Lp graphical and simplexx892

Quantitative Analysis
for Management
LP:
Graphical solution Method
Simplex Method
Mohammad T. Isaai

Mathematical Statement of Linear Programming
In symbolic form, the linear programming model is:
Maximize Z  c1x1   cn xn  Objective Function
subject to
a11x1   a1n xn  b1
a21x1   a2n xn  b2
am1x1   amnxn  bm







 Functional Constraints
Quantitative Analysis for Management
and
x1  0,, xn  0  Nonnegativity Constraints
for known parameters c1, … , cn ; a11, … , amn ; b1, … , bm.

Basic Assumptions of Linear
Programming
• Certainty
• Proportionality (1-3, 10-30)
• Additivity (8,3,11)
• Divisibility (10.2)
• Nonnegativity

Graphical Solution
Solve the following problem. Find the optimal solution.
Max Z = 300 x1 +500 x2
Subject to
x1  4 Resource 1
2x2  12 Resource 2
3 x1 + 2 x2  18 Resource 3
x1  0, x2  0
Product 1 needs 1 unit of resource 1, and 3 units of resource 3.
Product 2 needs 2 units of resource 2 and 2 units of resource 3
There are 4 units of resource 1, 12 units of resource 2, and 18 units of
resource 3

Solution - Constraints
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
Max
Z = 300 x1 +500 x2
Subject to
x1  4
2x2  12
3 x1 + 2 x2  18
x1  0, x2  0

Solution – Objective Function
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
Max
Z = 300 x1 +500 x2
Subject to
x1  4
2x2  12
3 x1 + 2 x2  18
x1  0, x2  0
Isoprofit Line
Or
Isocost Line

Terms Used in LP
• Solution
• Feasible Solution (Infeasible Solution)
• Feasible Region
• Optimal Solution
• Boundary Equation (Constraint Line)
• Corner Point Solution
• Adjacent Corner Point Solutions

Property of an Optimal Solution
• If there is only one optimal solution, then it lies on a
corner point solution.
• If there are more than one optimal solution, then at
least two adjacent corner point solutions are optimal
as well as every point on the line connecting them.

The Concept of Simplex Method
• Initial Step: Start from a feasible corner point solution
• Iterative Step: Move from the existing corner point
solution to a better, feasible adjacent one
• Stopping Rule: If the existing corner point solution is
better than all of its adjacent feasible corner point
solutions, it is optimal

Simplex Algorithm Movement

Standard L.P. Model
• Simplex alg. Must be applied to the standard form in which:
– Objective Function in Maximization Form
– Constraints are less than or equal .
– Decision Variables are nonnegative
In different books standards are defined differently

Standard L.P. Model
Maximize Z c1 x1 cn xn  Objective Function
subject to
a11x1 a1n xn =b1
a21x1 a2n xn =b2
M
am1x1 amnxn =bm







 Functional Constraints
and
x1 0,, xn 0  Nonnegativity Constraints

How to generate the standard form?
• Slack variables
• Surplus variables
• Artificial variables
• Min Z = 300 x1 +500 x2
Subject to
• x1 + 3x2  4
• - x1 + x2 = 12
• 3 x1 + 2 x2  18
• x1  0, x2 Unrestricted in
sign

Sensitivity Analysis
post optimality analysis
• The impact of changing parameters
• “What-if” questions.
• Sensitive Parameters.
• In the real world, real data is not certain and we
use our best estimate. However, they may change.

Sensitivity Analysis
• Changes in Resources (RHS)
• Changes in the Objective Function Coefficients
• Changes in Technological Coefficients

Sensitivity Example
• How much are you willing to pay for each
additional Resource Unit?
• How many additional units do you buy?
assume that we can get 1 extra unit of res. 2.
Let’s start with 1 additional unit of resource 2.
Then 2, 3 and more additional units.

1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
2x2 = 12
3 x1 + 2 x2 = 18
x1 = 4
Z = 300 (2) +500(6)
Z = 3600
Main Problem

1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
2x2 = 12 + 1
3 x1 + 2 x2 = 18
x1 =4
x2 = 6.5
x1 = 5/3
Z = 300 (5/3)+500(6.5)
Z = 3750
3750-3600 = 150
One unit increase in b2

Two units increase in b2
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
2x2 = 12 + 2
3 x1 + 2 x2 = 18
x1 =4
x2 = 7
x1 = 4/3
Z = 300 (4/3) +500(7)
Z = 3900
3900-3750 = 150

1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
2x2 = 12 + a
3 x1 + 2 x2 = 18
x1 =4
x2 = 6+a/2
x1 = 2-a/3
Z = 300 (2-a/3)
+500(6+a/2)
Z = 600-100a
+3000+250a
Z = 3600 +150a
a unit increase in b2

1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
3 x1 + 2 x2 = 18
x1 = 0
x2 = 9
x1 = 0
Z = 300 (0)+500(9)
Z = 4500
we buy at most 6
additional units

Shadow Price
• Since the resources are limited, the profit is also limited.
• If we increase the limited resources the profit also
increases.
• How much are we ready to pay for one extra unit of each
resource?
• Shadow price for each resource is the maximum amount
one is willing to pay for one additional unit of that
resource.
• Clearly, Shadow price is different from the market price

Shadow Price-2
• If a resource is not limited, then its Shadow price is
zero.
• It means, if the value of the slack variable for a
constraint is positive, then its Shadow price is zero.
• In the simplex method for standard case, shadow
prices are shown on the objective function row and
under the slack variables.

Problem
Given the following model
Min Z = 40 x1 + 50 x2
Subject to
(C1) 2x1 + 3x2  30
(C2) x1 + x2  12
(C3) 2 x1 + x2  20
x1  0, x2  0

Min
Z = 40 x1 +50 x2
Subject to
2x1 + 3x2  30
x1+ x2  12
2 x1 + x2  20
x1  0, x2  0
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
feasible region

the Optimal Solution
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
Min
Z = 40 x1 +50 x2
Subject to
2x1 + 3x2  30
x1+ x2  12
2 x1 + x2  20
x1  0, x2  0
2x1 + 3x2 = 30
2 x1 + x2 = 20
x2 = 5
x1 = 7.5
Z = 40 (7.5)+50
(5)
Z = 550

What if the objective function is changed
from 40 x1 +50 x2 to 40 x1 +70 x2
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
Min
Z = 40 x1 +70 x2
Subject to
2x1 + 3x2  30
x1+ x2  12
2 x1 + x2  20
x1  0, x2  0
2x1 + 3x2 = 30
x2 = 0
x1 = 15
Z = 40 (15)+70 (0)
Z = 600, changed from 550 to 600 (O.F. is Min)

Min
Z = 40 x1 +50 x2
Subject to
2x1 + 3x2  30
x1+ x2  12
2 x1 + x2  20
x1  0, x2  0
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
again the original problem
What if the green
constraint is
changed to
2 x1 + x2  15

Min Z= 40 x1+50 x2
Subject to
2x1 + 3x2  30
x1+ x2  12
2 x1 + x2  15
x1  0, x2  0
1 2 3 4 5 6 7 8 9 10
x2
10
9
8
7
6
5
4
3
2
1
x1
2x1 + 3x2 = 30
x1+ x2 = 12
x2 = 6
x1 = 6
Z = 540 from 550 to 540 (O.F. is Min)

Special Cases in LP
• Infeasibility
• Redundancy
• More Than One Optimal Solution
• Unbounded Solutions
• Degeneracy

A Problem with No Feasible Solution
X2
X1
8
6
4
2
0
2 4 6 8
Region Satisfying
the 3rd Constraint
Region Satisfying First 2 Constraints

A Problem with a Redundant Constraint
X2
2X1 + X2 < 30
X1 + X2 < 20
X1
30
25
20
15
10
5
0
Feasible
Region
Redundant
Constraint
X1 < 25
5 10 15 20 25 30

An Example of Alternate Optimal Solutions
Optimal Solution Consists of All
Combinations of X1 and X2 Along the AB
Segment
A Isoprofit Line for $8
Isoprofit Line for $12
Overlays Line Segment
B
AB

A Solution Region That is Unbounded to the Right
X1 > 5
Feasible Region
X2
X1
15
10
5
0
5 10 15
X2 < 10
X1 + 2X2 > 10

• Having selected the pivot column, one divides each
quantity column no. (RHS) to the corresponding
pivot column no., if all ratios are negative or
undefined, it indicates that the problem is
unbounded.

Degeneracy
• Having selected the pivot column, one divides each
quantity column no. (RHS) to the corresponding
pivot column no., if there is a tie for the smallest
ratio, this is a signal that degeneracy exists.
• Cycling may result from degeneracy.

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