Feasible Solution) DUALITY METHOD Introduction The term ‘dual’ in a general sense implies two or double. The concept of duality is very useful in mathematics, physics, statistics, engineering and managerial decision making. For example, in two-person game theory, one competitor’s problem is the dual of the opponent’s problem In the context of linear programming duality implies that each linear programming problem can be analyzed in to different ways but having equivalent solutions. Each LP problem (Maximization and Minimization) stated in the original form has associated with another linear programming problem (called dual linear programming or in short dual), which is unique , based on the same data dual in general, it is immaterial which of the two problems is called primal or dual, since the dual of the dual is primer. Thus the main focus of dual is to find for each resource its best marginal value of shadow price.
• 146. Linear Programming Problem 146 Shadow price= change in optimal objective function value Unit change in the availability of resource Rules for constructing the dual from primal The rules for constricting the dual from for the primal or primal or primal from the dual when using the symmetrical form are: 1. If the objective function of the primal is to be maximized, the objective function of the dual becomes minimized ition and vice versa. 2. For a maximization primal with all ≤ type constraints, their exist a minimization dual problem with all ≥ type constraints and vice versa. Thus, the inequality sing is reversed in all the constraints expect the non-negativity condition. 3. Each constants in the primal correspond to a dual variable in the dual an vice versa. Thus given primal problem with M constraints and N variable, there exist a dual problem with M variable and N constraints. 4. The right hand side constants b1, b2…….bm of the primal become the coefficients of the dual variable y1, y2,……,Yn in the dual objective function Zy Also the coefficients C1,C2,…Cn of the primal variable X1, X2, ….. xn in the objective function become the right hand side constants in the dual. 5. The matrix of the coefficients of variables in the constraints of dual is the transpose of the matrix of variables in the constraints of primal and vice versa. That is coefficients of the primal variables x1,x2…..xn in the constraints of primal LP problem are the coefficients of dual variables in first, second,…nth, constraints for the dual problem respectively. 6. If a variable say J in the primal LP problem is unrestricted in sign, Jth dual constraint is=(equality) type and vice versa. The primal dual relationship may also be remembered conveniently by using the following table: Dual variables Primal variables X1 X2 ……..XN Maximum zx Y1 A11 A12………….A1n ≤ b1 Y2 . A21 A22……….A22n . ≤ b2 Ym Am1 Am2……...Amn ≤bm Minimum Zy ≥ C1 ≥C2……… ≥Cn The primal constraints should be read across the row and dual constraints should be read across the columns
• 147. Linear Programming Problem 147 ILLUSTRATIONS: 1. Max Z = 30x1+20x2 Sub. To 2x1+7x2 ≤ 25 5x1+x2 ≤ 15 x1,x2 ≥0 Soln- 30 20 minZD ≤ ≤ 2 7 25 5 1 15 minZD= 25w1+15w2 sub.to 7w1+w2 ≤ 20-----------------------(1) 2w1+5w2 ≤ 30---------------------(2) W1,w2 ≥0 Balancing the eqn- 7w1+w2+s1=20 2w1+5w2+s2=30 NOW solve the problem using simplex method. 2. MaxZ= 30x1+20x2 Sub.to 2X1+7X2 ≤ 25 5x1+x2= 15 x1 , x2 ≥0 Ans. 2x1+7x2 ≤ 25-----------------------------------------------(1) 5x1+x2 ≥ 25 or -5x1-x2≤ 15--------------------------(2) 5x1+x2 ≤ 15------------------------------------------------(3)
• 148. Linear Programming Problem 148 30 20 minZD ≤ ≤ 2 7 25 - 5 -1 15 5 1 15 minZD=25w1+15w2+15w3 2w1-5w2+5w3 ≤ 30 7w1-w2+w3 ≤ 20 Balancing these eqn 2w1-5w2+5w3 +s1= 30 7w1-w2+w3 +s2= 20 NOW Solve the problem using simples method. 3. MaxZ= 30x1+20x2 Sub.to 2X1+7X2 ≤ 25 5x1+x2= 15 x1 ≥0, x2 = URWS IMAGINE X2= (Y1-Y2) 2X1+7 (Y1-Y2) ≤ 25---------------------(1) 5x1+ (Y1-Y2) = 15------------------------(2) NOW Solve the problem using simples method. Unsolved Questions: Graphical method Q.1 Maximize(Profit) Z=90X1+60X2 Sub to
• 149. Linear Programming Problem 149 5X1+8X2≤2000 X1≤175 X2≤225 7X1+4X2≤1400 X1,X2≥O Sol: Profit Rs.19666X2/3, X1=800/9 & X2=1750/9 Q.2. Minimize(Cost) Z=200X1+400X2 Sub to X1+X2≥200 X1+3X2≥400 X1+2X2≥350 X1,X2≥0 Sol: Cost Rs.60,000,X1=100&X2=100 Q.3 Maximize (Profit) Z=80X1+120X2 Sub to X1+X2≤9 X1≥2 X2≥3 X1,X2≥0 Sol: Profit Rs.960, X1=3&X2=6 Simplex method Q.1 Maximize (Profit) Z =3X1+2X2 Sub to 5X1+3X2≤16000 3X1+3X2≤14000 2X1+4X2≤12000 X1,X2 ≥0 Sol: Profit Rs. 10,000, X1=2000,&X2=2000
• 150. Linear Programming Problem 150 Q.2 Minimize (cost) Z=4X1+3X2 Sub to 200X1+100X2≥4000 X1+2X2≥50 40X1+40X2≥1400 X1,X2≥0 Sol: Cost Rs.110,X1=5,&X2=30 Q.3 Maximize (Profit) Z=2X1+3X2+4X3 Sub to 3X1+X2+6X3≤600 2X1+4X2+2X3≥480 2X1+3X2+3X3=540 X1,X2,X3≥0 Sol: Maximum Z =624,X1=0,X2=96 &X3=84
• 151. Bibliography 151 BIBLIOGRAPHY  Render, B., Ralph, M.S., Michal, E.H. Quantitative Analysis for Management Edition 2008. Arora, M. N., Cost and Management Accounting, edition 2006.  Sharma, J.K. Quantitative Techniques for Managerial Decision, Edition 2009.  Khandelwal, R.S., Gupta, B.L., Agarwal, S., Ahmad, T. Quantitative Analysis for Management, Edition 2009.