one problem with linear programming (lp) model with excel solver, one problem with average method, one revenue and cost question

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Operations Management - 5th Edition

Chapter 13 Supplement

Roberta Russell & Bernard W. Taylor, III

Linear Programming

Supplement 13-*

Lecture Outline

Model Formulation
Graphical Solution Method
Linear Programming Model
Solution
Solving Linear Programming Problems with Excel
Sensitivity Analysis

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A model consisting of linear relationships

representing 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

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Types of LP

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Types of LP (cont.)

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Types of LP (cont.)

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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

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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 variables

bi = constraint levels

cj = objective function coefficients

aij = constraint coefficients

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LP Model: Example

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 variables

x1 = number of bowls to produce

x2 = number of mugs to produce

RESOURCE REQUIREMENTS

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LP Formulation: Example

Maximize Z = $40 x1 + 50 x2

Subject to

x1 + 2x2 40 hr (labor constraint)

4x1 + 3x2 120 lb (clay constraint)

x1 , x2 0

Solution is x1 = 24 bowls x2 = 8 mugs

Revenue = $1,360

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Graphical Solution Method

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

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Graphical Solution: Example

4 x1 + 3 x2 120 lb

x1 + 2 x2 40 hr

Area common to

both constraints

50 –

40 –

30 –

20 –

10 –

0 –

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Computing Optimal Values

x1 + 2x2 = 40

4x1 + 3x2 = 120

4x1 + 8x2 = 160

-4x1 - 3x2 = -120

5x2 = 40

x2 = 8

x1 + 2(8) = 40

x1 = 24

4 x1 + 3 x2 120 lb

x1 + 2 x2 40 hr

40 –

30 –

20 –

10 –

0 –

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

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Extreme Corner Points

x1 = 224 bowls

x2 =8 mugs

Z = $1,360

x1 = 30 bowls

x2 =0 mugs

Z = $1,200

x1 = 0 bowls

x2 =20 mugs

Z = $1,000

40 –

30 –

20 –

10 –

0 –

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4x1 + 3x2 120 lb

x1 + 2x2 40 hr

40 –

30 –

20 –

10 –

0 –

Z = 70x1 + 20x2

Objective Function

Optimal point:

x1 = 30 bowls

x2 =0 mugs

Z = $2,100

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Minimization Problem

Minimize Z = $6x1 + $3x2

subject to

2x1 + 4x2  16 lb of nitrogen

4x1 + 3x2  24 lb of phosphate

x1, x2  0

CHEMICAL CONTRIBUTION

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

Gro-plus 2 4

Crop-fast 4 3

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14 –

12 –

10 –

8 –

6 –

4 –

2 –

0 –

Graphical Solution

x1 = 0 bags of Gro-plus

x2 = 8 bags of Crop-fast

Z = $24

Z = 6x1 + 3x2

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Simplex Method

A mathematical procedure for solving linear programming problems according to a set of steps
Slack variables added to ≤ constraints to represent unused resources

x1 + 2x2 + s1 =40 hours of labor

4x1 + 3x2 + s2 =120 lb of clay

Surplus variables subtracted from ≥ constraints to represent excess above resource requirement. For example

2x1 + 4x2 ≥ 16 is transformed into

2x1 + 4x2 - s1 = 16

Slack/surplus variables have a 0 coefficient in the objective function

Z = $40x1 + $50x2 + 0s1 + 0s2

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Solution Points with

Slack Variables

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Solution Points with

Surplus Variables

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Solving LP Problems with Excel

Click on “Tools”

to invoke “Solver.”

Objective function

Decision variables – bowls

(x1)=B10; mugs (x2)=B11

=C6*B10+D6*B11

=C7*B10+D7*B11

=E6-F6

=E7-F7

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Solving LP Problems with Excel (cont.)

After all parameters and constraints have been input, click on “Solve.”

Objective function

Decision variables

C6*B10+D6*B11≤40

C7*B10+D7*B11≤120

Click on “Add” to insert constraints

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Solving LP Problems with Excel (cont.)

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Sensitivity Analysis

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Sensitivity Range for Labor Hours

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Sensitivity Range for Bowls

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Copyright 2006 John Wiley & Sons, Inc.
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