Linear programming

1. CHAPTER 2
LINEAR PROGRAMMING
PROBLEMS FORMULATIONAND
SOLUTION
4/30/2023 1

2. 2.1.DEFINITION:
 Linear programming is widely applied mathematical
technique that helps manager in decision making & planning
for optimal allocation of limited resources.
 Linear Programming is a mathematical process that has
been developed to help management in decision-making
involving allocation of scares resources
 In other words linear programming is quantitative science that
deals with reconciliation of scarce business resource and
unlimited operation wants of the firms.
4/30/2023 2

3. 2.2. COMPONENTS OFLPP
1.Decision variables - mathematical symbols representing levels of activity
of a firm
2.Objective function - in terms of decision variables (e.g., maximize the
profit or minimize the cost) Optimization in linear programming implies
either
maximization (max) Profit, revenue, market share , resource utilization
, production rate or
minimization (min) Cost, time, distance, or a certain objective
function, labour turnover, overtime, risk
2.Constraints - restrictions or limitations placed on the firm by the operating
environment stated in linear relationships of the decision variables
(restriction on decision making), scarce resources (such as labor supply,
RMs, production capacity, machine time, storage space etc
3.Parameters - numerical coefficients, values and constants used in the
objective function and constraint (are fixed values that specify the impact
that one unit of each decision variable will have on the objective)
4/30/2023 3

4. 2.3.ASSUMPTIONS OF LINEAR PROGRAMMING
A. Linearity/Proportionality/
The Objective Function and the constraints must be linear in nature in order to deal with
a Linear Programming Problems (LPP). Here the term linearity implies proportionality
and additively.
With linear programs, we assume that the contribution of individual
variables in the objective function and constraints is proportional to their
value.
That is, if we double the value of a variable, we double the contribution of
that variable to the objective function and each constraint in which the
variable appears.
B. Certainty
It is assumed that the decision maker here is completely certain (i.e., deterministic
conditions) regarding all aspects of the situation, i.e., availability of resources, profit
contribution of the Products, courses of action and their consequences etc.
C. Divisibility
It is assumed that the decision variables are continuous. It means that companies
manufacture products in fractional units.
For example, a company manufactures 2.5 vehicles, 3.2 barrels of oil etc.
4/30/2023 4

5. D.Additivity
Additivity means that the total value of the objective function
and each constraint function is obtained by adding up the
individual contributions from each variable.
E. Non- Negativity
Indicate all variables are restricted to non-negative values (i.e.,
their numerical value will be ≥ 0) .i.e. negative values of
variables are unrealistic or meaningless.
4/30/2023 5

6. 2.4. FORMULATION OFLPPM:
4/30/2023 6
General form of LPM of Maximization:
Max z = c1x1 + c2x2 + ... + cnxn---objectivefunction
subject to:
a11x1 + a12x2 + ... + a1nxn (≤) b1
a21x1 + a22x2 + ... + a2nxn (≤) b2
:
an1x1 + an2x2 + ... + annxn (≤) bn
x1,x2,x3… xn ≥ 0 non-negativity
Where: aij = constraintcoefficients
xj = decision variables
bi = constraint levels
cj = objective function coefficients

7. General form of LPM of Minimization:
Min z = c1x1 + c2x2 + ...+ cnxn---objectivefunction
subject to:
4/30/2023 7
a11x1 + a12x2 + ...+ a1nxn ( ≥) b1
a21x1 + a22x2 + ...+ a2nxn ( ≥) b2
:
an1x1 + an2x2 + ...+ annxn ( ≥)bn
≥ 0 non-negativity
x1,x2,x3… xn
Where:aij = constraintcoefficients
xj = decision variables
bi = constraintlevels
cj= objective functioncoefficients

8. Steps:
1.Identify the decision variables (X1, X 2….. )
2.Determine the objective function
 First decide whether the problem is maximization or minimization
 Second identify the coefficients of each decision variable.
 If the problem is a maximization problem, the profit per unit for each
variable must be determined and cost per unit must be determined for
minimization.
3. Identify the constraints
 First, express each constraint in words.
Second identify the coefficients of the decision variables in the constraints;
and the RHS values of the constraints.
Determine the limits for the constraints i.e. see whether the constraint is of
the form (≤), (≥)
4. Write the model in the standard form
 Using the above information (step 1 to 3), build the model.
4/30/2023 8

9. Example 1:
A firm that assembles computers and computer equipment is about to start
production two new microcomputers.
Each type of microcomputers will require assembly time, inspection time
and storage space.
The amount of each of these resources that can be devoted to the
production of these microcomputers is limited.
The manager of the firm would like to determine the quantity of each
microcomputer to produce in order to maximize the profit generated by
sales of these microcomputers. Additionalinformation
In order to develop a suitable model of the problem, the manager has met
with design and manufacturing personnel. As a result of these meetings the
manager has obtained the following information.
4/30/2023 9

10. The manager has also acquired information on the
availability of company resources. These weekly
resources are
The manager also met with the firm‟s marketing manager and
learned that demand for the microcomputers was such that what ever
combination of these two types of microcomputers is produced, all of
the outputs can be sold.
4/30/2023 10

11. Required: Formulate the LPPM of the problem.
Solution
Step 1: Identify the decision variable
The quantity/amount/ units of each microcomputer (microcomputer type 1
and microcomputer type 2) to be produced
produced
Let X1=represent quantity of microcomputer type 1 to be
X2=represent quantity of microcomputer type 2 to be produced
Step 2: Identify the objective function
The problem is maximization problem, as indicated in the problem
To write the model both the objective function and the constraints
summarize the given information in tabular form accordingly
4/30/2023 11

12.  Hence,the objective function is
Max Z = 60X1 + 50X2
Step 3:identify each constraints and write the equation
4X1 + 10X2 ≤ 100 ….........Assembly time constraint
2X1 + X2 ≤ 22 ……………Inspection time constraint
3X1+ 3X2 ≤ 39 ……………Storage spaceconstraint
Non negativity constraint X1 and X2>0
 The mathematicalmodel of the microcomputer problem is:
Max Z= 60X1 + 50X2
Subject to
4X1 + 10 X2 ≤100
2X1 + X2 ≤22
3X1+ 3 X2 ≤ 39
And X1 and X2 > 0
4/30/2023 12

13. 2.5.Method for solving LPM
4/30/2023 13
 Basically there are two methods for solving LPM
Graphical method and Simplex method
2.5.1.Graphical method of solving LPP
used to find solutions for LPP when the decision variables are only limited to two
Steps in solving LPM Using GraphicalApproach
1.Convert each inequality of the constraints in to equality to determine intercepts
(Set X1 = 0, and find the value for X2, Set X2 = 0, and find the value for X1)
2.Plot each of the constraint on the graph using intercepts (consider non negativity
assumption)
3.Determine the area of feasibility (feasible region: a region which satisfies all the
constraints in common) by shading using the inequalities of the constraints
 For minimization case (≥), the feasible region is the area that
lies above constraint line (to the right sides of the constraint
lines)
 For maximization case (≤), the feasible region is the area that
lies below constraint line (to the left sides of constraintlines)
4.Determine the corner points bounding the feasible region (feasible solution space)
and identify coordinates of each corner points

14. 5.Find optimum solution (that optimize the objective
function) by substituting coordinates of the corner points
in the objective function
4/30/2023 14
6.After all corner points have been evaluated in a similar
fashion, select the one with the
highest value of the objective function (for a
maximization problem) or
lowest value (for a minimization problem) as
the optimal solution.

15. Example: Consider the following Maximization problem and
solve by using graphical method
Max Z = 4X1 + 3X2
Subject to
4X1+ 3X2 ≤ 24
X1 ≤ 4.5
X2 ≤ 6
X1 ≥ 0 ,X2 ≥ 0
4/30/2023 15

16. Sketch graph
4/30/2023 16

17. MINIMIZATION LPP:
General form of LPM of Minimization:
Min z = c1x1 + c2x2 + ...+ cnxn---objective function
subject to:
a11x1 + a12x2 + ...+ a1nxn ( ≥) b1
a21x1 + a22x2 + ...+ a2nxn ( ≥) b2
:
an1x1 + an2x2 + ...+ annxn ( ≥) bn
x1,x2,x3… xn ≥ 0 non-negativity
Where: aij= constraint coefficients
xj= decisionvariables
bi = constraintlevels
cj= objective functioncoefficients
4/30/2023 17

18. Example1: Suppose that a machine shop has two different types of machines;
machine 1 and machine 2, which can be used to make a single product.
These machine vary in the amount of product produced per hr., in the
amount of labor used and in the cost of operation.
Assume that at least a certain amount of product must be produced and that
we would like to utilize at least the regular labor force to meet the
requirement.
How much should we utilize each machine in order to utilize total costs and
still meets the requirement?
Required:
a) Model the problem of this company
b) Solve the model graphically
4/30/2023 18

19. a). Model the problem of thiscompany
Zmin=25X1+30X2
Subject to:
20X1+15X2≥100
2X1+3X2≥15
X1, X2≥0
4/30/2023 19
Minimum required hours
Machine 1 (X1) Machine (X2)
Product produced/hr
Labor/hr
Operation Cost
20
2
$25
15
3
$30
100
15

20. b) Solve the model graphically
4/30/2023 20
Steps in solving LPM Using GraphicalApproach
1.Convert each inequality of the constraints in to equality to determine
intercepts
2.Plot each of the constraint on the graph using intercepts (consider non
negativity assumption)
3.Determine feasible region: a region which satisfies all the constraints in
common) by shading using the inequalities of the constraints
For minimization case (≥), the feasible region is the area that lies
above constraint line (to the right sides of the constraintlines)
NB. Important terminologies
A feasible solution is a solution for which all the constraints are
satisfied.
An infeasible solution is a solution for which at least one constraint
is violated.
An optimal solution is a feasible solution that has the most
favorable value of the objective function.
The most favorable value is the largest value if the objective
function is to be maximized, whereas it is the smallest value if the
objective function is to be minimized.

21. 4.Determine the corner points bounding the feasible region (feasible
solution space) and identify coordinates of each cornerpoints
5.Find optimum solution (that optimize the objective function) by
substituting coordinates of the corner points in the objectivefunction
6.After all corner points have been evaluated in a similar fashion, select the
one with the
lowest value (for a minimization problem) as the optimal solution.
4/30/2023 21

22. 4/30/2023 22

23. Example 2. Minimization Case (Fertilizer case):
4/30/2023 23
The agricultural research institute recommended a farmer to spread out at
least 5000 kg of phosphate fertilizer and not less than 7000 kg of nitrogen
fertilizer to raise the productivity of his crops on thefarm.
There are two mixtures A and B, weighs 100 kg each, from which these
fertilizers can be obtained.
The cost of each Mixture A and B is Birr 40 and 25respectively.
Mixture A contains 40 kg of phosphate and 60 kg of nitrogen while the
Mixture B contains 60 kg of phosphate and 40 kg ofnitrogen.
This problem can be represented as a linear programming problem to find
out how many bags of each type a farmer should buy to get the desired amount
of fertilizers at the minimum cost.

24. Step 1. Suppose x1 and x2 are the number of bags of mixture A
and mixture B.
Step 2. The cost of both the mixture is 40X1 + 25X2 and thus,
the objective function will be: Minimize Z = 40X1+25X2
BagA content
4/30/2023 24
Bag B content Minimum
Requirement
Phosphate 40kg 60kg 5000kg
Nitrogen 60kg 40kg 7000kg
Cost per bag $40 $25
Step 3. Constraints
40X1 + 60X2 ≥ 5000…………………. phosphate constraint
60X1 + 40X2 ≥ 7000………………… nitrogen constraint
Where, Bag A contains 60 kg of nitrogen and Bag B contains 40
kg of nitrogen, and the minimum requirement of nitrogen is
7000 kg.

25. Step 4.Develop linear programming problem (model)
4/30/2023 25
Minimize Z = 40X1+25X2 (cost)
Subject to:
40x1
60x1
+ 60x2
+ 40x2
≥ 5000 (Phosphate
≥ 7000 (Nitrogen
Constraint)
Constraint)
x1, x2 ≥ 0 (Non-negativity Restriction)

26. Practice, Practice,Practice!
4/30/2023 26
Thank you for Your
Attention & Participation!