Chapter Two linear programming in business mathematics .pdf

Chapter 2
Linear
programming
22
CH 1 introduction to OR

2.1 Introduction
Linear programming(LP) is a mathematical model in which
the objective function and constraint functions are linear in the
decision variables.
LP is a problem solving approach that has been developed to
make decisions.
23

2.2. Components of LP models
 There are three major components of LP models
including:
1.Objective function- maximization or
minimization
2. Decision variables- unknown quantities to be
solved for.
3.Constraints –availability of resource (restriction)
24

Con’t
25

2.3 Assumptions of LP Models
The assumptions reveal the condition under which the model is
valid
1. Linearity
2. Continuity/ Divisibility
3. Additives
4. Certainty
5. Choices
6. Non- negativity assumptions
26

2.4 Formulating Linear Programming Models (LPM)
 The formulation of LPM involves the following steps :
1. Define the problem: - this involves the determination of our
specific objectives. The word expression of our objective.
Example: to determine the number of units of P1 and P2 to be
produced per month so as to maximize profit given the
restrictions (constraints).
2. Determine the decision variables: - this involves the
representation of the unknown quantities by letters.
27

Con’t
Example: let x1 and x2 represent the number of units of P1 and P2
to be produced per month respectively.
3. Formulate the objective function:- in developing the
objective function make sure that:
 All the decision variables are represented in the objective
function
 The unit of measurement of all the coefficients in the objective
function must be the same. Example, if we use kg for x1, we
should use kg for x2 too.
 All the terms in the objective must include a variable.
28

Con’t
4. Formulate the constraints:-decide on the constraints and
formulate the mathematical relationship that is used to express
the limitations.
5. Determine the non-negativity assumptions
29

Con’t
 The general linear programming problem can be represented in
the following mathematical forms.
Maximization case
Max Z= C1X1 + C2X2 + C3X3 + C4X14… CnXn
subject to
A11 x1 + A12x2+ A13x3+ A14x4+ A15x5+ A16x6+ A17x7+ …. + A1nxn <R1
A21 x1 + A22x2+ A23x3+ A24x4+ A25x5+ A26x6+ A27x7+ …. + A2nxn <R2
Ai1 x1 + Ai2x2+ Ai3x3+ Ai4x4+ Ai5x5+ Ai6x6+ Ai7x7+ …. + Ainxn < Rn
Xij > o for J= 1, 2,3,4… n
30

Con’t
Minimization case
Min Z= C1X1 + C2X2 + C3X3 + C4X14… CnXn
subject to
A11 x1 + A12x2+ A13x3+ A14x4+ A15x5+ A16x6+ A17x7+ …. + A1nxn >R1
A21 x1 + A22x2+ A23x3+ A24x4+ A25x5+ A26x6+ A27x7+ …. + A2nxn >R2
.Ai1 x1 + Ai2x2+ Ai3x3+ Ai4x4+ Ai5x5+ Ai6x6+ Ai7x7+ …. + Ainxn > Rn
Xij > o for J= 1, 2,3,4… n
31

2.4 Application Area of LP
 There is a wide range of problems that lend themselves to
solution by linear programming techniques. Some of these
include:
 Production management (product mix, blending problems,
production planning, Assembly line balancing.),
Marketing management (media selection, traveling sales man
problem, physical distribution),
32

Con’t
Financial management (portfolio selection, profit planning)
Agricultural application
Military applications
Personnel management (staffing problem, Determination of
equitable salary and etc.
33

Example1:
A firm that assembles computer equipments is about to start production of two
new computers (type 1 and type2). Each type of micro computer will
require assembly time, inspection time and storage space. The amount of
each of these resources that can be devoted to the production of the micro
computers is limited. The manager of the firm would like to determine the
quantity of each micro computer to produce in order to maximize the profit
generated by sale of the micro computers. In order to develop a suitable
model of the problem the manager has met with the design and
manufacturing personnel. As a result of those meetings the manager has
obtained the following information:
34

Con’t
Type1 Type 2
Unit profit $60 $50
Assembly time/unit (A) 4 10
Inspection time/unit (I) 2 1
Storage space/unit (S) 3 3
The manager has also information on the availability of company
resources:
Resources Amount available
Assembly time 100hrs
Inspection time 22hrs
Storage space 39cubic feet
35

Con’t
Required
Formulate the linear programming model
Solution
Step1. Define the problem
To determine the quantity of type I and type II micro computers to be
produced per day so as to maximize profit, given the restrictions.
Step2. Identify the decision variables
Let x1 = the quantity of Type I micro computers to be produced & sold
per day.
x2 = the quantity of Type II micro computers to be produced& sold per
day.
z = daily profit
36

Con’t
Step 3: Develop the Objective Function
Max Z= 60x1 + 50x2
Step 4: Formulate the Constraints
A: 4x1 + 10x2 < 100
I: 2x1 + x2 <22
S: 3x1 + 3x2 <39
Step 5: non-negativity
X1, X2 > 0
37

Con’t
The complete LPM is:
Max Z = 60x1 + 50x2
Subject to:
4x1 + 10x2 < 100
2x1 + x2 <22
3x1 + 3x2 < 39
x1, x2 > 0
38

Example 2:
Two machines C1 and C2 produce two grades of tyres A and B. In
one hour of operation, machine C1 produces 30 units of grade A
and 40 units of grade B tyres while machine C2 produces 30
units of grade A and 40 units of grade B tyres. The machines
are required to meet a production schedule of at least 1400
units of grade A and 1200 units of grade B tyres. The cost of
operating machine C1 is $50 per hour and the cost of operating
machine C2 is $80 per hour.
Required
 Formulate the LPM if the objective is to minimize the cost of
operating the machines.
39

Solution
Step 1: Define the Problem
to determine the number of hours the two machines operate in
order to minimize cost, meeting the production requirement.
Step 2:- Determine the decision variables
Let x1= be the number of hours machine C1 should operate
x2= be the number of hours machine C2 should operates
Z= be the total cost
Step 3: Formulate the objective function
Min Z = 50x1 + 80x2
40

Con’t
Step 4:- List the constraints
Grade A: 30x1 + 30x2 > 1400
Grade B: 40x1 + 40x2 > 1200
Step 5:- non-negativity
x1, x2 > 0
Therefore ,The complete LPM is
Min Z = 50x1 + 80x2
Subject to: 30x1 + 30x2 > 1400
40x1 + 40x2 > 1200
x1, x2 > 0
41

2.4 Solving LP Models
 There are two methods to solve LPM:
1. Graphical Method
2.Simplex Method
42

2.4.1 Graphical Method
The graphic method is applicable when we have two decision
variables and the steps include the following:
Step 1: Formulate the mathematical model of the problem
Step 2: Graph the constraints in the plane
Step 3: Identify the area that satisfies the entire set of constraints,
determine corners and their coordinate either from graphing
procedure or by the elimination procedure
43

Con’t
 Step 4: Evaluate the objective function at each corner. The
largest value is the maximum and the smallest value is the
minimum. If two corners have the same optimal value, then the
optimum occurs at every point on the line segment joining the
respective corners.
44

The Maximization Problem
Example1:- a furniture manufacturing company plans to
make two products hardly, chairs and tables, from its
available resources which consist of 400 cubic feet of
mahogany timber and 450 man hours of labor. It knows
that to make a chair it requires 5 cubic feet of timber and
10 man hours and yields a profit of Birr 45/chair. To
manufacture a table, it requires 20 cubic feet of timber and
15 man hours and yields a profit of Birr 80/table. The
problem is to determine how many chairs and tables the
company can make keeping within the resources
constraints so that it maximizes the profit. Formulate
L.P.P. model and provide its graphical solution.
45

Solution
 LPP model
46
Resource Chair Table Amount Availability
Wood(sqft) 5 20 400
Labor (man hour) 10 15 450

Con’t
To solve LPP by graphic method
 Step 1: Formulate the mathematical model of the
problem
Max θ = 45x1 + 80x2
Subject to: 5X1 + 20X2 < 400
10X1 + 15X2 < 450
x1, x2 > 0
47

Con’t
Step 2: Graph the constraints in the plane
5X1 + 20X2 < 400 i.e. X1 + 4X2 < 80 - - - (a)
10X1 + 15X2 < 450 i.e. 2X1 + 3X2 < 90 - - - (b)
X1, X2 > 0
Converting
(a) and (b) to equality, we get
X1 + 4X2 = 80 . . . . (1) equation 1 (80,20)
2X1 + 3X2 = 90 . . . (2) equation 2 (45,30)
48

Con’t
49
X1 + 4X2 =80
2X1 + 3X2 =90

Con’t
 Two straight lines intersect at point B. Its coordinates are
calculated as follows:
X1 + 4X2 = 80 (1)
2X1+ 3X2 = 90 (2)
Multiply equation (1) by 2
2X1 + 8X2 = 160
2X1 + 3X2 = 90
Then subtract (2) from (1)
5X2 = 70 /5, X2 = 14
X1 + 4 × 14 = 80
X1 = 24 i.e. point B in the graph
50

Con’t
 Step 3: Identify the area that satisfies the entire set of
constraints, determine corners and their coordinate either from
graphing procedure or by the elimination procedure.
 Step 4: Evaluate the objective function at each corner. The
largest value is the maximum and the smallest value is the
minimum. If two corners have the same optimal value, then the
optimum occurs at every point on the line segment joining the
respective corners.
51

Con’t
Coordinate of corner point Objective function
45X1 + 80X2 = Z max
Value
(0, 0) 45 ×0 + 80×0 = 0 0
(0, 20) 0 + 1600 1600
(24, 14) 45 × 24 + 80 × 14 2220
(45, 0) 45 × 45 + 80 × 0 2025
52
Hence, this maximum profit is obtained at B i.e. 24 chairs and 14 tables and is
equal to 2220.

The Minimization Problem
Example : Two machines C1 and C2 produce two grades of tyres
A and B. In one hour of operation, machine C1 produces 50
units of grade A and 30 units of grade B tyres while machine
C2 produces 30 units of grade A and 40 units of grade B tyres.
The machines are required to meet a production schedule of at
least 1400 units of grade A and 1200 units of grade B tyres.
The cost of operating machine C1 is $50 per hour and the cost
of operating machine C2 is $80 per hour.
Required
How many hours should each machine operate so that the cost of
production is minimized?
53

Con’t
 Grade of Tyre Production Production
capacity/hour Schedule
C1 C2
A 50 30 1400
B 30 40 1200
Cost of operation/hr $50 $80
54

Con’t
Step 1:- to determine the number of hours the two machines operate in order to
minimize cost, meeting the production requirement.
Step 2:- represent the decision variables
Let x1 be the number of hours machine C1 should operate
x2 be the number of hours machine C2 should operates
θ be the total cost
Step 3:- formulate the objective function
Min θ = 50x1 + 80x2
Step 4:- list the constraints
Grade A: 50x1 + 30x2 > 1400
Grade B: 30x1 + 40x2 > 1200
Step 5:- non-negativity x1, x2 > 0
55

Con’t
The complete LPM is
Min θ = 50x1 + 80x2
Subject to:
50x1 + 30x2 > 1400
30x1 + 40x2 > 1200
x1, x2 > 0
Hence : draw a graph and select optimal solution
56

2.4.1.1 Graphical Solutions for the Special Cases of LP
1. Unboundedness
2. Infeasibility
3. Redundancies
4. Multiple optimal solutions
57

1. Unboundedness
Unboundedness occurs when the decision variable increased
indefinitely without violating any of the constraints. The reason for it
may be concluded to be wrong formulation of the problem such as
incorrectly maximizing instead of minimizing and/or errors in the
given problem.
Example : Max Z = 10X1 + 20X2
Subject to
2X1 + 4X2 > 16
X1 + 5X2 > 15
X1, X2 > 0
58

2. Infeasibility
 Infeasibility is a condition that arises when there is no solution
to a LP problem that satisfies all the constraints. This problem
occurs when the problem was formulated with conflicting
constraints.
Example: Max Z = 3X1+2X2
Subject to:
2X1 + X2 < 2
3X1 + 4X2 > 12
X1, X2 > 0
59

3. Redundancies
 In some cases, a constraint does not form a unique boundary of the
feasible solution space. Such a constraint is called a redundant constraint. A
constraint is redundant if its removal would not alter the feasible solution
space. Redundancy of any constraint does not cause any difficulty in solving
an LP problems graphically. Constraints appear redundant when it may be
more binding (restrictive) than others.
Example :Max Z=x1 + x2
Subject to
x1+x2<20
2X1+x2<3 0
X1<25
X1, x2>0
60

4. Multiple optimal solutions
 Recall the optimum solution is that extreme point for which the
objective function has the largest value. It is of course possible that in a
given problem there may be more than one optimal solution.
Example : Max Z = 8X1+16X2
Subject to:
X1 + X2 < 200
3X1 + 6X2 < 900
X2 < 125
X1, X2 > 0
61

2.4.2 The Simplex Method
Simplex method examines corner points of
the feasible region, using matrix row
operations until an optimal solution is
reached.
62

Simplex method
1. Maximization case
The solution steps of the simplex method can be outlined as follows:
Step1:Formulate the linear programming model of the real world
problem, i.e., obtain a mathematical representation of the problem's
objective function and constraints.
Step2: Express the mathematical model of L.P. problem in the standard
form by adding slack variables in the left-hand side of the constraints
and assign a zero coefficient to these in the objective function.
CH 1 introduction to OR 63

Con’t
Example : Maximize Z = C1X1+ C2X2 + ... +CnXn + 0S1 + 0S2 +...
+0Sm
Subject to
a11X1+a12X2+... + a1nxn+s1=b1
a21X1+a22X2+... + a2nXn+S2 = b2
am1Xl + am2 X2 +... + amnXn + Sm = bm
where X1, X2... Xn and S1, S2 ... Sm are non-negative.
Step 3: Design the initial feasible solution. An initial basic feasible
solution is obtained by setting the decision variables to zero.
X1= X2 = ... = Xn = 0. Thus, we get S1 = b1, S2 = b2 ... Sm = bm.
64

Con’t
 Step 4: Set up the initial simplex tableau. For
computational efficiency and simplicity, the initial
basic feasible solution, the constraints of the standard
LPP as well as the objective function can be displayed
in a tabular form, called the simplex tableau
65

Con’t
Cj C1 C2 Cn,…. 00..0 Quantity Ratio
XB/amn
Basic va. CB X1, X2, ... Xn S1 S2 ... Sm
S1 CB1 a11 a12 ... a1n 1 0 ... 0 b1
S2 CB2 a21 a22 ... a2n 0 1 ... 0 b2
S3 CB3 am1 am2.amn 0 0 ... 1 b3
Zj=ΣCBiXj 0 0 ... 0 0 0 ... 0 ΣCBiXj
Cj-Zj C1-Z1 C2-Z2 ... Cn-Zn
66

Con’t
Cj--coefficients of the variables in the objective function.
X1, X2. Xn----Non-basic variables
S1, S2... Sm-------basic variables
Zj entries represent the decrease in the value of objective
function that would result if one of the variables not
included in the solution were brought into the solution.
Cj-Zj--index row or net evaluation row, is used to
determine whether or not the current solution is optimal or
not
67

Con’t
Step 5. We test if the current solution is optimum or not. If all the
elements or entries in the Cj- Zj row (i.e., index row) are
negative or zero, then the current solution is optimum. If there
exists some positive number, the current solution can be further
improved by removing one basic variable from the basis and
replacing it by some non-basic one. So start trying to improve
the current solution in line with the following steps.
Step 6. Further, iterate towards an optimum solution. To improve
the current feasible solution, we replace one current basic
variable (called the departing variable) by a new non-basic
variable (called the entering variable).
68

Con’t
Step 7. Evaluate the new solution by constructing a second
simplex tableau. After identifying the entering and departing
variable, all that remains is to find the new basic feasible
solution by constructing a new simplex tableau from the
current one.
Step 8. If any of the numbers in Cj - Zj row are positive, repeat
the steps (6-7) again until an optimum solution has been
obtained.
Therefore : A simplex solution in a maximization problem is
optimal if the Cj-Zj row consists of entirely zeros and
negative numbers, i.e. there are no positive values in the row
69

Example (Maximization Case)
Max Z = 60X1+50X2
Subject to
4X1+10X2 < 100
2X1+ X2 < 22
3X1+ 3X2 <39
X1, X2 > = 0
70

Solution
 Step 1: Formulate the linear programming model of
the real world problem i.e mathematical
representation .
 Step 2: standard form
Max Z = 60X1+50X2+0S1+0S2+0S3
Subject to
4X1+10X2+S1 = 100
2X1+ X2+S2 = 22
3X1+ 3X2+S3 = 39
X1, X2 > = 0
71

Con’t
Step 3: Design the initial feasible solution.
X1= X2=0 so S1= 100 S2= 22 S3=39
Step 4: Set up the initial simplex tableau
72

Con’t
73
Cj
Basic V
60
X1
50
X2
0
S1
0
S2
0
S3
Quantity
S1 0 4 10 1 0 0 100
S2 0 2 1 0 1 0 22
S3 0 3 3 0 0 1 39
Zj
Cj-Zj
0 0 0 0 0 0
60 50 0 0 0

Gauss-Jordan Computations
CH 1 introduction to OR 74

Developing the Second Simplex Tableau
Cj 60 50 0 0 0 Quantity
X1 X2 S1 S2 S3
S1 0 4 10 1 0 0 100 100/4= 25
S2 0 2 1 0 1 0 22 22/2= 11
S3 0 3 3 0 0 1 39 39/3=13
Zj 0 0 0 0 0 0
Cj-Zj 60 50 0 0 0
Pivot Row Pivot Column Pivot Element
75
2

Second table
X1 X2 S1 S2 S3
S1 0 0 8 1 -2 0 56
X1 60 1 1/2 0 1/2 0 11
S3 0 0 3/2 0 -3/2 1 6
Zj 60 30 0 30 0 660
Cj-Zj 0 20 0 -30 0
76

Third table
X1 X2 S1 S2 S3
S1 0 0 0 1 6 -16/3 24
X1 60 1 0 0 1 -1/3 9
X2 50 0 1 0 -1 2/3 4
Zj 60 50 0 10 40/3 740
Cj-Zj 0 0 0 -10 -40/3
Therefore :S1 = 24, X1 = 9, and X2 = 4 producing a
maximum profit of $740.
77

2. Minimization case ( Big M-method)
The various steps involved in using simplex method for minimization
problems are:
Step1: Formulate the linear programming model, and express the
mathematical model of L.P. problem in the standard form by introducing
surplus and artificial variables in the left hand side of the constraints. Assign
a 0 (zero) and +M as coefficient for surplus and artificial variables
respectively in the objective function. M is considered a very large number
so as to finally drive out the artificial variables out of basic solution.
Step 2: Next, an initial solution is set up. Just to initiate the solution procedure,
the initial basic feasible solution is obtained by assigning zero value to
decision variables. This solution is now summarized in the initial simplex
table. Complete the initial simplex table by adding two final rows Z, and Cj-
Zj. These two rows help us to know whether the current solution is optimum
or not.
78

Con’t
Step 3: Now; we test for optimality of the solution. If all the entries of Cj - Zj,
row are positives and zeros, then the solution is optimum. However, this
situation may come after a number of iterations. But if at least one of the
Cj - Zj values is less than zero, the current solution can be further
improved by removing one basic variable from the basis and replacing
it by some non-basic one.
Step 4: (i) Determine the variable to enter the basic solution. To do this, we
identify the column with the largest negative value in the Cj - Zj row of the
table.
(ii) Next we determine the departing variable from the basic solution. If an
artificial variable goes out of solution, then we discard it totally and even
this variable may not form part of further iterations. Same procedure, as in
maximization case, is employed to determine the departing variable.
79

Con’t
Step 5: We update the new solution now. We evaluate the entries
for next simplex table in exactly the same manner as was
discussed earlier in the maximization case.
Step 6: Step (3-5) are repeated until an optimum solution is
obtained.
So the following are the essential things to observe in solving for
minimization problems:
 The entering variable is the one with the largest negative value
in the Cj-Zj row while the leaving variable is the one with the
smallest non-negative ratio.
 The optimal solution is obtained when the Cj-Zj row contains
entirely zeros and positive values.
80

Example 1:
Min Z = 7X1+9X2
Subject to
3X1+6X2 > 36
8X1+4X2 > 64
X1, X2 > 0
81

Solution
Step 1: standard form
Min Z = 7X1+9X2+0S1+0S2+MA1+MA2
Subject to
3X1+6X2-S1+A1 = 36
8X1+4X2-S2+A2 = 64
X1, X2 > 0
82

Con’t
Step 2: Initial Simplex Tableau
Cj 7 9 0 0 M M Quantity
X1 X2 S1 S2 A1 A2
A1 M 3 6 -1 0 1 0 36
A2 M 8 4 0 -1 0 1 64
Zj 11M 10M -M -M M M 100M
Cj-Zj 7-11M 9-10M M M 0 0
83

Con’t
 Second Simplex Tableau
Cj 7 9 0 0 M Quantity
X1 X2 S1 S2 A1
A1 M 0 9/2 -1 3/8 1 12
X1 7 1 ½ 0 -1/8 0 8
Zj 7 7/2+9/2M -M 3/8M-7/8 M 56+12M
Cj-Zj 0 11/2-9/2M M 7/8-3/8M 0
84

Con’t
Third Simplex Tableau
Cj 7 9 0 0 Quantity
X1 X2 S1 S2
X2 9 0 1 -2/9 1/12 8/3
X1 7 1 0 1/9 -1/6 20/3
Zj 7 9 -11/9 -5/12 212/3
Cj-Zj 0 0 11/9 5/12
Therefore, the optimal solution is: X1 = 20/3 and X2 = 8/3 and value
of objective function is 212/3.
85

Assignment
1. Duality in LPP
2. Sensitivity Analysis
86

Chapter Two linear programming in business mathematics .pdf

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