MT 2351 Chapter 2 An Introduction to Linear Programming

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Chapter 2

An Introduction to Linear Programming

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• Courtesy of NPR:

“The Mathematician Who Solved Major Problems”

http://www.npr.org/dmg/dmg.php?prgCode=WESAT&showDate=21-May-2005&segNum=14&

George Dantzig

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General Form of an LP Model

Maximize (or Minimize)

Subject to

c x c x c x

a x a x a x b

a x a x a x b

a x a x a x b

x x x

n n

n n

n n

m m mn n m

n

1 1 2 2

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

1 2 0

( , , )

( , , )

( , , )

, , ,

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General Form of an LP Model where the c’s, a’s and b’s are constants

determined from the problem and the x’s are the decision variables

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Components of Linear Programming

An objective Decision variables Constraints Parameters

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Assumptions of the LP Model Divisibility - basic units of x’s are divisible Proportionality - a’s and c’s are strictly

proportional to the x’s Additivity - each term in the objective

function and constraints contains only one variable

Deterministic - all c’s, a’s and b’s are known and measured without error

Non-Negativity (caveat)

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Sherwood Furniture CompanyRecently, Sherwood Furniture Company has been interested in developing a new line of stereo speaker cabinets. In the coming month, Sherwood expects to have excess capacity in its Assembly and Finishing departments and would like to experiment with two new models. One model is the Standard, a large, high-quality cabinet in a traditional design that can be sold in virtually unlimited quantities to several manufacturers of audio equipment. The other model is the Custom, a small, inexpensive cabinet in a novel design that a single buyer will purchase on an exclusive basis. Under the tentative terms of this agreement, the buyer will purchase as many Customs as Sherwood produces, up to 32 units. The Standard requires 4 hours in the Assembly Department and 8 hours in the Finishing Department, and each unit contributes $20 to profit. The Custom requires 3 hours in Assembly and 2 hours in Finishing, and each unit contributes $10 to profit. Current plans call for 120 hours to be available next month in Assembly and 160 hours in Finishing for cabinet production, and Sherwood desires to allocate this capacity in the most economical way.

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Sherwood Furniture Company – Linear Equations

21 1020 xx Maximize

Subject to12034 21 xx

16028 21 xx

322 x02,1 xx

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Sherwood Furniture Company – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

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Sherwood Furniture Company – Graph Solution Constraint 1

Set

then

4 3 120

0 40

0 30

1 2

1 2

2 1

x x

x x

x x

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Sherwood Furniture Company – Graph Solution Constraint 1

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

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Sherwood Furniture Company – Graph Solution Constraint 2

Set

then

8 2 160

0 80

0 20

1 2

1 2

2 1

x x

x x

x x

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Sherwood Furniture Company – Graph Solution Constraint 1 & 2

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

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Sherwood Furniture Company – Graph Solution Constraint 3

32

Set

2 x

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Sherwood Furniture Company – Graph Solution Constraint 1, 2 & 3

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

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Sherwood Furniture Company – Graph Solution

Set

then

20 10 400

0 40

0 20

1 2

1 2

2 1

x x

x x

x x

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Sherwood Furniture Company – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

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Sherwood Furniture Company – Solve Linear Equations

Set

4 3 120

8 2 160

1 2

1 2

x x

x x

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Sherwood Furniture Company – Solve Linear Equations

20

804

)16028(

)12034(2

Set

2

2

21

21

x

x

xx

xx

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Sherwood Furniture Company – Solve Linear Equations

500)20(10)15(20

function objectiveThen

15

120)20(34

1

1

x

x

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Sherwood Furniture Company – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

Optimal Point(15, 20)

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Sherwood Furniture Company – Slack Calculation

20,15 21 xx

0

120 120

120 3(20) 4(15)

hours;assembly for slack calulateThen

slack

0

160 160

160 2(20) 8(15)

hours; finishingfor slack calculateThen

slack

12

32 20

Contract; Customfor slack calculateThen

slack

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Sherwood Furniture Company - Slack Variables

Max

20x1 + 10x2 + 0S1 + 0S2 + 0S3

s.t.

4x1 + 3x2 + 1S1 = 120

8x1 + 2x2 + 1S2 = 160

x2 + 1S3 = 32

x1, x2, S1 ,S2 ,S3 >= 0

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Sherwood Furniture Company – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

2

3

1

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Sherwood Furniture Company – Slack Calculation Point 1

32

160

120

0

0

2

2

1

2

1

s

s

s

Then

x

x

If

Point 1

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Sherwood Furniture Company – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

2

3

1

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Sherwood Furniture Company – Slack Calculation Point 2

12

0

40

0

20

2

2

1

2

1

s

s

s

Then

x

x

If

Point 2

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Sherwood Furniture Company – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

2

3

1

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Sherwood Furniture Company – Slack Calculation Point 3

12

0

0

20

15

2

2

1

2

1

s

s

s

Then

x

x

If

Point 3

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Sherwood Furniture Company – Slack Calculation Points 1, 2 & 3

12

0

40

0

20

2

2

1

2

1

s

s

s

Then

x

x

If

12

0

0

20

15

2

2

1

2

1

s

s

s

Then

x

x

If

32

160

120

0

0

2

2

1

2

1

s

s

s

Then

x

x

If

Point 1 Point 2 Point 3

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Sherwood Furniture Company – Slack Variables

For each ≤ constraint the difference between the RHS and LHS (RHS-LHS). It is the amount of resource left over. Constraint 1; S1 = 0 hrs.

Constraint 2; S2 = 0 hrs.

Constraint 3; S3 = 12 Custom

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Pet Food CompanyA pet food company wants to find the optimal mix of ingredients, which will minimize the cost of a batch of food, subject to constraints on nutritional content. There are two ingredients, P1 and P2. P1 costs $5/lb. and P2 costs $8/lb. A batch of food must contain no more than 400 lbs. of P1 and must contain at least 200 lbs. of P2. A batch must contain a total of at least 500 lbs. What is the optimal (minimal cost) mix for a single batch of food?

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Pet Food Company – Linear Equations

21 8P 5P

Minimize,

To,Subject

40001 21 PP

20010 21 PP50011 21 PP

02,1 PP

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Pet Food Company – Graph Solution

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2

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Pet Food Company – Graph Solution Constraint 1

400

then

4001

Set

1

1

P

P

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Pet Food Company – Graph Solution Constraint 1

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2

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Pet Food Company – Graph Solution Constraint 2

200

then

2001

Set

2

2

P

P

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Pet Food Company – Graph Solution Constraint 1 & 2

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2

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Pet Food Company – Graph Solution Constraint 3

50011

Set

21 PP

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Pet Food Company – Graph Solution Constraint 1, 2 & 3

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2

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Pet Food Company – Solve Linear Equations

8000

5000

then

400085

Set

12

21

21

PP

PP

PP

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Pet Food Company – Graph Solution

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2

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Pet Food Company – Solve Linear Equations

300

3001

500)200(11

50011

200

Set

1

1

1

21

2

P

P

P

Then

PP

P

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Pet Food Company – Solve Linear Equations

3100)200(8)300(5

85

function objectiveThen

200,300

21

21

PP

PP

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Pet Food Company – Graph Solution

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2

Optimal Point(300, 200)

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Pet Food Company – Slack/ Surplus Calculation

200 P 300; P 2 1

100

400 300

;Pfor slack calculateThen 1

slack

0

200 200

;Pfor surplus calculateThen 2

surplus

0

500 200 300

Batch;for surplus calculateThen

surplus

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Pet Food Co. – Linear Equations Slack/ Surplus Variables

Min

5P1 + 8P2 + 0S1 + 0S2 + 0S3

s.t.

1P1 + 1S1 = 400

1P2 - 1S2 = 200

1P1 + 1P2 - 1S3 = 500

P1, P2, S1 ,S2 ,S3 >= 0

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Pet Food Co. – Slack Variables For each ≤ constraint the difference

between the RHS and LHS (RHS-LHS). It is the amount of resource left over. Constraint 1; S1 = 100 lbs.

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Pet Food Co. – Surplus Variables For each ≥ constraint the difference

between the LHS and RHS (LHS-RHS). It is the amount bt which a minimum requirement is exceeded. Constraint 2; S2 = 0 lbs.

Constraint 3; S3 = 0 lbs.

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Special Cases Alternate Optimal Solutions No Feasible Solution Unbounded Solutions

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Alternate Optimal Solutions

0,

186

142

204

S.T.

24Min

21

21

21

21

21

xx

xx

xx

xx

xx

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Alternate Optimal Solutions

0

5

10

15

20

25

0 5 10 15 20

x1

x2

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Alternate Optimal Solutions

50

200

Then

204

Set

12

21

21

xx

xx

xx

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Alternate Optimal Solutions

0

5

10

15

20

25

0 5 10 15 20

x1

x2

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Alternate Optimal Solutions

70

140

Then

142

Set

12

21

21

xx

xx

xx

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Alternate Optimal Solutions

0

5

10

15

20

25

0 5 10 15 20

x1

x2

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Alternate Optimal Solutions

180

30

Then

186

Set

12

21

21

xx

xx

xx

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Alternate Optimal Solutions

0

5

10

15

20

25

0 5 10 15 20

x1

x2

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Alternate Optimal Solutions

100

200

Then

4024

Set

12

21

21

xx

xx

xx

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Alternate Optimal Solutions

0

5

10

15

20

25

0 5 10 15 20

x1

x2

A

B

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Alternate Optimal Solutions

28)8(2)3(4

Then

8 and 20=x+4(3)

3 and 62

)142(

204

SetA Point At

22

11

21

21

x

xx

xx

xx

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Alternate Optimal Solutions

28)2(2)6(4

Then

2 and 14)6(2

6 and 6611

)186(

)142(6

Set BPoint At

22

11

21

21

xx

xx

xx

xx

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Special Cases Alternate Optimal Solutions No Feasible Solution Unbounded Solutions

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No Feasible Solution

0,

1243

22

S.T.

23Max

21

21

21

21

xx

xx

xx

xx

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No Feasible Solution

0

1

2

3

4

0 1 2 3 4 5

x1

x2

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No Feasible Solution

0

1

2

3

4

0 1 2 3 4 5

x1

x2

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Special Cases Alternate Optimal Solutions No Feasible Solution Unbounded Solutions

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Unbounded Solutions

0,

402

10

S.T.

2Max

21

21

21

21

xx

xx

xx

xx

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Unbounded Solutions

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 5 10 15 20 25

x1

x2

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Unbounded Solutions

200

100

Then

202

Set

12

21

21

xx

xx

xx

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Unbounded Solutions

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 5 10 15 20 25

x1

x2

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