The document defines linear programming and its components. It provides examples of linear programming models for profit maximization and cost minimization. It discusses the assumptions and methods for solving linear programming problems, including the graphical method. The graphical method involves plotting the feasible region defined by the constraints and determining the optimal solution by evaluating the objective function at the corner points of the feasible region, choosing the point that maximizes profit or minimizes cost.

1. CHAPTER THREE
LINEAR PROGRAMMING
Definition:
Linear programming is a technique applied for
the optimizing a linear objective function, subject to linear
equality and linear inequality constraints.
Is a particular technique used for economic allocation of
‘scarce’ or ‘limited’ resources.
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2. Cont…
e.g. labour, material, machine, time,
warehouse space, capital, energy, etc.:
several competing activities e.g. products,
services, jobs, new equipment, projects, etc. on
the basis of a given criterion of optimality.
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3. Components of LPP
The linear programming model consists of the
following components:
Decision variables: mathematical symbols representing levels
of activity of an operation.
Objective function: a linear relationship reflecting the objective
of an operation.
Constraint: a linear relationship representing a restriction on
decision making.
Non-negativity: values of the decision variables are zero or
non-zero positives.
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4. General form of LP
Z = c1x1 + c2x2 +…………. +cnxn
Subject to restrictions
a11x1 + a12x2 + …..........+a1nxn (≤ or ≥) b1
a21x1 + a22x2 + …………..+a2nxn (≤ or ≥) b2
.
am1x1 + am2x2 + ………. +amnxn (≤ or ≥) bm &
x1 ≥ 0, x2 ≥ 0,…, xn ≥ 0
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5. Cont…
Where
 Z = value of overall measures of performance
xj = level of activity (for j = 1, 2, ...,n)
cj = a unit profit/cost increase in Z that would result from
each unit increase in the decision variables (j=1,2, . . .,n)
bi = amount of resource i that is available for allocation to
each decision variables (for i = 1,2, …, m)
aij = amount of resource i consumed by each unit of activity j
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6. Cont. . .
Resource
Resource usage per unit of activity Amount of
resource
available
Activity
1 2 …………………….. n
1
2
.
.
.
M
a11 a12 …………………….a1n
a21 a22 …………………….a2n
.
.
.
am1 am2 …………………amn
b1
b2
.
.
.
Bm
Contribution to Z per
unit of activity
C1, c2 ……………………….cn
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7. Linear Programming Model Construction
General steps taken are as under:
 Define the decision variables .
Identify the coefficient of each decision variables.
Formulate the objective function.
Identify the coefficients of each decision variables in the obj fun.
Identify the constrained resources or RHS values.
Formulate the suitable mathematical constraints related to each
respective resource.
Mention the non-negativity constraints associated with the
decision variables.
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8. Assumptions of Linear Programming
Proportionality
Additive
Divisibility
Deterministic
Finiteness
Optimality
One objective
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9. Profit Maximization case
Example 1: Problem Definition
1. A company manufactures two products, tables and chairs by
using three machines A, B, and C so as to maximize a profit.
Machine A has 10 hrs of capacity available during the coming
week. Similarly, the available capacity of machine B and C
during the coming week is 36 hrs and 30 hrs, respectively. One
unit of a table requires 2 hrs of machine A, 3 hrs of machine B
and 3 hrs of machine C. Similarly, one unit of a chair requires 1
hr, 12 hrs and 6 hrs of machine A, B, and C respectively. When
one unit of table is produced and sold, it yields a profit of Birr
5 and that of a chair is Birr 7. Given the above information,
formulate the linear programming model.
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10. Solution
I. Decision variable
how tables and chairs can the managers produce?
the quantities to be produced can be denoted as:
X1: number of tables to be produced
X2: number of chairs to be produced
II. The Objective function
It is to max profit (sum of each)
Then total profit = $5x1+ 7x2 i.e.
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11. Cont…
III. Model Constraints
three resources are used for production are hours of
machine A, B, and C respectively.
 hours of machine A constraint is 2x1+ 1x2 ≤ 10
hours of machine B constraint is 3x1 + 12x2 ≤36
 hours of machine C constraint is 3x1 + 6x2 ≤ 30
IV. Non-negativity constraint:
x1 ≥ 0, x2 ≥ 0.
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12. Cont. . .
The complete linear programming model for
this problem can now be summarized
bellow as
Max Z = $5x1+ 7x2
ST: 2x1 + 1x2 ≤ 10
5x1 + 2x2 ≤ 36
3x1 + 6x2 ≤ 30
x1, x2 ≥ 0
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13. Example 2: Problem Definition
• Example1: Assume ‘XY’ company is a small craft operation run by a
Native American tribal council. The company employs skilled artisans
to produce clay bowls and mugs with authentic Native American
designs and colours. The company also employed inspectors who will
check the products after finished. The three primary resources used
by the company are special pottery clay, skilled labour and inspectors.
Given these limited resources, the company desires to know how
many bowls and mugs to produce each day in order to max profit.
The two products have the following resource requirements for
production and profit per item produced (i.e. the model parameters).
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14. Example 1: Problem Definition . . .
Resource requirement
Product Labor Inspector Clay Profit
Hr/unit Hr/Unit Ib/unit $/unit
Bowl 2 1 5 50
Mug 2 4 2 35
There are 40 hours of labour 50 hours of inspector and 60 pounds of
clay available each day for production.
Required: Formulate the LP model of this problem.
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15. Example 1: Problem Definition . . .
The final complete LP model
Max Z = $50x1 + 35x2
ST: 2x1 + 2x2 ≤ 40
5x1 + 2x2 ≤ 50
x1 + 4x2 ≤ 60
x1, x2 ≥ 0
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16. Exercise 1
A firm manufactures two types of products A and B and sells them at a profit of
$2 on type A and $3 on type B. Each product is processed on two machines G
and H. Type A requires 1 minute of processing time on G and 2 minutes on H;
type B requires 1 minute on G and 1 minute on H. The machine G is available for
not more than 6 hours 40 minutes while machine H is available for 10 hours
during any working day.
Required: a) Formulate the problem as a linear programming problem
b) Solve the problem using graphical method
Answer for que a)
Maximize Z = 2x1 + 3x2
ST X1 + x2 ≤ 400
2x1 + 3x2 ≤ 600
X1 ≥ 0, x2 ≥ 0

17. Cost Minimization case – Example 1
A company owns 2 oil mills A and B which have
different production capacities for low, high and
medium grade oil. The company enters into a
contract to supply oil to a firm every week with 12, 8,
24 barrels of each grade respectively. It costs the
company $1000 and $800 per day to run the mills A
and B. On a day A produces 6, 2, 4 barrels of each
grade and B produces 2, 2, 12 barrels of each grade.
Required: Develop the LP model.
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18. Solution
Let x1 be the number of days a week the mill A has to work
x2 be the number of days per week the mill B has to work
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Grade A B Minimum requirement
Low 6 2 12
High 2 2 8
Medium 4 12 24
Cost per day $1000 $800

19. Cont…
Therefore, the complete LP model is;
Minimize Z = 1000x1 + 800 x2
Subject to
6x1 + 2x2 ≥ 12
2x1 + 2x2 ≥ 8
4x1+12x2 ≥ 24
x1 ≥ 0, x2 ≥ 0
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20. Optimal Solution Methods
There are two approaches to solve linear
programming problems:
The Graphical Algorithm
The Simplex Algorithm
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21. Graphical Method
A graphical approach is limited to linear
programming problems with only two decision
variables.
The complexity of its application increases as
the number of constraint functions increases.
It is a relatively straightforward method for
determining the optimal solution to certain linear
programming problems.
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22. The following are the steps to be followed in solving LP
problems graphically:
Formulate the problem in terms of mathematical
constraints and an objective function.
Convert the inequalities into equation and compute the
coordinate points
Plot the model constraints on the graph
Identify the feasible region i.e. the area which satisfies all
the constraints simultaneously
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23. Cont…
Plot the objective function then move the line out from the
origin till it touches the extreme maximum points of the
feasible solution area.
Plot the objective function then move the line down till it
touches the extreme minimum point of the feasible solution
area.
Substitute these values into the objective function to find the
set of values that result in the maximum profit or the
minimum cost.
Interpreting the results
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24. Cont. . .
Case 1: Graphical Solution Method for Maximization Problem
Example 1: Max Z = $50x1+ 35x2
ST: 2x1 + 2x2 ≤ 40
5x1 + 2x2 ≤ 50
x1 + 4x2 ≤ 60
x1, x2 ≥ 0
Required: Compute the optimal solution using graphical
method. Apply all possible steps.
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25. Solution:
First develop the problem into mathematical linear
programing if not given.
The structured model is given so we need to proceed
into change the inequalities into equations for
computing the coordinate points.
2x1 + 2x2 = 40
5x1 + 2x2 = 50
x1 + 4x2 = 60
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26. Cont. . .
x1 0 20
x2 20 0
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x1 0 60
x2 15 0
x1 0 10
x2 25 0
) When x1 = 0, x2 = 20:
(0, 20)
When x2 = 0, x1 = 20:
(20,0)
When x1 = 0, x2 = 25:
(0, 25)
When x2 = 0, x1 = 10
(10, 0)
When x1 = 0, x2 = 15:
(0, 15)
When x2 = 0, x1 = 60:
(60, 0)
2x1 + 2x2 = 40 5x1 +2x2 = 50 x1 + 4x2 = 60

27. Cont…
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A

28. Cont…
In this graph, the FSA is represented by the shaded
region.
The feasible region has four corner points i.e. O, A, B,
and C.
From these corner points at least one will give us
maximum profit
The one that generate the highest profit would be
regarded as optimal corner point/s.
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29. Cont…
Corner
points
Coordinate
Points
Profits
O (0,0) Z = $50(0)+ 35(0) = $0
A (0,15) Z = $50(0)+ 35(15) = $525
B (40/9,125/9) Z = $50(40/9)+ 35(125/9) = $6375/9
C (10,0) Z = $50(10)+ 35(0) = $500
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30. Cont…
To compute the coordinates for B, use simultaneous
method as below
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 ,
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31. Cont…
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32. Cont…
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The optimal solution has been reached at corner point B
x1 = 40/9
Unique Solution
x2 = 125/9 why?
Max Z = $6375/9 in profit

33. Cont…
Interpretation
Form this result; we can conclude that the
manager has different alternatives to make
decisions but since the objective of the
problem is to maximize profit, she has to
select the best alternative i.e. alternative
corner point B at x1 = 40/9 and x2 = 125/9
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34. Case 2: Graphical Solution Methods for Minimization
Problem
Example 1:
Minimize Z = 1000x1 + 800x2
Subject to
6x1 + 2x2 ≥ 12
2x1 + 2x2 ≥ 8
4x1 +12x2 ≥ 24
X1 ≥ 0, x2 ≥ 0
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35. Cont…
Required:
A. Sketch the graph and identify the feasible solution
area
B. How many numbers of products type x1 and type x2
the company has to produce to minimize cost?
C. What is the z value of the problem?
D.Does the problem have unique solution? Why?
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36. Ans for A.
To sketch the graph, determine the CPs of each constraint by
changing the ≥ into =
6x1 + 2x2 = 12
Let x1 = 0, x2 = 6 coordinate points = (0, 6)
Let x2 = 0, x1 = 2 coordinate points = (2, 0)
2x1 + 2x2 = 8
Let x1 = 0, x2 = 4 coordinate points = (0, 4)
Let x2 = 0, x1 = 4 coordinate points = (4, 0)
4x1 +12x2 = 24
• Let x1 = 0, x2 = 2 coordinate points = (0, 2)
•Let x2 = 0, x1 = 6 coordinate points = (6, 0)
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37. Cont…
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38. Ans for B.
CP Coordinates Profits
A (0,6) Z = $1000(0) + 800(6) = $4800
B (1,3) Z = $1000(1)+ 800(3) = $3400
C (3,1) Z = $1000(3)+ 800(1) = $3800
D (6,0) Z = $1000(6)+ 800(0) = $6000
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 Mill A has to work 1 day a week and
 Mill B has to work 3 days a week

39. Ans for C.
The optimal solution or the minimum possible cost of
the problem is $3400.
Ans for D
The answer is yes because the company has no
alternatives days to work more or less for both mill A
and mill B without changing the minimum possible
cost.
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40. Graphical Solution of Irregular Types of Linear
Programming Problems:
If a linear programming problem has a unique
solution, then it must occur at a corner point of the
feasible set.
However, there are several special types of typical
linear programming problems (linear programming
models for which the general rules do not always
apply).
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41. These special types of linear programming problems
include the following:
A. Multiple optimal solutions
B. Infeasible problem
C. Unbounded problem
D. Redundancy
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42. A. Multiple optimal solutions
If the objective function is optimized at two adjacent
corner points of the feasible area, it is optimized at
every point on the line segment joining these corner
points, in which case there are infinitely many solutions
to the problem.
Multiple optimal solutions can benefit the decision
maker since the number of decision options is enlarged.
Multiple optimal solutions provide greater flexibility
to the decision maker.
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43. Con…
Example 1:
Max Z = 4x1 + 3x2
Subject to
4x1+ 3x2 ≤ 24
x1 ≤ 4.5
x2 ≤ 6
x1 ≥ 0, x2 ≥ 0
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44. Solution
The first constraint 4x1+ 3x2 ≤ 24, written in a form of
equation
4x1+ 3x2 = 24
Put x1 =0, then x2 = 8
Put x2 =0, then x1 = 6
The coordinates are (0, 8) and (6, 0)
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45. Cont…
The second constraint x1 ≤ 4.5, written in a form of
equation
x1 = 4.5
The third constraint x2 ≤ 6, written in a form of
equation
x2 = 6
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46. Cont…
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47. Cont…
The corner points of feasible region are A, B, C and D.
So the coordinates for the corner points are
 A (0, 6)
B (1.5, 6) (Solve the two equations 4x1+ 3x2 = 24 and
x2 = 6 to get the coordinates)
C (4.5, 2) (Solve the two equations 4x1+ 3x2 = 24 and
x1 = 4.5 to get the coordinates)
D (4.5, 0)
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48. Cont…
We know that Max Z = 4x1 + 3x2
At A (0, 6)
Z = 4(0) + 3(6) = 18
At B (1.5, 6)
Z = 4(1.5) + 3(6) = 24
At C (4.5, 2)
Z = 4(4.5) + 3(2) = 24
At D (4.5, 0)
Z = 4(4.5) + 3(0) = 18
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49. Cont…
Max Z = 24, which is achieved at both B and C
corner points. It can be achieved not only at
B and C but every point between B and C.
Hence the given problem has multiple
optimal solutions.
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50. B. No Optimal Solution/infeasible solution
An infeasible solution has no feasible solution
area; every possible solution point violates one
or more constraints. In other words, if a
solution violates at least one constraint, the
solution is said to be infeasible.
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51. Cont . . .
Example1: Solve graphically
Max Z = 3x1 + 2x2
Subject to
x1+ x2 ≤ 1
x1+ x2 ≥ 3
x1 ≥ 0 , x2 ≥ 0
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52. Cont…
Solution
The first constraint x1+ x2 ≤ 1 is written in the
form of x1+ x2 = 1
Put x1 =0, then x2 = 1
Put x2 =0, then x1 = 1
The coordinates are (0, 1) and (1, 0)
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53. Cont…
The second constraint x1+ x2 ≥ 3, written in a form of
equation
x1+ x2 = 3
Put x1 =0, then x2 = 3
Put x2 =0, then x1 = 3
The coordinates are (0, 3) and (3, 0)
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54. Cont…
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55. Cont…
There is no common feasible region generated by
two constraints together i.e. we cannot identify
even a single point satisfying the constraints.
Hence there is no optimal solution.
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56. C. Unbounded Solution
In an unbounded problem, the feasible solution area
formed by the model constraints is not closed.
In unbounded problem the objective function can
increase indefinitely without reaching a maximum
value.
Unlimited profits are not possible in the real world;
and an unbounded solution, like an infeasible
solution, typically reflects an error in defining the
problem or in formulating the model.
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57. Cont…
Example1: Solve by graphical method
Max Z = 3x1 + 5x2
Subject to
2x1+ x2 ≥ 7
x1+ x2 ≥ 6
x1+ 3x2 ≥ 9
x1 ≥ 0 , x2 ≥ 0
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58. Cont…
Solution
The first constraint 2x1+ x2 ≥ 7, written in a
form of equation
2x1+ x2 = 7
Put x1 =0, then x2 = 7
Put x2 =0, then x1 = 3.5
The coordinates are (0, 7) and (3.5, 0)
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59. Cont…
The second constraint x1+ x2 ≥ 6, written in a
form of equation
x1+ x2 = 6
Put x1 =0, then x2 = 6
Put x2 =0, then x1 = 6
The coordinates are (0, 6) and (6, 0)
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60. Cont…
The third constraint x1+ 3x2 ≥ 9, written in a
form of equation
x1+ 3x2 = 9
Put x1 =0, then x2 = 3
Put x2 =0, then x1 = 9
The coordinates are (0, 3) and (9, 0)
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61. Cont…
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62. Cont…
The corner points of feasible region are A, B, C and D. So the
coordinates for the corner points are:
A (0, 7)
B (1, 5) (Solve the two equations 2x1+ x2 = 7 and x1+ x2 = 6
to get the coordinates)
C (4.5, 1.5) (Solve the two equations x1+ x2 = 6 and x1+ 3x2 =
9 to get the coordinates)
D (9, 0)
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63. D. Redundancy
A redundant constraint is simply one that does not
affect the feasible solution region.
One constraint may be more binding or restrictive than
the other and thereby negate its need to be considered.
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64. Con…
Example 1:
Maximize Z = X + 2Y
Subject to: X + Y ≤ 12
2X + Y ≤ 30
X ≤ 25
X, Y ≥ 0
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