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Lecture Day 3

Solving Linear Programming Problems

Using the Graphical Method Special Cases in Using the Graphical Method

Solving Linear Programming Problems

Using the Simplex Method

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(-4)(0.5X1+ X2= 81)

1.2X1+ 4X2= 240

X2= 81 0.5X1

1.2X1+ 4(81 0.5X1) = 240

1.2X1+ 324 - 2X1= 240

-0.8X1= -84

-0.8 -0.8

X1= 105

X2= 81 0.5X1

= 81 0.5(105)

= 81 52.5

X2= 28.5

-2X1- 4X2= -324

1.2X1+ 4X2= 240

-0.8X1 = -84

-0.8 -0.8

X1= 105

Z = 200X1+ 500X2

Point C:

= 200(105) + 500(28.5)

= 21,000 + 14,250

Z = 35,250

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X1

X2

50

50

100

100

150

150

2000

X2= 40

1.2X1+ 4X2= 240

0.5X1+ X2= 81

AB

C

D

Techniques to DetermineWhich Extreme Point

Gives the Optimal Solution

1. Use the objective function to single out theextreme point that is the optimum.

2. Find the coordinates of the extreme points andcompute the profit associated with each

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X1

X2

50

50

100

100

150

150

2000

X2= 40

1.2X1+ 4X2= 240

0.5X1+ X2= 81

AB

C

D

Point A

X1= 0

X2= 40

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

Z = 200X1+ 500X2

= 200(0) + 500(40)

= 20,000

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X1

X2

50

50

100

100

150

150

2000

X2= 40

1.2X1+ 4X2= 240

0.5X1+ X2= 81

AB

C

D

Point B

1.2X1+ 4X2= 240

X2= 40

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

Z = 200X1+ 500X2

= 33,333.33

1.2X1+ 4(40) = 240

1.2X1+ 160 = 240

1.2X1= 240 - 160

= 200(66.6) + 500(40)

1.2X1= 80

1.2 1.2

X1= 66.6

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X1

X2

50

50

100

100

150

150

2000

X2= 40

1.2X1+ 4X2= 240

0.5X1+ X2= 81

AB

C

D

Point D

0.5X1+ X2= 81

X2= 0

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

0.5X1= 81

0.5 0.5

X1= 162

Z = 200X1+ 500X2

= 32,400

= 200(162) + 500(0)

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Sample Problem # 2

The Reed Pump Co. manufactures pumping apparatus for thepetroleum industry. It is trying to determine an optimal production

and distribution plan for its new submersible pump. They have onemain plant and two regional warehouses. Warehouse 1 has demandfor at least 80 pumps this month, and warehouse 2 has demand forat least 100 units. The following are the combined costs of producingand shipping to the two warehouses:

To achieve economies of scale, the company wants to produce atleast 300 pumps this month. The only resource limitation involvesmanufacturing resource hours. Pumps sent to regional warehouse 1need an extra-fine filter and require five hours of manufacturing time.Those pumps sent to warehouse 2 require only four hours ofmanufacturing time. The company has 2,000 hours of manufacturing-

resource time available per month.

Warehouse 1 Warehouse 2

Cost of Producing & Shipping

P240 P280From Plant

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Let: X1= number of pumps to bedelivered to Warehouse 1

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2= number of pumps to be

delivered to Warehouse 2

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X2= 300

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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5X1+ 4X2= 2,000

if X2= 0

5X1= 2,000

5 5

X1= 400

if X1= 0

4X2= 2,000

4 4

X2= 500

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X2= 300

5X1+ 4X2= 2,000

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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240X1+ 280X2 = 100,000

240X1= 100,000

X1= 416.67

if X1= 0

280X2= 100,000

X2= 357.14

if X2= 0

240 240

280 280

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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240X1+ 280X2 = 50,000

240X1= 50,000

X1= 208.33

if X1= 0

280X2= 50,000

X2= 178.57

if X2= 0

240 240

280 280

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X

2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X

2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

Point D

X1+ X2 = 300

X2 = 100

X1+ 100= 300

X1= 300 - 100

X1= 200

Z = 240X1+ 280X2

= 240(200) + 280(100)

= 48,000 + 28,000

Z = 76,000

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X

2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

Point A

X1+ X2 = 300

X1 = 80

80 + X2 = 300

X2= 300 - 80

X2= 220

Z = 240X1+ 280X2

= 240(80) + 280(220)

= 19,200 + 61,600

Z = 80,800

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X

2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

Point B

5X1+ 4X2 = 2,000

X1 = 80

5(80) + 4X2 = 2,000

400 + 4X2 = 2,000

4X2 = 2,000 - 400

4X2 = 1,6004 4

X2 = 400

Z = 240X1+ 280X2

= 240(80) + 280(400)

= 19,200 + 112,000

Z = 131,200

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X1

X2

500

500

100

100

400

400

2000 300

200

300

X1= 80

X2= 100

X1+ X

2= 300

5X1+ 4X2= 2,000

A

B

C

D

Min. Z = 240X1+ 280X2

Subject to:

X1> 80

X2> 100

X1+ X2> 300

5X1+ 4X2< 2,000

Point C

5X1+ 4X2 = 2,000

X2 = 100

5X1+ 4(100)= 2,000

5X1 = 1,6005 5

X1 = 320

Z = 240X1+ 280X2

= 240(320) + 280(100)

= 76,800 + 28,000

Z = 104,800

5X1+ 400= 2,000

5X1= 2,000 - 400

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Special Cases in the Graphical Solutionto Linear Programming Problems

It is possible that an adjacent extreme point will yieldthe same value. Graphically, this happens wheneverthe slope of the objective function equals the slope

of a constraint equation that passes through anoptimal extreme point.

Alternative Optima

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X1

X2

50

50

100

100

150

150

2000

X2= 40

1.2X1+ 4X2= 240

0.5X1+ X2= 81

A

D

C

B

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

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Special Cases in the Graphical Solutionto Linear Programming Problems

It is possible for an LP problem to have a non-emptyset of feasible solutions and yet have no finite optimalsolution.

This can occur whenever the feasible region extendsinfinitely in the direction of improvement for theobjective function.

Having this scenario implies that the model has beenformulated incorrectly. Usually, a constraint has beenomitted or the signs on some of the coefficients havebeen reversed.

Unbounded Solution

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Max. Z = X1+ X2

Subject to:

5X1 - X2> 10

3X1- 2X2< 9

X1, X2> 0

X1

X2

5

5

1

1

4

4

20 3

23

-1-2-3-4-5-6-7-8-9

-10

5X1X2 = 10

3X12X2= 9

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Special Cases in the Graphical Solutionto Linear Programming Problems

It is also possible to have an LP problem in which nofeasible solution exists.

This situation corresponds to a problem that is

formulated incorrectly or has conflicting restrictionswithin the constraint set.

No Feasible Solution

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Min. Z = X1+ X2

Subject to:

X1 - X2> 1

-X1+ X2> 1

X1, X2> 0

X1

X2

1

1

2 3

23

-1-2-3

-2-3

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SolvingLinear Programming Problems

Using the Simplex Method

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GeorgeDantzig developed the simplex methodin 1947. It is the most widely used method for

solving LP problems.

Thesimplex method is a solution algorithm. Analgorithmis an iterative procedure withspecific computational rules that solves aproblem in a finite number of steps.

Thesimplex algorithm is a systematic algebraicprocedure for examining the extreme points ofthe LP feasible region. The extreme points areexamined in a sequence such that eachsuccessive point yields a solution that is at leastas effective as the previous point.

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Thesimplex method requires that all constraints be

expressed as equations, or that the problem betransformed in standard form.

That is, there is a need to convert less-than-or-equal-to and greater-than-or-equal-to

constraints into equations. This can be done by adding dummy

variablesto the inequalities.

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Sample Problem In Standard Form

0X1 +1X2 +1S1 = 40

1.2X1 +4X2 +0S1+1S2 = 240

Let : X1= number of

power amplifiersX2= number of

preamplifiers

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

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Let : X1= number of

power amplifiersX2= number of

preamplifiers

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

Sample Problem In Standard Form

0X1 +1X2 +1S1 = 40

1.2X1 +4X2 +0S1

+

1S2 = 240

0S2

+

0.5X1 +1X2 +0S1+0S2+1S3 = 81

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Let : X1= number of

power amplifiersX2= number of

preamplifiers

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

Sample Problem In Standard Form

0X1 +1X2 +1S1 = 40

1.2X1 +4X2 +0S1

+

1S2 = 240

0S2

+

0.5X1 +1X2 +0S1+0S2+1S3 = 81

+0S3

+0S3

200X1 +500X2 +0S1 +0S2 +0S3

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Canonical Form of the Simplex Tableau

The table consisting of the system equations and other relevantinformation of an LP problem is called a Simplex Tableau.

00

.

.

.

0

S1S2.

.

.

Sm

a11a21.

.

.

am1

a12a22.

.

.

am2

a1na2n.

.

.

amn

10

.

.

.

0

01

.

.

.

0

00

.

.

.

1

b1b2.

.

.

bm

SolutionValues

. . .

. . .

. . .

BasicVariables

X1 X2 Xn

. . .

. . .

. . .

CB . . . S2 Sm. . .

Cj C1 C2 Cn. . . 0 0 0. . .

Decision Variables Slack Variables

Zj

Cj- Zj

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Let : X1= number of

power amplifiersX2= number of

preamplifiers

Max. Z = 200X1+ 500X2

Subject to:

X2< 40

1.2X1+ 4X2< 240

0.5X1+ X2< 81

X1, X2> 0

Sample Problem In Standard Form

0X1 +1X2 +1S1 = 40

1.2X1 +4X2 +0S1

+

1S2 = 240

0S2

+

0.5X1 +1X2 +0S1+0S2+1S3 = 81

+0S3

+0S3

200X1 +500X2 +0S1 +0S2 +0S3

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Cj

BasicVariables

CB SolutionValues

X1 X2 S1 S2 S3

Zj

Cj- Zj

S1

S2

S3

200 500 0 0 0

0 1 1 0 0 40

1.2 4 0 1 0 240

0.5 1 0 0 1 81

0

0

0

0 0 0 0 0 0

200 500 0 0 0

Tableau 1

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End of Lecture Day 3