Linear Programming Fundamentals

10/16/2010

B Linear Programming
Why Use Linear Programming?
Outline

Requirements of a Linear
PowerPoint presentation to accompany
p p y Programming Problem
g g
Heizer and Render
Operations Management, 10e Formulating Linear Programming
Principles of Operations Management, 8e
Problems
PowerPoint slides by Jeff Heyl
Shader Electronics Example

© 2011 Pearson Education, Inc. publishing as Prentice Hall B-1 © 2011 Pearson Education, Inc. publishing as Prentice Hall B-2

Outline – Continued Outline – Continued

Graphical Solution to a Linear Sensitivity Analysis
Programming Problem Sensitivity Report
Graphical Representation of Changes in the Resources of the
g
Constraints
C t i t Right-Hand-Side Values
Iso-Profit Line Solution Method Changes in the Objective Function
Coefficient
Corner-Point Solution Method Solving Minimization Problems


Outline – Continued Learning Objectives
When you complete this module you
Linear Programming Applications should be able to:
Production-Mix Example
1. Formulate linear programming
p
Diet Problem Example models,
models including an objective
Labor Scheduling Example function and constraints
The Simplex Method of LP 2. Graphically solve an LP problem with
the iso-profit line method
3. Graphically solve an LP problem with
the corner-point method


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10/16/2010

Learning Objectives Why Use Linear Programming?
When you complete this module you
should be able to: A mathematical technique to
help plan and make decisions
4. Interpret sensitivity analysis and relative to the trade-offs
shadow prices necessary to allocate resources
5. Construct and solve a minimization Will find the minimum or
problem maximum value of the objective
6. Formulate production-mix, diet, and
labor scheduling problems Guarantees the optimal solution
to the model formulated


LP Applications LP Applications
1. Scheduling school buses to minimize 4. Selecting the product mix in a factory
total distance traveled to make best use of machine- and
labor-hours available while maximizing
2. Allocating police patrol units to high
the firm’s profit
p
crime areas in order to minimize
response time to 911 calls 5. Picking blends of raw materials in feed
mills to produce finished feed
3. Scheduling tellers at banks so that
combinations at minimum costs
needs are met during each hour of the
day while minimizing the total cost of 6. Determining the distribution system
labor that will minimize total shipping cost

© 2011 Pearson Education, Inc. publishing as Prentice Hall B-9 © 2011 Pearson Education, Inc. publishing as Prentice Hall B - 10

LP Applications Requirements of an
LP Problem
7. Developing a production schedule that
will satisfy future demands for a firm’s 1. LP problems seek to maximize or
product and at the same time minimize minimize some quantity (usually
total production and inventory costs p
profit or cost) expressed as an
) p
8. Allocating space for a tenant mix in a objective function
new shopping mall
so as to maximize 2. The presence of restrictions, or
revenues to the constraints, limits the degree to
leasing company which we can pursue our
objective

© 2011 Pearson Education, Inc. publishing as Prentice Hall B - 11 © 2011 Pearson Education, Inc. publishing as Prentice Hall B - 12

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10/16/2010

Requirements of an Formulating LP Problems
LP Problem
The product-mix problem at Shader Electronics
3. There must be alternative courses Two products
of action to choose from
1. Shader x-pod, a portable music
4. The bj ti
4 Th objective and constraints in
d t i t i player
linear programming problems 2. Shader BlueBerry, an internet-
must be expressed in terms of connected color telephone
linear equations or inequalities
Determine the mix of products that will
produce the maximum profit


Formulating LP Problems Formulating LP Problems
Hours Required Objective Function:
to Produce 1 Unit Maximize Profit = $7X1 + $5X2
x-pods BlueBerrys Available Hours
Department (X1) (X2) This Week There are three types of constraints
Electronic 4 3 240 Upper limits where the amount used is ≤
Assembly 2 1 100 the amount of a resource
Profit per unit $7 $5 Lower limits where the amount used is ≥
Table B.1 the amount of the resource
Decision Variables:
X1 = number of x-pods to be produced Equalities where the amount used is =
X2 = number of BlueBerrys to be produced the amount of the resource

Formulating LP Problems Graphical Solution
First Constraint: Can be used when there are two
Electronic Electronic decision variables
time used is ≤ time available 1. Plot the constraint equations at their
limits by converting each equation
y g q
4X1 + 3X2 ≤ 240 (hours of electronic time) to an equality
Second Constraint: 2. Identify the feasible solution space
Assembly Assembly 3. Create an iso-profit line based on
time used is ≤ time available the objective function

2X1 + 1X2 ≤ 100 (hours of assembly time) 4. Move this line outwards until the
optimal point is identified

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Graphical Solution Graphical Solution
X2 X2
Iso-Profit Line Solution Method
100 – 100 –
– Choose a–possible value for the
objective –
80 function Assembly (Constraint B)
eBerrys

eBerrys
80 – Assembly (Constraint B)
– –
$210 = 7X1 + 5X2
Number of Blue

Number of Blue
60 – 60 –
– –
40 – Solve for – axis intercepts of the function
40 the
– Electronics (Constraint A) and plot the line
– Electronics (Constraint A)
20 – Feasible 20 – Feasible
–
region
– X2 =
region 42 X1 = 30
|– | | | | | | | | | | X1 |– | | | | | | | | | | X1
0 20 40 60 80 100 0 20 40 60 80 100
Figure B.3 Number of x-pods Figure B.3 Number of x-pods

Graphical Solution Graphical Solution
X2 X2

100 – 100 –
– – $350 = $7X1 + $5X2
eBerrys

eBerrys
80 – 80 –
– – $280 = $7X1 + $5X2
Number of Blue

Number of Blue

60 – 60 –
$210 = $7X1 + $5X2 $210 = $7X1 + $5X2
– –
(0, 42)
40 – 40 –
– –
20 – (30, 0) 20 –
$420 = $7X1 + $5X2
– –
|– | | | | | | | | | | X1 |– | | | | | | | | | | X1
0 20 40 60 80 100 0 20 40 60 80 100

Graphical Solution Corner-
Corner-Point Method
X2 X2

100 – 100 –
– Maximum profit line 2 –
eBerrys

eBerrys

80 – 80 –
– –
Number of Blue

Number of Blue

60 – 60 –
Optimal solution point
– –
(X1 = 30, X2 = 40) 3
40 – 40 –
– –
20 –
$410 = $7X1 + $5X2 20 –
– –
|– | | | | | | | | | | X1 |– | | | | | | | | | | X1
1
0 20 40 60 80 100 0 20 40 60 80 100
4

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10/16/2010

Corner-
Corner-Point Method Corner-
Corner-Point Method
The optimal value will always be at a The optimal value will always be at a
corner point corner point intersection of two constraints
Solve for the

Find the objective function value at each 4X1 + 3X2 ≤ 240 (electronics time)
Find the objective function value at each
corner point and choose the one with the corner 2X1 + 1X2 ≤ 100 (assemblyone with the
point and choose the time)
highest
hi h t profit
fit highest
hi h t profit
fit
4X1 + 3X2 = 240 4X1 + 3(40) = 240
- 4X1 - 2X2 = -200 4X + 120 = 240
Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 1 : (X1 = 0, X2 = 0) Profit 1
$7(0) + $5(0) = $0
+ 1X2 = 40 X1 = 30
Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400
Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350


Corner-
Corner-Point Method Sensitivity Analysis
The optimal value will always be at a
corner point How sensitive the results are to
parameter changes
Find the objective function value at each
corner point and choose the one with the Change in the value of coefficients
highest
hi h t profit
fit Change in a right-hand-side value of
a constraint
Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0
Trial-and-error approach
Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400
Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Analytic postoptimality method
Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410


Sensitivity Report Changes in Resources

The right-hand-side values of
constraint equations may change
as resource availability changes
The shadow price of a constraint is
the change in the value of the
objective function resulting from a
one-unit change in the right-hand-
side value of the constraint

© 2011 Pearson Education, Inc. publishing as Prentice Hall Program B.1 B - 29 © 2011 Pearson Education, Inc. publishing as Prentice Hall B - 30

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10/16/2010

Changes in Resources Sensitivity Analysis
X2
Shadow prices are often explained –
Changed assembly constraint from
as answering the question “How 100 –
2X1 + 1X2 = 100
–
much would you pay for one 80 – 2
to 2X1 + 1X2 = 110

additional unit of a resource?” –
Corner point 3 i still optimal, but
C i t is till ti l b t
60 –
Shadow prices are only valid over a –
values at this point are now X1 = 45,
X2 = 20, with a profit = $415
particular range of changes in right- 40 –

hand-side values –
Electronics constraint
20 –
3 is unchanged
Sensitivity reports provide the –
|– | | | | | | | | | |
upper and lower limits of this range 1
0 20 40 4 60 80 100 X1 Figure B.8 (a)


Sensitivity Analysis Changes in the
X2
Objective Function
–
Changed assembly constraint from
100 –
– 2X1 + 1X2 = 100 A change in the coefficients in the
80 –
to 2X1 + 1X2 = 90 objective function may cause a
2 –
Corner point 3 i still optimal, but
C i t is till ti l b t
different corner point to become the
60 –
values at this point are now X1 = 15, optimal solution
– 3
X2 = 60, with a profit = $405
40 – The sensitivity report shows how
20 –
–
Electronics constraint much objective function
–
is unchanged coefficients may change without
1 |– | | | | | | | | | | changing the optimal solution point
0 20 40 4 60 80 100 X1 Figure B.8 (b)


Solving Minimization Minimization Example
Problems X1 = number of tons of black-and-white picture
chemical produced
Formulated and solved in much the
X2 = number of tons of color picture chemical
same way as maximization produced
problems
Minimize total cost = 2,500X1 + 3,000X2
In the graphical approach an iso-
Subject to:
cost line is used
X1 ≥ 30 tons of black-and-white chemical
The objective is to move the iso- X2 ≥ 20 tons of color chemical
cost line inwards until it reaches X1 + X2 ≥ 60 tons total
the lowest cost corner point X1, X2 ≥ $0 nonnegativity requirements


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10/16/2010

Minimization Example Minimization Example
Table B.9
X2
60 – X1 + X2 = 60
Total cost at a = 2,500X1 + 3,000X2
50 – = 2,500 (40) + 3,000(20)
Feasible = $160,000
40 – g
region

30 –
Total cost at b = 2,500X1 + 3,000X2
b = 2,500 (30) + 3,000(30)
20 – = $165,000
a
10 – X1 = 30
X2 = 20
Lowest total cost is at point a
|
– | | | | | |
X1
0 10 20 30 40 50 60


Production-
Production-Mix Example X1 = number of units of XJ201 produced
Department X2 = number of units of XM897 produced
Product Wiring Drilling Assembly Inspection Unit Profit X3 = number of units of TR29 produced
X4 = number of units of BR788 produced
XJ201 .5 3 2 .5 $ 9
XM897 1.5 1 4 1.0 $12 Maximize profit = 9X1 + 12X2 + 15X3 + 11X4
TR29 1.5
15 2 1 .5
5 $15
BR788 1.0 3 2 .5 $11 subject to .5X1 + 1.5X2 + 1.5X3 + 1X4 ≤ 1,500 hours of wiring
3X1 + 1X2 + 2X3 + 3X4 ≤ 2,350 hours of drilling
Capacity Minimum 2X1 + 4X2 + 1X3 + 2X4 ≤ 2,600 hours of assembly
Department (in hours) Product Production Level .5X1 + 1X2 + .5X3 + .5X4 ≤ 1,200 hours of inspection
Wiring 1,500 XJ201 150 X1 ≥ 150 units of XJ201
Drilling 2,350 XM897 100 X2 ≥ 100 units of XM897
Assembly 2,600 TR29 300 X3 ≥ 300 units of TR29
Inspection 1,200 BR788 400 X4 ≥ 400 units of BR788


Diet Problem Example X1 = number of pounds of stock X purchased per cow each month
X2 = number of pounds of stock Y purchased per cow each month
Feed X3 = number of pounds of stock Z purchased per cow each month

Product Stock X Stock Y Stock Z Minimize cost = .02X1 + .04X2 + .025X3
A 3 oz 2 oz 4 oz
B 2 oz 3 oz 1 oz Ingredient A requirement: 3X1 + 2X2 + 4X3 ≥ 64
C 1 oz 0 oz 2 oz Ingredient B requirement: 2X1 + 3X2 + 1X3 ≥ 80
D 6 oz 8 oz 4 oz Ingredient C requirement: 1X1 + 0X2 + 2X3 ≥ 16
Ingredient D requirement: 6X1 + 8X2 + 4X3 ≥ 128
Stock Z limitation: X3 ≤ 80
X1, X2, X3 ≥0

Cheapest solution is to purchase 40 pounds of grain X
at a cost of $0.80 per cow

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10/16/2010

Labor Scheduling Example Minimize total daily
= $75F + $24(P1 + P2 + P3 + P4 + P5)
manpower cost
Time Number of Time Number of
Period Tellers Required Period Tellers Required F + P1 ≥ 10 (9 AM - 10 AM needs)
9 AM - 10 AM 10 1 PM - 2 PM 18 F + P1 + P2 ≥ 12 (10 AM - 11 AM needs)
10 AM - 11 AM 12 2 PM - 3 PM 17 1/2 F + P1 + P2 + P3 ≥ 14 (11 AM - 11 AM needs)
11 AM - Noon 14 3 PM - 4 PM 15 1/2 F + P1 + P2 + P3 + P4 ≥ 16 (
(noon - 1 PM needs))
Noon - 1 PM 16 4 PM - 5 PM 10 F + P2 + P3 + P4 + P5 ≥ 18 (1 PM - 2 PM needs)
F + P3 + P4 + P5 ≥ 17 (2 PM - 3 PM needs)
F = Full-time tellers F + P4 + P5 ≥ 15 (3 PM - 7 PM needs)
P1 = Part-time tellers starting at 9 AM (leaving at 1 PM) F + P5 ≥ 10 (4 PM - 5 PM needs)
P2 = Part-time tellers starting at 10 AM (leaving at 2 PM) F ≤ 12
P3 = Part-time tellers starting at 11 AM (leaving at 3 PM)
P4 = Part-time tellers starting at noon (leaving at 4 PM) 4(P1 + P2 + P3 + P4 + P5) ≤ .50(10 + 12 + 14 + 16 + 18 + 17 + 15 + 10)
P5 = Part-time tellers starting at 1 PM (leaving at 5 PM)


Minimize total daily
= $75F + $24(P1 + P2 + P3 + P4 + P5) There are two alternate optimal solutions to this
manpower cost
problem but both will cost $1,086 per day
F + P1 ≥ 10 (9 AM - 10 AM needs)
F + P1 + P2 ≥ 12 (10 AM - 11 AM needs)
1/2 F + P1 + P2 + P3 ≥ 14 (11 AM - 11 AM needs)
First Second
1/2 F + P1 + P2 + P3 + P4 ≥ 16 (
(noon - 1 PM needs)) Solution Solution
F + P2 + P3 + P4 + P5 ≥ 18 (1 PM - 2 PM needs) F = 10 F = 10
F + P3 + P4 + P5 ≥ 17 (2 PM - 3 PM needs) P1 = 0 P1 = 6
F + P4 + P5 ≥ 15 (3 PM - 7 PM needs)
P2 = 7 P2 = 1
F + P5 ≥ 10 (4 PM - 5 PM needs)
F ≤ 12 P3 = 2 P3 = 2
P4 = 2 P4 = 2
4(P1 + P2 + P3 + P4 + P5) ≤ .50(112) P5 = 3 P5 = 3
F, P1, P2, P3, P4, P5 ≥ 0


The Simplex Method
Real world problems are too
complex to be solved using the
graphical method
The simplex method is an algorithm
for solving more complex problems
All rights reserved. No part of this publication may be reproduced, stored in a retrieval

Developed by George Dantzig in the system, or transmitted, in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the publisher.
late 1940s Printed in the United States of America.

Most computer-based LP packages
use the simplex method

8

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