Supplement 14- 1 Robert S. Russell, Bernard W. Taylor III, Ignacio Castillo,  - PowerPoint Presentation

Slide1

Supplement 14-1

Robert S. Russell, Bernard W. Taylor III, Ignacio Castillo, 

Navneet

VidyarthiCHAPTER 14 SUPPLEMENT Linear Programming

OPERATIONS MANAGEMENT: Creating Value Along the Supply Chain, Canadian EditionSlide2

Lecture Outline

Model FormulationGraphical Solution MethodLinear Programming Model SolutionSolving Linear Programming Problems with ExcelSensitivity AnalysisSupplement 14-2Slide3

Linear Programming (LP)

A model consisting of linear relationships representing a firm’s objective and resource constraintsA mathematical modeling technique which determines a level of operational activity in order to achieve an objective, subject to restrictions called constraintsSupplement 14-

3Slide4

Types of LP

Supplement 14-4Slide5

Types of LP

Supplement 14-5Slide6

Types of LP

Supplement 14-6Slide7

LP Model Formulation

Decision variablessymbols representing levels of activity of an operationObjective functionlinear relationship for

the objective of an operationmost frequent business objective is to maximize profit

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

linear relationship representing a restriction on decision makingSupplement 14-7Slide8

LP Model Formulation

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

cnxnsubject to:

a11x1

+ a12x2 + ... + a1nxn (≤, =, ≥) b1 a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2

: an1x1 + an2x2 + ... + annxn (≤, =, ≥) b

n xj = decision variables

b

i

= constraint levels

c

j

= objective function coefficients

a

ij

= constraint coefficients

Supplement 14-

8

ConstraintsSlide9

Highlands Craft Store

Supplement 14-9

Labor Clay Revenue

Product (hr/unit) (lb/unit) ($/unit) Bowl 1 4 40 Mug 2 3 50 There are 40 hours of labor and 120 pounds of clay available each day

Decision variablesx1 = number of bowls to producex2 = number of mugs to produce

Resource

RequirementsSlide10

Highlands Craft Store

Supplement 14-10

Maximize Z = $40 x1 + 50

x2

Subject to x1 + 2x2 40 hr (labor constraint) 4x1 + 3x

2 120 lb (clay constraint) x1 , x2 0

Solution is x

1

= 24 bowls

x

2

= 8 mugs

Revenue = $1,360Slide11

Graphical Solution Method

Plot model constraint on a set of coordinates in a planeIdentify the feasible solution space on the graph where all constraints are satisfied simultaneouslyPlot objective function to find the point on boundary of this space that maximizes (or minimizes) value of objective function

Supplement 14-11Slide12

Graphical Solution Method

Supplement 14-12

4

x

1 + 3 x2 120 lb

x1 + 2 x2 40 hr

Area common toboth constraints

50 –

40 –

30 –

20 –

10 –

0 –

|

10

|

60

|

50

|

20

|

30

|

40

x

1

x

2

Objective functionSlide13

Computing Optimal Values

Supplement 14-13

40 –30 –

20 –10 –

0 – x1 + 2x2 = 40

4x1 + 3x2 = 120 4x1 + 8x

2 = 160 -4x1

- 3

x

2

= -120

5

x

2

= 40

x

2

= 8

x

1

+ 2(8) = 40

x

1

= 24

4

x

1

+ 3

x

2

=

120 lb

x

1

+ 2

x

2

=

40 hr

|

10

|

20

|

30

|

40

x

1

x

2

Z

= $40(24) + $50(8) = $1,360

24

8Slide14

Extreme Corner Points

Supplement 14-14

x

1 = 224 bowls

x2 =8 mugsZ = $1,360x1 = 30 bowlsx2 =

0 mugsZ = $1,200

x1 = 0 bowlsx2

=



20 mugs

Z

= $1,000

A

B

C

|

20

|

30

|

40

|

10

x

1

x

2

40 –

30 –

20 –

10 –

0 –Slide15

Objective Function

Supplement 14-1540 –

30 –20 –10 –0 –

4

x1 + 3x2 =120 lbx

1 + 2x2 =40 hr

B

|

10

|

20

|

30

|

40

x

1

x

2

C

A

Z

= 70

x

1

+ 20

x

2

Optimal point:

x

1

= 30 bowls

x

2

=



0 mugs

Z

= $2,100Slide16

Minimization Problem

Supplement 14-16

CHEMICAL CONTRIBUTION

Brand Nitrogen (lb/bag) Phosphate (lb/bag)

Gro-plus 2 4Crop-fast 4 3

Minimize Z = $6x

1 + $3x2

subject to

2

x

1

+ 4

x

2

 16 lb of nitrogen

4

x

1

+ 3

x

2

 24 lb of phosphate

x

1

,

x

2

 0Slide17

Graphical Solution

Supplement 14-17

14 –

12 –

10 –8 –

6 –4 –2 –0 –|2

|4

|

6

|

8

|

10

|

12

|

14

x

1

x

2

A

B

C

x

1

= 0 bags of

Gro

-plus

x

2

= 8 bags of Crop-fast

Z

= $24

Z = 6

x

1

+ 3

x

2Slide18

Simplex Method

Mathematical procedure for solving LP problemsFollow a set of steps to reach optimal solutionSlack variables added to ≤ constraints to represent unused

resources x1 + 2x2 + s1 = 40

hours of labor4x1 + 3x2 + s

2 = 120 lb of claySurplus variables subtracted from ≥ constraints to represent excess above resource requirement. 2x1 + 4x2 ≥ 16 is transformed into2x1 + 4x2 - s1 = 16

Slack/surplus variables have a 0 coefficient in the objective functionZ = $40x1 + $50x2 + 0s1 + 0s2Supplement 14-18Slide19

Solution Points With Slack Variables

Supplement 14-19Slide20

Solution Points With Surplus Variables

Supplement 14-20Slide21

Solving LP Problems with Excel

Supplement 14-21

Objective function

=C6*B10+D6*B11

=E6-F6=E7-F7=C7*B10+D7*B11

Decision variables bowls (X1) = B10mugs (x2

) = B11

Click on “Data”

to invoke “Solver”Slide22

Solving LP Problems with Excel

Supplement 14-22

After all parameters and constraints

have been input, click on “Solve”

Objective functionDecision variablesC6*B10+D6*B11≤40andC7*B10+D7*B11≤120

Click on “Add” toinsert constraints

Click on “Options” to add

non-negativity and linear conditionsSlide23

LP Solution

Supplement 14-23Slide24

Sensitivity Analysis

Supplement 14-24

Sensitivity range for labor;

30 to 80 lbs.

Sensitivity range for clay;60 to 160lbs.Shadow prices – marginalvalues – for labor and clay.Slide25

Sensitivity Range for Labor Hours

Supplement 14-25Slide26

COPYRIGHT

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