linear programming adapted from russell taylor€¦ · max 7t + 5c (profit) subject to the...

Linear programming (Adapted from Chapter 13 Supplement, Operations and Management, 5Chapter 13 Supplement, Operations and Management, 5thth edition edition by by Roberta Russell & Bernard W. Taylor, III., Copyright 2006 John Wiley & Sons, Inc. This presentation also contains material of Pearson, Prentice hall

Dr. Arturo S. Leon, BSU (Spring 2010)

1© Arturo S. Leon, BSU, Spring

2010

Copyright 2006 John Wiley & Sons, Inc.

Supplement 13-2

Lecture Outline

� Model Formulation

� Graphical Solution Method

� Linear Programming Model

� Solution

� Solving Linear Programming Problems with Excel

� Sensitivity Analysis

Copyright 2006 John Wiley & Sons, Inc. Supplement 13-3

A model consisting of linear relationshipsrepresenting a firm’s objective and resource constraints

Linear Programming (LP)

LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints


A model consisting of linear relationshipsrepresenting a firm’s objective and resource constraints

Linear Programming

LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints


http://video.google.com/videoplay?docid=754226565202103395#docid=8211036228894039768

Video of Linear Programming (LP)


LP Model Formulation

� Decision variables– mathematical symbols representing levels of activity of an

operation

� Objective function– a linear relationship reflecting the objective of an operation

– most frequent objective of business firms is to maximize profit

– most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost

� Constraint– a linear relationship representing a restriction on decision

making


LP Model Formulation (cont.)

Max/min z = c1x1 + c2x2 + ... + cnxn

subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1

a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2

:am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm

xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients


LP Model: Example

LaborLabor ClayClay RevenueRevenuePRODUCTPRODUCT (hr/unit)(hr/unit) (lb/unit)(lb/unit) ($/unit)($/unit)

BowlBowl 11 44 4040

MugMug 22 33 5050

There are 40 hours of labor and 120 pounds of clay There are 40 hours of labor and 120 pounds of clay available each dayavailable each day

Decision variablesDecision variables

xx11 = number of bowls to produce= number of bowls to produce

xx22 = number of mugs to produce= number of mugs to produce

RESOURCE REQUIREMENTSRESOURCE REQUIREMENTS

LP Formulation: Example

Maximize Maximize ZZ = $40 = $40 xx11 + 50 + 50 xx22

Subject toSubject to

xx11 ++ 22xx22 ≤ ≤ 40 hr40 hr (labor constraint)(labor constraint)

44xx11 ++ 33xx22 ≤ ≤ 120 lb120 lb (clay constraint)(clay constraint)

xx1 1 , , xx22 ≥ ≥ 00

Quick solution with Excel solverQuick solution with Excel solver

Using solver for the Bowl/mug example

Decision variables

X1 -5 Maximize -250

x2 -1

constraint 1: -7 <= 40

constraint 2: -23 <= 120

constraint 3: -5 >= 0

constraint 4: -1 >= 0

Initial conditions for Solver

LP Example (Cont.)

Solution is Solution is xx11 = 24 bowls = 24 bowls xx2 2 = 8 mugs= 8 mugs

Revenue or benefit = Revenue or benefit = $1,360$1,360

Using solver for the Bowl/mug example

Decision variables

X1 24 Maximize 1360

x2 8

constraint 1: 40 <= 40

constraint 2: 120 <= 120

constraint 3: 24 >= 0

constraint 4: 8 >= 0

Solution


Graphical Solution Method

1.1. Plot model constraint on a set of coordinates Plot model constraint on a set of coordinates in a planein a plane

2.2. Identify the feasible solution space on the Identify the feasible solution space on the graph where all constraints are satisfied graph where all constraints are satisfied simultaneouslysimultaneously

3.3. Plot objective function to find the point on Plot objective function to find the point on boundary of this space that maximizes (or boundary of this space that maximizes (or minimizes) value of objective functionminimizes) value of objective function


LP Formulation: Example

Maximize Maximize ZZ = $40 = $40 xx11 + 50 + 50 xx22

Subject toSubject to

xx11 ++ 22xx22 ≤ ≤ 40 hr40 hr (labor constraint)(labor constraint)

44xx11 ++ 33xx22 ≤ ≤ 120 lb120 lb (clay constraint)(clay constraint)

xx1 1 , , xx22 ≥ ≥ 00

Solution is Solution is xx11 = 24 bowls = 24 bowls xx2 2 = 8 mugs= 8 mugs

Revenue = $1,360Revenue = $1,360


Graphical Solution: Example

4 4 xx11 + 3 + 3 xx2 2 ≤ ≤ 120 lb120 lb

xx11 + 2 + 2 xx2 2 ≤ ≤ 40 hr40 hr

Area common toArea common toboth constraintsboth constraints

50 50 –

40 40 –

30 30 –

20 20 –

10 10 –

0 0 – |1010

|6060

|5050

|2020

|3030

|4040 xx11

xx22


Computing Optimal Valuesxx11 ++ 22xx22 == 4040

44xx11 ++ 33xx22 == 120120

44xx11 ++ 88xx22 == 160160

--44xx11 -- 33xx22 == --120120

55xx22 == 4040

xx22 == 88

xx11 ++ 2(8)2(8) == 4040

xx11 == 2424

4 4 xx11 + 3 + 3 xx2 2 ≤ ≤ 120 lb120 lb

xx11 + 2 + 2 xx2 2 ≤ ≤ 40 hr40 hr

40 40 –

30 30 –

20 20 –

10 10 –

0 0 – |1010

|2020

|3030

|4040

xx11

xx22

ZZ = $50(24) + $50(8) = $1,360= $50(24) + $50(8) = $1,360

248


Extreme Corner Points

xx11 = 224 bowls= 224 bowls

xx2 2 == 8 mugs8 mugs

ZZ = $1,360= $1,360 xx11 = 30 bowls= 30 bowls


ZZ = $1,200= $1,200



ZZ = $1,000= $1,000

AA

BB

CC|2020

|3030

|4040

|1010 xx11

xx22

40 40 –

30 30 –

20 20 –

10 10 –

0 0 –


44xx11 + 3+ 3xx2 2 ≤ ≤ 120 lb120 lb

xx11 + 2+ 2xx2 2 ≤ ≤ 40 hr40 hr

40 40 –

30 30 –

20 20 –

10 10 –

0 0 –

BB

|1010

|2020

|3030

|4040 xx11

xx22

CC

AA

ZZ = 70= 70xx11 + 20+ 20xx22

Optimal point:Optimal point:



ZZ = $2,100= $2,100

Objective Function


Minimization Problem

CHEMICAL CONTRIBUTIONCHEMICAL CONTRIBUTION

BrandBrand Nitrogen (lb/bag)Nitrogen (lb/bag) Phosphate (lb/bag)Phosphate (lb/bag)

GroGro--plusplus 22 44

CropCrop--fastfast 44 33

Minimize Minimize ZZ = $6x= $6x11 + $3x+ $3x22

subject tosubject to

22xx11 ++ 44xx22 ≥≥ 16 lb of nitrogen16 lb of nitrogen

44xx11 ++ 33xx22 ≥≥ 24 lb of phosphate24 lb of phosphate

xx11, , xx22 ≥≥ 00


14 14 –

12 12 –

10 10 –

8 8 –

6 6 –

4 4 –

2 2 –

0 0 – |22

|44

|66

|88

|1010

|1212

|1414 xx11

xx22

A

B

C

Graphical Solution

x1 = 0 bags of Gro-plusx2 = 8 bags of Crop-fastZ = $24

Z = 6x1 + 3x2

Another example:

Max 7T + 5C (profit)

Subject to the constraints:

3T + 4C < 2400 (carpentry hrs)

2T + 1C < 1000 (painting hrs)

C < 450 (max # chairs)

T > 100 (min # tables)

T, C > 0 (nonnegativity)

Graphical Solution

� Graphing an LP model helps provide insight into LP models and their solutions.

� While this can only be done in two dimensions, the same properties apply to all LP models and solutions.

Carpentry

Constraint Line

3T + 4C = 2400

Intercepts

(T = 0, C = 600)

(T = 800, C = 0)

0 800 T

C

600

0

Feasible

< 2400 hrs

Infeasible

> 2400 hrs

Painting

Constraint Line

2T + 1C = 1000

Intercepts

(T = 0, C = 1000)

(T = 500, C = 0)

0 500 800 T

C

1000

600

0

0 100 500 800 T

C

1000

600

450

0

Max Chair Line

C = 450

Min Table Line

T = 100

Feasible

Region

0 100 200 300 400 500 T

C

500

400

300

200

100

0

Objective Function Line

7T + 5C = ProfitOptimal Point

(T = 320, C = 360)

0 100 200 300 400 500 T

C

500

400

300

200

100

0

Additional Constraint

Need at least 75 more chairs than tables

C > T + 75

Or

C – T > 75

T = 320

C = 360

No longer feasible

New optimal point

T = 300, C = 375

LP Characteristics

� Feasible Region: The set of points that satisfies all constraints

� Corner Point Property: An optimal solution must lie at one or more corner points

� Optimal Solution: The corner point with the best objective function value is optimal

Special Situation in LP

1. Redundant Constraints - do not affect the feasible region

Example: x < 10

x < 12

The second constraint is redundant because it is less restrictive.


2. Infeasibility – when no feasible solution exists (there is no feasible region)

Example: x < 10

x > 15

Special Situation in LP3. Alternate Optimal Solutions – when there is

more than one optimal solution

Max 2T + 2CSubject to:

T + C < 10T < 5

C < 6T, C > 0

0 5 10 T

C

10

6

0

All points onRed segment are optimal


4. Unbounded Solutions – when nothing prevents the solution from becoming infinitely large

Max 2T + 2CSubject to:

2T + 3C > 6T, C > 0

0 1 2 3 T

C

2

1

0

linear programming adapted from russell taylor€¦ · max 7t + 5c (profit) subject to the...

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