Your SlideShare is downloading. ×
Master of Computer Application (MCA) – Semester 4  MC0079
Upcoming SlideShare
Loading in...5

Thanks for flagging this SlideShare!

Oops! An error has occurred.


Introducing the official SlideShare app

Stunning, full-screen experience for iPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Master of Computer Application (MCA) – Semester 4 MC0079


Published on

Master of Computer Application (MCA) – Semester 4 …

Master of Computer Application (MCA) – Semester 4

Published in: Technology

1 Like
No Downloads
Total Views
On Slideshare
From Embeds
Number of Embeds
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

No notes for slide


  • 1. 1MC0079-Computer based Optimization MethodsQuestion 1- Describe the structure of Mathematical Model in your own words.A mathematical model is a description of a system using mathematical concepts andlanguage. The process of developing a mathematical model is termed mathematicalmodeling. Mathematical models are used not only in the natural sciences (such asphysics, biology, earth science, meteorology) and engineering disciplines (e.g. computerscience, artificial intelligence), but also in the social sciences (suchas economics, psychology, sociology and politicalscience); physicists, engineers, statisticians, operations research analystsand economists use mathematical models most extensively. A model may help to explain asystem and to study the effects of different components, and to make predictions aboutbehavior.Mathematical models can take many forms, including but not limited to dynamicalsystems, statistical models, differential equations, orgame theoretic models. These and othertypes of models can overlap, with a given model involving a variety of abstract structures. Ingeneral, mathematical models may include logical models, as far as logic is taken as a partof mathematics. In many cases, the quality of a scientific field depends on how well themathematical models developed on the theoretical side agree with results of repeatableexperiments. Lack of agreement between theoretical mathematical models and experimentalmeasurements often leads to important advances as better theories are developed.Many mathematical models can be classified in some of the following ways:1. Linear vs. nonlinear: Mathematical models are usually composed by variables, which areabstractions of quantities of interest in the described systems, and operators that act onthese variables, which can be algebraic operators, functions, differential operators, etc. If allthe operators in a mathematical model exhibit linearity, the resulting mathematical model is
  • 2. 2defined as linear. A model is considered to be nonlinear otherwise. The question of linearityand nonlinearity is dependent on context, and linear models may have nonlinear expressionsin them. For example, in a statistical linear model, it is assumed that a relationship is linear inthe parameters, but it may be nonlinear in the predictor variables. Similarly, a differentialequation is said to be linear if it can be written with linear differential operators, but it can stillhave nonlinear expressions in it. In a mathematical programming model, if the objectivefunctions and constraints are represented entirely by linear equations, then the model isregarded as a linear model. If one or more of the objective functions or constraints arerepresented with a nonlinear equation, then the model is known as a nonlinear model.Nonlinearity, even in fairly simple systems, is often associated with phenomena such aschaos and irreversibility. Although there are exceptions, nonlinear systems and models tendto be more difficult to study than linear ones.2. Deterministic vs. probabilistic (stochastic): A deterministic model is one in which every setof variable states is uniquely determined by parameters in the model and by sets of previousstates of these variables. Therefore, deterministic models perform the same way for a givenset of initial conditions. Conversely, in a stochastic model, randomness is present, andvariable states are not described by unique values, but rather by probability distributions.3. Static vs. dynamic: A static model does not account for the element of time, while a dynamicmodel does. Dynamic models typically are represented with difference equations ordifferential equations.4. Discrete vs. Continuous: A discrete model does not take into account the function of timeand usually uses time-advance methods, while a Continuous model does. Continuousmodels typically are represented with f(t) and the changes are reflected over continuous timeintervals.5. Deductive, inductive, or floating: A deductive model is a logical structure based on atheory. An inductive model arises from empirical findings and generalization from them. Thefloating model rests on neither theory nor observation, but is merely the invocation ofexpected structure. Application of mathematics in social sciences outside of economics hasbeen criticized for unfounded models. Application of catastrophe theory in science has beencharacterized as a floating model.Question 2 - Explain Erlang family of distributions of service times.The Erlang distribution is a continuous probability distribution with wide applicabilityprimarily due to its relation to the exponential and Gamma distributions. The Erlangdistribution was developed by A. K. Erlang to examine the number of telephone calls whichmight be made at the same time to the operators of the switching stations. This work ontelephone traffic engineering has been expanded to consider waiting times in queuingsystems in general. The distribution is now used in the fields of stochastic processes andof biomathematics.Probability density functionThe probability density function of the Erlang distribution is
  • 3. 3The parameter is called the shape parameter and the parameter is called the rateparameter. An alternative, but equivalent, parameterizations (gamma distribution) uses thescale parameter which is the reciprocal of the rate parameter (i.e., ):When the scale parameter equals 2, the distribution simplifies to the chi-squareddistribution with 2k degrees of freedom. It can therefore be regarded as a generalized chi-squared distribution, for even degrees of freedom.Because of the factorial function in the denominator, the Erlang distribution is only definedwhen the parameter k is a positive integer. In fact, this distribution is sometimes called theErlang-k distribution (e.g., an Erlang-2 distribution is an Erlang distribution with k=2). TheGamma distribution generalizes the Erlang by allowing to be any real number, using thegamma function instead of the factorial function.Cumulative distribution function (CDF)The cumulative distribution function of the Erlang distribution is:where is the lower incomplete gamma function. The CDF may also beexpressed asQuestion 3 - Explain the algorithm for solving a linear programming problem by graphicalmethod.Linear programming (LP) is a mathematical method for determining a way to achieve thebest outcome (such as maximum profit or lowest cost) in a given mathematical model forsome list of requirements represented as linear equations. More formally, linearprogramming is a technique for the optimization of a linear objective function, subject tolinear equality and linear inequality constraints. Given a polytope and a real-valued affinefunction defined on this polytope, a linear programming method will find a point on thepolytope where this function has the smallest (or largest) value if such point exists, bysearching through the polytope vertices.An Algorithm for solving a linear programming problem by Graphical Method:(This algorithmcan be applied only for problems with two variables).Step – I: Formulate the linear programming problem with two variables (if the given problemhas more than two variables, then we cannot solve it by graphical method).Step – II: Consider a given inequality. Suppose it is in the form a1x1 + a2x2 <= b (or a1x1 +a2x2 >= b). Then consider the relation a1x1+ a2x2= b. Find two distinct points (k, l), (c, d)that lie on the straight line a1x1+ a2x2= b. This can be found easily: If x1= 0, then x2 = b /
  • 4. 4a2.If x2=0, then x1 = b / a1. Therefore (k, l) = (0, b / a2) and (c, d) = (b / a1, 0) are two pointson the straight line a1x1+a2x2= b.Step – III: Represent these two points (k, l), (c, d) on the graph which denotes X–Y-axisplane. Join these two points and extend this line to get the straight line which representsa1x1+ a2x2= b.Step – IV: a1x1 + a2x2= b divides the whole plane into two half planes, which are a1x1+a2x2 <= b (one side) and a1x1+ a2x2 >= b (another side). Find the half plane that is relatedto the given inequality.Step – V: Do step-II to step-IV for all the inequalities given in the problem. The intersection ofthe half-planes related to all the inequalities and x1 >= 0,x2 >= 0 , is called the feasibleregion (or feasible solution space). Now find this feasible region.Step – VI: The feasible region is a multisided figure with corner points A, B,C, … (say). Findthe co-ordinates for all these corner points. These corner points are called as extremepoints.Step – VII: Find the values of the objective function at all these corner/extreme points.Step – VIII: If the problem is a maximization (minimization) problem, then the maximum(minimum) value of z among the values of z at the corner/extreme points of the feasibleregion is the optimal value of z. If the optimal value exists at the corner/extreme point, say A(u, v), then we say that the solution x1= u and x2= v is an optimal feasible solution.Step – IX:Write the conclusion (that include the optimum value of z, and the co-ordinates ofthe corner point at which the optimum value of z exists).Question 4 - Determine optimal solution to the problem given below. Obtain the initialsolution by VAM. Write down the differences between PERT and CPM.Warehouses Stores AvailabilityI II III IVA 5 1 3 3 34B 3 3 5 4 15C 6 4 4 3 12D 4 -1 4 2 19Requirement 21 25 17 17 80Since the aggregate supply is 220 units and the aggregate demand is 200 units, we shallintroduce a dummy market, M5, for an amount equal to 20 (the difference between theaggregate supply and demand), with all cost elements equal to zero. The solution is given ina one table. The initial solution obtained by VAM is degenerate, since it contains only sevenbasic variables (since there are only 7 cells occupied and not 8 (= 4 + 5 -1) required for non-degeneracy).
  • 5. 5Here empty cells P1M3, P2M3, P3M1, P3M2, P3M4, P3M5, and P4M3 are independentwhile others are not.For removing degeneracy place in the cell P3M5and then test it for optimality. This isdone in the following table.This solution is found to be non-optimal.Therefore the total cost is:Rs. (6 x 30) + (1 x 10) + (3 x 20) + (0 x 50) + (7 x 60) + (1 x 30) = 700.CPM was developed by Du Pont and the emphasis was on the trade-off between thecost of the project and its overall completion time (e.g. for certain activities it may be possibleto decrease their completion times by spending more money - how does this affect theoverall completion time of the project?)Definition: In CPM activities are shown as a network of precedence relationships usingactivity-on-node network construction– Single estimate of activity time– Deterministic activity timesUSED IN: Production management - for the jobs of repetitive in nature where theactivity time estimates can be predicted with considerable certainty due to the existence ofpast experience.PERT was developed by the US Navy for the planning and control of the Polarismissile program and the emphasis was on completing the program in the shortest possibletime. In addition PERT had the ability to cope with uncertain activity completion times (e.g.for a particular activity the most likely completion time is 4 weeks but it could be anywherebetween 3 weeks and 8 weeks).Basic difference between PERT and CPM:Though there are no essential differences between PERT and CPM as both of themshare in common the determination of a critical path and are based on the networkrepresentation of activities and their scheduling that determines the most critical activities tobe controlled so as to meet the completion date of the project.PERT:1. Since PERT was developed in connection with an R and D work, therefore it had tocope with the uncertainties which are associated with R and D activities. In PERT,total project duration is regarded as a random variable and therefore associatedprobabilities are calculated so as to characterize it.
  • 6. 62. It is an event-oriented network because in the analysis of network emphasis is givenan important stages of completion of task rather than the activities required to beperformed to reach to a particular event or task.3. PERT is normally used for projects involving activities of non-repetitive nature in whichtime estimates are uncertain.4. It helps in pinpointing critical areas in a project so that necessary adjustment can bemade to meet the scheduled completion date of the project.CPM:1. Since CPM was developed in connection with a construction project which consistedof routine tasks whose resources requirement and duration was known with certainty,therefore it is basically deterministic.2. CPM is suitable for establishing a trade-off for optimum balancing between scheduletime and cost of the project.3. CPM is used for projects involving activities of repetitive nature.Project scheduling by PERT-CPM:It consists of three basic phases: planning, scheduling and controlling.1. Project Planning.2. Scheduling.3. Project Control.Question 5 - Explain the use of finite queuing tables.There will be cases, where the possible number of arrivals is limited and is relatively small.In a production shop, if the machines are considered as customers requiring service fromrepair crews or operators, the population is restricted to the total number of machines in theshop. In a hospital ward, the probability of the doctors or nurses being called for service isgoverned by the number of beds in the ward. Similarly, in an aircraft the number of seats isfinite and the number of stewardesses provided by the airlines will be based on theconsideration of the maximum number of passengers who can demand service. As in thecase of a queuing system with infinite population, the efficiency of the system can beimproved in tens of reducing the average length of queues, average waiting time and timespent by the customer in the system by increasing the number of service channels.However, such increases mean additional cost and will have to be balanced with the benefitslikely to accrue. If the queuing system in a machine shop is under study, the cost ofproviding additional maintenance crews or operators can be compared with the value ofadditional production possible due to reduced downtime of the machines` In cases where itis not possible to quantify the benefits, the management will have to base its decisions onthe desired standards for customer serviceThe queue discipline in a finite queuing process can be:i) First come-first servedii) Priority e.g.: Machines of high cost may be given priority for maintenance whileothers may be kept waiting even if they had broken down before.
  • 7. 7iii) Random e.g.; in a machine shop if a single operator is attending to severalmachines and several machines call for his attention at a time, he may attend firstto the one nearest to him.The analysis of Finite Queuing Models is more complex than those with infinite populationalthough the approach is similar.Notations: Notations used are different and are given below:N - Population (machines, customers etc)M - Service channels (repairmen, telephone lines etc.)T - Average service time (repair time length of conversation on a telephone etc.)W - Average waiting timeU - Average running time (of machines) or mean time between calls for service per unit H -Average number of units being servicedL - Average number of units waiting for serviceJ - Average number of units in operationX - Service FactorProbability that if a unit calls for service, it will have to wait,Let us consider a machine shop with N machines. The inter breakdown time of thesemachines follows a negativel exponential distribution with mean U. The number of breakdowns follows Poissondistribution with mean À. It is assumed that machines are kept running (or in operation)except when they are under repairs or waiting for repair crew to attend. If M repair crews areavailable, the time taken by any crew follows a negative exponential distribution with meanT. Naturally a machine which has broken down will have to wait for repairs if all the repaircrews are busy.The tables give the values of F and D for different values of N, M and X. They are arrangedin the ascending order of the values of the population. For each N, the value of X increasesfrom .001 to .950. For a given service factor X, several values of M can be found. For eachvalue of X and M, values of D and F are tabulated,The steps in the use of Finite Queuing Tables may be summarized as follows.(i) Find mean service timeTand mean running tirne U.(ii) Compute the service factor(iii) Select the table corresponding to the population N.(iv) For the given population, locate the service factor value.(v) Read off from tables, values of D and F for the number of service crews M.If necessary, these values may be interpolated between relevant values of X. (vi) Calculatethe other measures L W, H, and J from the formulae given.The overall efficiency F of the system will increase with the number of service channels (M)provided. As mentioned earlier, addition of service crews involves cost, which should bejustified by the increase in the efficiency of the system i.e. additional running time of
  • 8. 8machines possible. However it will be seen from the tables that as M increase the rate 0fincrease in efficiency decreases. The practical significance is that beyond a certain value ofM, it is not worthwhile increasing M as there would be no appreciable increase in theefficiency of the system.Question 6 - Customers arrive at a small post office at the rate of 30 per hour. Service bythe clerk on duty takes an average of 1 minute per customera) Calculate the mean customer time.(i) Spent waiting in line(ii) Spent receiving or waiting for service.b) Find the mean number of persons(i) in line(ii) Receiving or waiting for service.Mean arrival rate λ = 30 customers per hour = ½ customer per minute.Mean service rate λ = 1 per minuteTraffic intensityP = λ/µ= ½a) Mean Customer Timea. Spent waiting in lineE(w) = λ/( µ(µ - λ))= 1 minuteb. Spent receiving or waiting for serviceW(v) = 1/( µ - λ)= 2 minutesb) Find the mean number of personsa. in lineE(m) = λ2/( µ(µ - λ))= ½ Customerb. Receiving or waiting for serviceE(n) = λ /( µ - λ)= 1 Customer