OR is defined as a scientific approach to optimal decision making through modelling of deterministic and probabilistic systems that originate from real life. Scientific approach: LPP, PERT/CPM, Queueing model, NLP, DP,MILP, Game theory, heuristic programming. Deterministic system: - a system which gives the same result for a particular set of input, no matter how many times we recalculate it

1. OR Notes – Module 1
History of OR
• Operations research (OR) was first used in the late 1930’s to describe the process of
evaluation of radar as an essential tool in the air defense of great Britain. The ORigins
were in military applications.
• Lord Patrick Blackett was one of the four founders of the OR club in 1948, which in
1953 became the Operational Research (OR) Society. He was director of Naval
Operational Research at Admiralty and a recipient of the Nobel Prize in physics in 1948.
• The first OR journal – Operational Quarterly in 1950, which in 1978 became the Journal
of the Operational Research Society.
• George B Dantzing developed in 1947 simplex algorithm to solve LPP, which solved
problems involving thousands of variables and problem constraints
• P C Mahalonobis – established OR unit in 1953 at Indian statistical Institute, Kolkata
to apply OR methods in national planning and survey – (in preparing the draft of second
five-year plan).
• ORSI (Operations research society of India) – 1957.
Growth in size, specialization and complexity of organization increases, each department in
the organization may tend to develop autonomous goals but not meshing with the
organizational goals. As the complexity and specialization in an organization increases, it
becomes more & more difficult to allocate resources to various activities. Because of this, there
was an emergence of field called Operations research, to optimize this allocation of resources.
OR can be traced back to and attributed to world war II, where there was scarcity for resource.
With its success in the defense field, OR began to creep into other fields such industry,
agriculture, government policy making, business etc.

2. Definitions of OR:
OR is defined as a scientific approach to optimal decision making through modelling of
deterministic and probabilistic systems that originate from real life.
Scientific approach: LPP, PERT/CPM, Queueing model, NLP, DP,MILP, Game
theory, heuristic programming.
Deterministic system: - i.e a system which gives the same result for a particular set of
input, no matter how many times we recalculate it.
e.g.
Capacity used per unit
production rate
Available
capacity
Product
Plant
1 2
1 1 0 4
2 0 2 12
3 3 2 18
Probabilistic system: weather forecast, tossing of die etc.
Scope of OR:
OR has solved successfully many cases of research for military, govt. industry etc. like
Transportation, inventory planning, production planning, communication operations, computer
operations, financial assets, risk management, revenue management etc…..
It is scope is spread across many other fields where improving business productivity is
paramount.
Also, its scope to address social issues as below:
The basic problem in developing countries like Africa, is to remove poverty & hunger, so there
is lot of scope for economist, statisticians, administrators, politicians & technicians working in
a team to solve this problem by using OR approach.
With the explosion of population & consequent shortage of food, every country is facing the
problem of optimum allocation of land for various crops in accordance with climate conditions
& available facilities – OR approach can be used to solve
The problem of optimal distribution of water from resources like canal for irrigation – OR
approach can be used to solve

3. Applications of OR
• Time tabling
• Site selection for an industry
• Production planning, controlling and scheduling
• Optimal tariff design
• Design of civil engineering structures for minimum cost
• Design of aircraft and aerospace structures for minimum weight
• Design of electrical machinery such as motors, generators, transformers
• Design of pumps, turbines and heat exchangers for maximum efficiency
• Design of pipeline network for process industries
• Determining the trajectories for space vehicles
• Managing consumer credit delinquency in the US economy: a multi-billion dollar
management science application
• Control of the water distribution system under Irrigation scheme in Malaysia
• Formulating insurance polices by Life Insurance Company in India
• Application of OR for optimum utilization of urban facilities
• Optimization model for integrated municipal solid waste management in Mumbai,
India
Phases of OR
1. Defining the problem: Here we define the scope of the OR problem. This step should
lead to identify:
a. Objective of the study
b. Decision variables
c. Constraints under which we should address the problem
2. Construct the mathematical model
3. Solve the model by using various mathematical and statistical tools by using the input
data
4. Implementation
Models in OR
1. Physical Models
2. Mathematical Models: (i) Deterministic Models (ii) Probabilistic Models
3. Combined Analogue and Mathematical Models
4. Function Models
5. Heuristic Models
Physical models: These models represent the system as it is by scaling it up or down. Even
though use of these models in the area of management appears to be narrow, their usefulness
is seen in the field of engineering and science. For example, (a) In the field of R&D, prototype

4. of the product is developed and tested to know the workability of the new product development.
(b) Photographs, portraits, drawings are the good example of iconic types. These models help
in testifying the samples thus avoiding fullscale designing and probable loss.
Mathematical models:
a) Deterministic model: The exact statement of variables and their relationships are made
under these models. The coefficients used for the mathematical formulation are known
and are constant with certainty. With a given set of data the answer will always be the
same. For instance, determination of the break-even sales volume (BEP), the volume
where the total cost equals the total sales revenue earned pertaining to a product.
b) Probabilistic model: The risk involved and the state of uncertainty are covered by these
models. The decision variables take the form of a probability distribution and can
assume more than single values. In the presence of risk and uncertainty, these models
tend to yield different answers every time when attempted to. For example, ncertainty
over acquisition of raw material to execute customer orders during a certain period. The
purchaser has to consider both the sale and the delivery timing of the orders as they are
variables. A probability distribution can be developed for the instant period for both
sales delivery timings.
Heuristic models: When intuition guides a problem-solver to find solutions, heuristic models
are developed. Even though he would not be able to find an optimum solution to the problem,
with his past experience he arrives at the most advantageous solution. However, it depends on
how intuitive and creative the decision-maker
Limitation of OR:
1. DEPENDENCE ON AN ELECTRONIC COMPUTER: O.R. techniques try to find out
an optimal solution taking into account all the factors. In the modern society, these
factors are enormous and expressing them in quantity and establishing relationships
among these require voluminous calculations that can only be handled by computers. -
2. NON-QUANTIFIABLE FACTORS: O.R. techniques provide a solution only when all
the elements related to a problem can be quantified. All relevant variables do not lend
themselves to quantification. Factors that cannot be quantified find no place in O.R.
models
3. DISTANCE BETWEEN MANAGER AND OPERATIONS RESEARCHER: O.R.
being specialist's job requires a mathematician or a statistician, who might not be aware
of the business problems. Similarly, a manager fails to understand the complex working
of O.R. Thus, there is a gap between the two.
4. MONEY AND TIME COSTS: When the basic data are subjected to frequent changes,
incorporating them into the O.R. models is a costly affair. Moreover, a fairly good
solution at present may be more desirable than a perfect O.R. solution available after
sometime.
5. IMPLEMENTATION: Implementation of decisions is a delicate task. It must take into
account the complexities of human relations and behaviour.

5. Linear Programming Problem (LPP)
First conceived by George B Dantzig in 1947. His first paper was titled “Programming in linear
structure”. Koopman coined the term Linear programming in 1948.
Problem formulation:
Example 1: (Product mic problem)
Consider a small manufacturer making two products A & B.
Two resources R1 & R2 are required to make these products.
Each unit of product A requires 1 unit of R1 & 3 units of R2.
Each unit of product B requires 1 unit of R1 & 2 units of R2.
The manufacturer has 5 units of R1 and 12 Units of R2 available.
The manufacturer also makes a profit of:
Rs. 6 per unit of product A sold &
Rs. 5 per unit of product B sold
Formulate LPP
Solution:
The manufacture has to decide on the no. of units of product A & B to produce. He would like
to make as much profit as possible, accordingly he would decide on production quantities of A
& B. Manufacturer has to ensure that the resources needed to make these quantities are
available. Before we attempt to find out the decision of the manufacturer, let us redefine
problem in algebraic form.
The manufacture has to decide on the production quantities, let us call them as X1 & X2
Let X1 be the no. of units of A made
Let X2 be the no. of units of B made
Let Z be the profit made
Z = 6X1 + 5X2
Usage of resource R1 = X1 + X2 <= 5
Usage of resource R2 = 3X1 + 2X2 <= 12
Complete model for the manufacturer is:
Objective function (OF) Maximize Z = 6X1 + 5X2
Subjected to constraints
X1 + X2 <= 5
3X1 + 2X2 <= 12
Non negativity constraints
X1, X2 >=0

6. Example 2 (Product mix problem)
Reddy mikk’s produces both interior and exterior paints using two types of raw materials, M1
& M2. The following table provides the basic data of the problem:
Tons of raw material per
ton of
Maximum daily
availability in
tons
Exterior
paints
Interior
paints
Raw material 1, M1 6 4 24
Raw material 2, M2 1 2 6
Profits per ton (in
1000’s Rs.)
5 4
Market survey indicates that the daily demand for interior paint cannot exceed that for exterior
paint by more than 1 ton. Also, the maximum daily demand for interior paint is 2 tons. Reddy
Mikk’s wants to determine the optimum (best) product mix of interior and exterior paints that
maximizes the total daily profit.
Solution:
Reddy Mikk’s has to decide on the no. of units of Exterior paint and interior paint to produce.
He would like to make as much profit as possible, accordingly he would decide on production
quantities of Exterior paint and interior paint.
He has to ensure that the resource needed to make these quantities are available.
Before we attempt to find out the decision of the company, lets redefine the problem in an
algebraic form.
The reddy mikk’s has to decide on the production quantities of Exterior paint and interior paint.
Let us call them as X1, X2
i.e.
X1 = Tons produced daily of exterior paint
X2 = Tons produced daily of interior paint
Let Z represent the total daily profit in thousand of Rs.
So,
Z = 5X1 + 4X2
Objective is to Maximize Z = 5X1 + 4X2
Subjected to constraints (STC)
Usage of raw material 1, M1 per day = 6X1 + 4X2 < = 24
Usage of raw material 2, M2 per day = X1 + 2X2 <= 6

7. Demand constraints:
X2 – X1 < = 1
X2 < = 2
Non negativity constraints:
X1, X2 > = 0
Complete reddy mikks model:
Maximize Z = 5X1 + 4X2
STC
6X1 + 4X2 < = 24
X1 + 2X2 <= 6
X2 – X1 < = 1
X2 < = 2
Non negativity constraints
X1, X2 > = 0
Example 3 (Diet Problem)
A person wants to decide the constituents of a diet which will fulfill his daily requirements of
proteins, fats and carbohydrates at the minimum cost. The choice is to be made from four
different types of food. The yield per unit of these foods are given. Formulate linear
programming problem.
Yield per unit of consumption
Food Type Proteins Fats Carbohydrates
Cost per units
(Rs)
A 3 2 6 45
B 4 2 4 40
C 8 7 7 85
D 6 5 4 65
Minimum
requirement
800 200 700
The decision is to be made on the right diet mix,
Let us define the decision variables of the model:
X1 = grams of food type A in daily consumption
X2 = grams of food type B in daily consumption
X3 = grams of food type C in daily consumption

8. X4 = grams of food type D in daily consumption
Z= Cost of consuming these food types
Z = 45X1 + 40X2 + 85X3 + 65X4
Daily quantity of Proteins required = 3X1 + 4X2 + 8X3 + 6X4 <= 800
Daily quantity of Fats required = 2X1 + 2X2 + 7X3 + 5X4 <= 200
Daily quantity of Carbohydrates required = 6X1 + 4X2 + 7X3 + 4X4 <= 700
LPP Model:
Minimize Z = 45X1 + 40X2 + 85X3 + 65X4
STC
3X1 + 4X2 + 8X3 + 6X4 <= 800
2X1 + 2X2 + 7X3 + 5X4 <= 200
6X1 + 4X2 + 7X3 + 4X4 <= 700
Non negativity constraints
X1, X2, X3, X4 >=0
Example 4 (Production planning problem)
Let us consider a company is making a single product. The estimated demand for the product
for next four months are 1000, 800, 1200 and 900 units respectively. The company has a regular
time capacity of 800 hours per month and overtime capacity of 200 hours per month. The cost
of regular time production is Rs. 20 per unit & the cost of overtime production is Rs. 25 per
unit. The company can carry inventory to the next month & the inventory holding cost is Rs. 3
per unit per month. The demand has to be met every month. Formulate LPP for the above
situation.
Solution:
Data given:
Month 1 Month 2 Month 3 Month 4
Demand 1000 800 1200 900
Regular time capacity = 800 hrs /month
Over time capacity 200 hrs / month
Inventory cost = 3 per unit per month
Regular time cost = Rs. 20 per unit
Over time cost = Rs. 25 per unit

9. The decision is to be made qty to be produced using regular time, over time and inventory
Let us define the decision variables of the model:
Xj = Qty produced using Regular time in the month j, where j = 1,2,3,4
Yj = Qty produced using Over time in the month j, where j = 1,2,3,4
Ij = Qty of inventory carried in month j, where j = 1,2,3,4
Z = cost of total production using regular time, over time & inventory
Objective function
Minimize Z = 20 ∑ 𝑥𝑗
4
𝑗=1
+ 25 ∑ 𝑦𝑗
4
𝑗=1
+ 3 ∑ 𝐼𝑗
3
𝑗=1
Subjected to constraints
Qty produced using RT, OT and Inventory in Month 1 = X1 + Y1 > = 1000
Qty produced using RT, OT and Inventory in Month 2 = I1 + X2 + Y2 >= 800
Qty produced using RT, OT and Inventory in Month 3 = I2 + X3 + Y3 > = 1200
Qty produced using RT, OT and Inventory in Month 4 = I3 + X4 + Y4 >=900
Where I1, I2, I3 are inventories produced in month 1, 2, 3
From month 1 constraint, the qty produced more than 1000 is I1
I1 = X1 + Y1 – 1000
Similarly,
I2 = I1+X2 + Y2 – 800
= X1 + Y1 – 1000 +X2 + Y2 – 800
= X1 + Y1 + X2 + Y2 – 1800
Similarly,
I3 = I2+X3 + Y3 – 1200
= X1 + Y1 + X2 + Y2 – 1800 + X3 + Y3 – 1200
= X1 + Y1 + X2 + Y2 + X3 + Y3 – 3000
Substituting I1,I2,I3 in the above constraints, we can redefine them as below:
STC
X1 + Y1 > = 1000
X1 + Y1 + X2 + Y2 > = 1800
X1 + Y1 + X2 + Y2+ X3 + Y3 > = 3000
X1 + Y1 + X2 + Y2 + X3 + Y3 + X4 + Y4 >=3900

10. Let us redefine the objective function to include inventory carrying cost:
Minimize Z = 20 ∑ 𝑥𝑗
4
𝑗=1
+ 25 ∑ 𝑦𝑗
4
𝑗=1
+3 (I1 + I2+I3)
Minimize Z = 20 ∑ 𝑥𝑗
4
𝑗=1
+ 25 ∑ 𝑦𝑗
4
𝑗=1
+3 (X1 + Y1 – 1000 + X1 + Y1 + X2 + Y2 – 1800
+ X1 + Y1 + X2 + Y2 + X3 + Y3 – 3000)
Final LPP Model
Minimize Z = 20 ∑ 𝑥𝑗
4
𝑗=1
+ 25 ∑ 𝑦𝑗
4
𝑗=1
+ 3 (3X1 + 3Y1 + 2X2 + 2Y2
+
X3 + Y3 – 3800)
STC
X1 + Y1 > = 1000
X1 + Y1 + X2 + Y2 > = 1800
X1 + Y1 + X2 + Y2+ X3 + Y3 > = 3000
X1 + Y1 + X2 + Y2 + X3 + Y3 + X4 + Y4 >=3900
Non negativity constraint
Xj Yj > = 0, where j = 1 to 4

11. Example 5 (Cutting stock problem or Trim loss problems)
Consider a big sheet roll from which steel sheets of the same lengths but different widths have
to be cut. Let us assume that roll is 20 cm wide & the following sizes have to be cut.
9 cm – 511 nos.
8 cm – 301 nos.
7 cm – 263 nos.
6 cm – 383 nos.
Let us assume that all the cut sheets have the same length of 25 cm. Only one-dimensional
cutting is allowed. The problem is to cut the sheet roll in such a way as to minimize wastage
of material.
Solution:
4 sizes
25 * 9 - 511 nos.
25 * 8 - 301 nos.
25 * 7 - 263 nos.
25 * 6 - 383 nos.

12. Let’s define waste:
Before defining waste lets identify different cutting patters possible:
Sizes /
Pattern
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10
25 * 9 2 0 0 0 1 1 1 0 0 0
25 * 8 0 2 0 0 1 0 0 1 1 0
25 * 7 0 0 2 0 0 1 0 1 0 1
25 * 6 0 0 1 3 0 0 1 0 2 2
Waste
in cm
2 4 0 2 3 4 5 5 0 1
Explanation for P1 : - with pattern P1, we can cut 2 pieces of 25*9 along width 20 cm, i.e. 18
cm is used to cut 2 pieces along width. i.e. 2 cm waste
Explanation for P3 - with pattern P3, we can cut 2 pieces of 25*7 along width 20 cm, i.e. 14
cm is used to cut 2 pieces along width. And 1 piece of 25*6 along width 20 cm. total all 20
com is used. So waste is 0
Similarly the other columns are filled.
So, there are 10 different patterns possible
Let us define the decision variable:
Xj = no. of sheets cut using the pattern j, where j = 1 to 10
Z is waste
For the requirement of size 25*9
Patters possible are P1,P5,P6,P7
No. of sheets cut using these patterns for size 25 * 9 will be:
2X1 + X5 + X6 + X7 > = 511
For the requirement of size 25*8
Patters possible are P2,P5,P8,P9
No. of sheets cut using these patterns for size 25 * 8 will be:
2X2 + X5 + X8 + X9 > = 301
For the requirement of size 25*7
Patters possible are P3, P6, P8, P10
No. of sheets cut using these patterns for size 25 * 7 will be:
2X3 + X6 + X8 + X10 > = 263
For the requirement of size 25*6

13. Patters possible are P3,P4,P7,P9,P10
No. of sheets cut using these patterns for size 25 * 6 will be:
X3 + 3X4 + X7 + 2X9 +2X10> = 383
Final LPP Model:
Minimize Z = 2X1+4X2+0X3+2X4+3X5+4X6+5X7+5X8+0X9+X10
STC
2X1 + X5 + X6 + X7 > = 511
2X2 + X5 + X8 + X9 > = 301
2X3 + X6 + X8 + X10 > = 263
X3 + 3X4 + X7 + 2X9 +2X10> = 383
Non negativity constraint
Xj > = 0
Home work:
Example 1 (Diet problem)
Ozrak farms uses at least 36000 grams of special feed daily. The special feed is a mixture of
corn and soyabean meal with the following composition.
Feed stuff
grams per grams of feedstuff
Cost
Protein Fiber
Corn 4 1 30
Soyabean 27 2.7 90
The dietary requirements of the special feed are at least 30% protein and 5% fiber. Ozrak farm
wishes to determine the daily minimum cost feed limit.
Example 2 (Portfolio selection / Investment decisions)
An investor is considering investing in two securities 'A' and 'B'. The risk and return associated
with these securities is different. Security 'A' gives a return of 9% and has a risk factor of 5 on
a scale of zero to 10. Security 'B' gives return of 15% but has risk factor of 8. Total amount to
be invested is Rs. 5, 00, 000/- Total minimum returns on the investment should be 12%.
Maximum combined risk should not be more than 6. Formulate as LPP.
Example 3 (Farm planning)
A farmer has 200 acres of land. He produces three products X, Y & Z. Average yield per acre
for X, Y & Z is 4000, 6000 and 2000 kg. Selling price of X, Y & Z is Rs. 2, 1.5 & 4 per kg
respectively. Each product needs fertilizers. Cost of fertilizer is Rs. 1 per kg. Per acre need for
fertilizer for X, Y & Z is 200, 200 & 100 kg respectively. Labour requirements for X, Y & Z
is 10, 12 & 10 man hours per acre. Cost of labour is Rs. 40 per man hour. Maximum availability
of labour is 20, 000 man hours. Formulate as LPP to maximise profit.

14. Example 4 (Inspection problem)
A company has two grades of inspectors, I and II to undertake quality control inspection. At
least 1, 500 pieces must be inspected in an 8-hour day. Grade I inspector can check 20 pieces
in an hour with an accuracy of 96%. Grade II inspector checks 14 pieces an hour with an
accuracy of 92%. Wages of grade I inspector are Rs. 5 per hour while those of grade II inspector
are Rs. 4 per hour. Any error made by an inspector costs Rs. 3 to the company. If there are, in
all, 10 grade I inspectors and 15 grade II inspectors in the company, find the optimal assignment
of inspectors that minimise the daily inspection cost.
Example 5 (Blending problem)
A manufacturing company making castings uses electric furnace to melt iron which must have
the following specification:
Minimum Maximum
Carbon 3.10% 3.30%
Silicon 2.15% 2.25%
Specifications and costs of various raw materials used for this purpose are given below:
Material Carbon % Silicon % Cost (Rs.)
Steel scrap 0.42 0.12 850/ton
C 3.80 2.40 900/ton
Remelt from
foundry
3.40 2.30 500/ton
Carbon briquettes 100 0 7/kg
Silicon briquettes 0 100 10/kg
If the total charge of iron metal required is 4 tons, find the weight in kg of each row material
that must be used in the optimal mix at minimum cost.
Example 6 (Transportation problem)
A diary firm has two milk plants with daily milk production of 6 million litres and 9 million
litres respectively. Each day the firm must fulfil the needs of its three distribution centres which
have milk requirement of 7, 5 and 3 million litres respectively. Cost of shipping one million
litres of milk from each plant to each distribution centres is given, in hundreds of rupees below.
Formulate the LP model to minimize the transportation cost
Distribution
centres
Plants 1 2 3 Supply
A 5 4 11 6
B 5 9 6 9
Demand 7 5 3

15. Example 7 (Product mix problem)

16. Linear programming solutions – Graphical method
Example 1:
Maximize Z = 6X1 + 5X2
STC
X1 + X2 < = 5
3X1 + 2X2 <= 12
Non negativity constraints
X1, X2 > = 0
Solution:
The horizontal X-axis represents X1, the vertical Y- axis represents X2.
As X1, X2 > = 0, the solution space is restricted to first quadrant, i.e. above X1 axis and right to
X2 axis.
To account for the remaining constraints, first replace each inequality with an equation’s and
then graph the resulting straight line by locating the coordinates for each constraint.
For this constraint, X1 + X2 <= 5, Replace <= with =
X1 + X2 = 5
By setting X1 =0, obtain X2 = 5
By setting X2 =0, obtain X1 = 5
Thus the line passes through (5,0) & (0,5)
For this constraint, 3X1 + 2X2 <= 12, replace <= with =
3X1 + 2X2 = 12
By setting X1 =0, obtain X2 = 6
By setting X2 =0, obtain X1 = 4
Thus the line passes through (4,0) & (0,6)

17. The area inside ABCD is the feasible region, as it satisfies all the constraints
Every point in the feasible region is called feasible solution
The best solution is that which satisfies the objective is called optimal solution
Determination of optimal solution
Method 1
Find the coordinates of each corner of polygon
Corner A – (0,0), X1 = 0, X2= 0, substitute in the objective function, Z = 0
Corner B – (4,0) X1 = 4, X2= 0, substitute in the objective function, Z = 24
Corner C – (1.5,3.5) X1 = 1.5, X2= 3.5, substitute in the objective function, Z = 26.5
Corner D – (0,5) X1 = 0, X2= 5, substitute in the objective function, Z = 25
As the objective function is to maximize, we can observe that corner C is yielding maximum
value. So the optimal solution is
X1 = 1.5, X2= 3.5, Z = 26.5
Method 2
Determination of optimum solution requires identifying the direction in which Z increase, as
the objective function is maximization type. We can do so by assigning arbitrary increasing
values for Z, for example Z = 10, Z = 15
It would be equivalent to drawing two lines
6X1 + 5X2 = 10 & 6X1 + 5X2 = 15

18. Draw parallel lines to the above line in the direction of increasing Z value. The optimum
solution occurs at C, Which is the point in the solution space beyond which any further increase
in Z will put us outside the boundaries of ABCD.
The values of X1 = 1.5, X2 = 3 and Z = 26.5.
Example 2
Minimize Z = 7X1 + 5X2
STC
X1 + X2 >= 4
5X1 + 2X2>= 10
Non negativity constraints
X1, X2 >= 0

19. Determination of optimal solution
Method 1: Corner point method
The area above ABC is the feasible region, as it satisfies all the constraints
Every point in the feasible region is called feasible solution
The best solution is that which satisfies the objective is called optimal solution
A (0,5) Z = 25
B (1,3) Z = 22
C (4,0) Z = 28
As the objective is to minimize, the solution to the given problem is
X1 = 1, X2 = 3 & Z = 22
Method 2 :
Determination of optimum solution requires identifying the direction in which Z decreasing,
as the objective function is minimization type. We can do so by assigning arbitrary decreasing
values for Z, for example Z = 35
It would be equivalent to drawing two lines
7X1 + 5X2 = 35
This line passes (0,7) & (5,0)
Draw parallel lines to the above line in the direction of decreasing Z value. The optimum
solution occurs at B, which is the point in the solution space beyond which any further decrease
in Z will put us outside the boundaries of ABC.

20. The values of X1 = 1, X2 = 3 and Z = 22.
Home Work
Problem 1
Let’s solve the Reddy Mikks problem using graphical method
Maximize Z = 5X1 + 4X2
STC
6X1 + 4X2 < = 24
X1 + 2X2 <= 6
X2 – X1 < = 1
X2 < = 2
Non negativity constraints
X1, X2 > = 0
Step 1: Determination of the feasible solution space
Non negativity constraints X1, X1 > = 0
Problem 2
Solve the LPP using graphical method
Maximize Z = 50000X1 – 10000X2
Subjected to
200000X1+500000X2 <= 8000000
X1 + X2 < = 20
Z > = 600000
X1, X2 > = 0
Problem 3
Feasible zone ABCDE identified by a set of constraints is shown below. If a constraint
X <= 2Y is added, identify the new feasible zone. Also identify & state all redundant
constraints.
A (1,0), B (1,2), C (2,3), D (4,1), E (2,0)
Determine max & min value of Z if Z = 3X+5Y after inclusion of additional constraint
X <= 2Y
Problem 4
Max. Z = 160 X1 + 240 X2
Subject to constraints
X1 ≥ 4
X2 ≥ 6
X1 + X2 ≤ 18
60 X1 + 60 X2 ≤ 720
Non negativity constraints, X1, X2 ≥ 0

21. Problem 5
Min. Z = 50 X1 + 20 X2
Subject to constraints
X1 + X2 ≤ 600
X1 + X2 ≥ 300
6 X1 + 2 X2 ≥ 1200
Non negativity constraint, X1, X2 ≥ 0
Four difficulties that may arise sometimes when using the graphical approach to solve
LPP
1) Infeasible Solution (Infeasibility)
Infeasible means not possible. Infeasible solution happens when the constraints have
contradictory nature. It is not possible to find a solution which can satisfy all
constraints. In graphical method, infeasibility happens when we cannot find Feasible
region.
Example 1:
Max. Z = 5 X1 + 8 X2
Subject to constraints
4 X1 + 6 X2 ≤ 24
4 X1 + 8 X2 ≤ 40
X1, X2 ≥ 0
There is no common feasible region for line AB and CD. Hence, solution is infeasible.

22. 2) Unbounded Solution (Unboundedness)
Unbounded mean infinite solution. A solution which has infinity solution is called unbounded
solution. As can be seen in the following graph, there is no bounded region which astisfies all
constraint. This condition is called as unboundedness
Example:
Max Z = 6X1 + 10X2
Subjected to constraints
X1 > = 5
X2 < = 10
2X1 + 4X2 >=20
X1,X2>=0
3) Redundancy
A constraint is called redundant when it does not affect the solution. The feasible region does
not depend on that constraint. Even if we remove the constraint from the solution, the optimal
answer is not affected. Example
Max. Z = 5 X1 + 8 X2
Subject to Constraints 3
X1 + 2 X2 ≤ 24
X1 + 3 X2 ≤ 12
X1 ≤ 16
X1, X2 ≥ 0

23. Redundant constraint is X1< = 16
4) Alternate Optimal Solution: (Multiple Optimal Solution)
Alternate or multiple optima solution means a problem has more than one solution which gives
the optimal answer.
There are two or more sets of solution values which give maximum profit or minimum cost. In
graphical method, we come to know that there is optimal solution which is alternative when:
The iso-profit or iso-cost line is parallel to one of the boundaries of feasible region (they have
the same slope value).
Terminologies
Feasible region it is a common region which satisfies all the constraints
Optimum solution it is the best solution which satisfies the objective