Lp graphical and simplexx892

Quantitative Analysis

for Management

LP:

Graphical solution Method

Simplex Method

Mohammad T. Isaai

Quantitative Analysis for Management

In symbolic form, the linear programming model is:

Maximize Z c1x1 cnxn Objective Function

subject to

a11x1 a1nxn b1

a21x1 a2nxn b2

am1x1 amnxn bm

Functional Constraints

and

x1 0,, xn 0 Nonnegativity Constraints

for known parameters c1, … , cn ; a11, … , amn ; b1, … , bm.

Mathematical Statement of Linear Programming


Basic Assumptions of Linear Programming

• Certainty

• Proportionality (1-3, 10-30)

• Additivity (8,3,11)

• Divisibility (10.2)

• Nonnegativity


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

Graphical Solution

Solve the following problem. Find the optimal solution.

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


1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1

Max

Z = 300 x1 +500 x2

Subject to

x1 4

2x2 12

3 x1 + 2 x2 18

x1 0, x2 0

Solution - Constraints


1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

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

Isoprofit LineOr

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

a11 x1 a1n xn =b1

a21x1 a2n xn =b2

M

am1 x1 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 Analysispost 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

1

2

3

4

5

6

7

8

9

10

x2

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

1

2

3

4

5

6

7

8

9

10

x2

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


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1

2

3

4

5

6

7

8

9

10

x2

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

Two units increase in b2


1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

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

1

2

3

4

5

6

7

8

9

10

x2

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 iszero.

• 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.


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

Problem


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

1

2

3

4

5

6

7

8

9

10

x2

x1

feasible region


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1

2

3

4

5

6

7

8

9

10

x2

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

the Optimal Solution


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8

9

10

x2

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)

What if the objective function is changed

from 40 x1 +50 x2 to 40 x1 +70 x2


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

1

2

3

4

5

6

7

8

9

10

x2

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

1

2

3

4

5

6

7

8

9

10

x2

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

02 4 6 8

Region Satisfying

the 3rd Constraint

Region Satisfying First 2 Constraints


A Problem with a Redundant Constraint

X2

X1

30

25

20

15

10

5

05 10 15 20 25 30

Feasible

Region

2X1 + X2 < 30

X1 < 25

X1 + X2 < 20

Redundant

Constraint


An Example of Alternate Optimal Solutions

Optimal Solution Consists of All

Combinations of X1 and X2 Along the AB

Segment

Isoprofit Line for $12

Overlays Line Segment

Isoprofit Line for $8A

B

AB


A Solution Region That is Unbounded to the Right

X2

X1

15

10

5

0

5 10 15

Feasible Region

X1 > 5X2 < 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.

Lp graphical and simplexx892

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