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Introduction to Operations Management

6S. Linear Programming6S. Linear Programming

Hansoo Kim ( 金翰秀 )Dept. of Management Information Systems,

YUST

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OM Overview

Class Overview(Ch. 0)

Project Management

(Ch. 17)

Strategic Capacity Planning(Ch. 5, 5S)

Operations, Productivity, and Strategy

(Ch. 1, 2)

Mgmt of Quality/Six Sigma Quality

(Ch. 9, 10)

Supply Chain Management

(Ch 11)

Location Planning and Analysis

(Ch. 8)

Demand MgmtForecasting

(Ch 3)

Inventory Management

(Ch. 12)

Aggregated Planning

(Ch. 13)

Queueing/ Simulation

(Ch. 18)

MRP & ERP (Ch 14)

JIT & Lean Mfg System

(Ch. 15)

Term Project

Process Selection/

Facility Layout; LP(Ch. 6, 6S)X X X X X

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Today’s Outline

What is LP? How to formulate (Model)? How to solve?

Computing tools MS-Excel.Solver Lindo (or Lingo)

How to apply?

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What is LP (Linear Programming) ? Mathematical technique (AlgorithmAlgorithm)

Not computer programming Allocates limited resources to achieve

an objective ( 목적을 추구하기 위해 제약된 자원을 어떻게 할당하는가 하는 문제 )

Pioneered by George DantzigGeorge Dantzig in World War II Developed workable solution called

Simplex MethodSimplex Method in 1947

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Example Problem ( 예제 )

Assume: You wish to produce two products

(1) Walkman AM/FM/MP3 Player and (2) Watch-TV Walkman takes 4 hours of electronic work and 2

hours assembly Watch-TV takes 3 hours electronic work and 1 hour

assembly There are 240 hours of electronic work time and

100 hours of assembly time available Profit on a Walkman is $7; profit on a Watch-TV $5

How many Walkman and Watch-TV should be produced to maximize the profits?

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LP Problem Formulation

Let: (Decision Variables) X1 = number of Walkmans

X2 = number of Watch-TVs

Then: Maximize 7X1 + 5X2

4X1 + 3X2 240 electronics constraint

2X1 + 1X2 100 assembly constraint

X10, and X20 nonnegative constraints

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Software for solving LP

MS-Excel Solver Lindo® (www.lindo.com) WinQSB

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Resource Constraints

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

Number of Walkmans (X1)

Num

ber o

f Wat

ch-T

Vs (X

2)

Electronics(Constraint A)Assembly(Constraint B)

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Feasible Region

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

Number of Walkmans (X1)

Num

ber o

f Wat

ch-T

Vs (X

2)

FeasibleFeasibleRegionRegion

Electronics(Constraint A)Assembly(Constraint B)

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Objective Function

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

Number of Walkmans (X1)

Num

ber o

f Wat

ch-T

Vs (X

2)

7*X1 + 5*X

2 = 210

7*X1 + 5*X2 = 410

Electronics(Constraint A)Assembly(Constraint B)

Iso-profit line

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Extreme Points

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

Number of Walkmans (X1)

Num

ber o

f Wat

ch-T

Vs (X

2)

Iso-profit line

Electronics(Constraint A)Assembly(Constraint B)

Possible Corner Point Solution

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Optimal Solution

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

Number of Walkmans (X1)

Num

ber o

f Wat

ch-T

Vs (X

2)

Optimal solution

Iso-profit line

Electronics(Constraint A)Assembly(Constraint B)

Possible Corner Point SolutionX1 = X1 =

3030X2 = X2 = 4040

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Simplex Algorithm (Using Dictionary)

Maximize 7X1 + 5X2

4X1 + 3X2 240 electronics constraint

2X1 + 1X2 100 assembly constraint

X10, and X20 nonnegative constraints

4X1+3X2+X3= 240 => X3 = 240 - 4X1 - 3X2

2X1+1X2+X4= 100 => X4 = 100 - 2X1 - 1X2

Z = 7X1+5X2 => Max Z = 7X1 + 5X2

X1, X2, X3, X4 0

Basic variables Nonbasic variables

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Simplex Algorithm (Using Dictionary)X3 = 240 - 4X1 - 3X2

X4 = 100 - 2X1 - 1X2

Z = 7X1 + 5X2

(X1=0, X2=0, X3=240, X4=100, Z=0)

Deciding Entering Variable ( 투입변수 결정 ), How much Z can increased, when X1 or X2 are increased? Find X such that Max{7, 5} => X1

Hence Entering Variable is X1.Deciding Leaving Variable( 이탈변수 결정 ),

How much X1 can increased not to violate nonnegative constraint? Find X such that Min{240/4, 100/2} => X4

X4 = 100 - 2X1 - 1X2 = > X1 = 50 – X2/2 – X4/2 (1)Calculation,

Enter (1) to Problem

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Simplex Algorithm (Using Dictionary)X1 = 50 – X2/2 – X4/2X3 = 240 – 4(50 – X2/2 – X4/2 ) - 3X2

Z = 7(50 – X2/2 – X4/2 ) + 5X2

-----------------------------------------------X1 = 50 – X2/2 – X4/2X3 = 40 – X2 + 2X4

Z = 350 + 3X2/2 – 7X4/2(X1=50, X2=0, X3=40, X4=0, Z=350)-----------------------------------------------Deciding Entering Variable,

Find X such that Max{3/2, -7/2} => X2

Hence Entering Variable is X2.Deciding Leaving Variable,

Find X such that Min{50/(1/2), 40/(1)} => X3

X3 = 40 – X2 + 2X4 = > X2 = 40 – X3 + 2X4 (1)Calculation,

Enter (1) to Problem

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Simplex Algorithm (Using Dictionary)

X2 = 40 – X3 + 2X4

X1 = 50 – ½(40-X3+2X4) – X4/2

Z = 350 + 3/2(40-X3+2X4) -7X4/2--------------------------------------X2 = 40 – X3 + 2X4

X1 = 30 + X3/2 - 3X4/2

Z = 410 –3X3/2 - X4/2

(X1=30, X2=40, X3=0, X4=0, Z=410)-----------------------------------------

Deciding Entering Variable, Find X such that Max{-3/2, -1/2} < 0 No more improvement!No possible Entering Variable ** The current solution is optimal!

Z* = 410, X1* = 30, X2

* = 40

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Simplex Algorithm (Table Format)Maximize 7X1 + 5X2

s.t. 4X1 + 3X2 240 2X1 + 1X2 100 X10, X20

Min -7X1 - 5X2

s.t. 4X1 + 3X2 + X3 = 240 2X1 + 1X2 + X4 =100 X10, X20, X30, X40

Z X1 X2 X3 X4 RHS

Z 1 7 5 0 0 0

X3 0 4 3 1 0 240

X4 0 2 1 0 1 100

Current Solution: XCurrent Solution: X33 = 240, X = 240, X44 = 100 = 100 Z = 0Z = 0

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Simplex Algorithm

Z X1 X2 X3 X4 RHS

Z 1 7 5 0 0 0

X3 0 4 3 1 0 240

X4 0 2 1 0 1 100

Step1: Find Entering Variable among non-basic variableSince Max {7,5}, X1 is Entering Variable

Step2: Find Leaving Variable among basic variableSince Min {240/4=60,100/2=50}, X4 is Leaving Variable

Step3: Pivoting with X1

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Pivoting with X1

Z X1 X2 X3 X4 RHS

Z 1 7 5 0 0 0

X3 0 4 3 1 0 240

X4 0 2 1 0 1 100X1 0 1 1/2 0 1/2 50

X3 0 0 1 1 -2 40

Z 1 0 3/2 0 -7/2 -350

New Solution: XNew Solution: X11 = 50, X = 50, X33 = 40 = 40 Z = -350Z = -350

Step1: Find Entering Variable among non-basic variableSince Max {3/2}, X2 is Entering Variable

Step2: Find Leaving Variable among basic variableSince Min {40/1=40,50/(1/2)=100}, X3 is Leaving

Variable

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Pivoting with X2

Z X1 X2 X3 X4 RHS

Z 1 0 3/2 0 -7/2 -350

X3 0 0 1 1 -2 40

X1 0 1 1/2 0 1/2 50X1 0 1 0 -1/2 3/2 30

X2 0 0 1 1 -2 40

Z 1 0 0 -3/2 -1/2 -410

New Solution: XNew Solution: X11 = 30, X = 30, X22 = 40 = 40 Z = -410Z = -410

Step1: Find Entering Variable among non-basic variableBut, since all negative (-3/2, -1/2), this solution is optimal

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Solution Searching Path

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

Number of Walkmans (X1)

Num

ber o

f Wat

ch-T

Vs (X

2)

FeasibleFeasibleRegionRegion

Electronics(Constraint A)Assembly(Constraint B)

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Simplex Algorithm Step0:Tablet Formulation Step1:Find Entering Variable (Xk) among Nonbasic

Variables Xk such that Max Zj-Cj > 0

If there is no candidate, the current is optimal solution Step2:Find Leaving Variable among current Basic

Variables Xr such that Min {b-

i/yik: yik > 0} – Minimum Ratio Test

If yik 0, Optimal Solution is unbounded

Pivoting with Xk, and Repeat Step 1

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Unbounded Case

Max X1 + 3X2

s.t. X1 - 2X2 4

-X1 + X2 3

X1, X2 0

X2

X1

X1 - 2X2 = 4

-X1 + X2 = 3

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Unbounded Case

Z X1 X2 X3 X4 RHS

Z 1 1 3 0 0 0

X3 0 1 -2 1 0 4

X4 0 -1 1 0 1 3

Z X1 X2 X3 X4 RHS

Z 1 4 0 0 -3 -9

X3 0 -1 0 1 2 10

X2 0 -1 1 0 1 3

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Unbounded Case

Z X1 X2 X3 X4 RHS

Z 1 4 0 0 -3 -9

X3 0 -1 0 1 2 10

X2 0 -1 1 0 1 3

Since all Yik 0, unbounded solution

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Alternative Solution Case

Min -2X1 - 4X2

s.t. X1 + 2X2 + X3 = 4

-X1 + X2 + X4 = 3

X1, X2, X3, X4 0X2

X1

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Alternative Case

Z X1 X2 X3 X4 RHS

Z 1 2 4 0 0 0

X3 0 1 2 1 0 4

X4 0 -1 1 0 1 1

Z X1 X2 X3 X4 RHS

Z 1 6 0 0 -4 -4

X3 0 3 0 1 -2 2

X2 0 -1 1 0 1 1

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Alternative Case

Z X1 X2 X3 X4 RHS

Z 1 6 0 0 -4 -4

X3 0 3 0 1 -2 2

X2 0 -1 1 0 1 1

Z X1 X2 X3 X4 RHS

Z 1 0 0 -2 0 -8

X1 0 1 0 1/3 -2/3 2/3

X2 0 0 1 1/3 1/3 5/3

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Alternative Case

Z X1 X2 X3 X4 RHS

Z 1 0 0 -2 0 -8

X1 0 1 0 1/3 -2/3 2/3

X2 0 0 1 1/3 1/3 5/3

Z X1 X2 X3 X4 RHS

Z 1 0 0 -2 0 -8

X1 0 1 0 1 0 4

X4 0 0 3 1 1 5

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Modeling Examples-Product Mix

Product Wiring Drilling AssemblyInspection

Unit Profit

XJ201XM897TR29BR788

0.51.51.51.0

3123

2412

0.51.00.50.5

$9$12$15$11

Department Capacity (hr) ProductMinimum

Production Level

WiringDrilling

AssemblyInspection

1,5002,3502,6001,200

XJ201XM897TR29BR788

150100300400

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Product Mix Formulation

Max 9X1+12X2+15X3+11X4

s.t. 0.5X1+1.5X2+1.5X3+ 1X4 1500 3X1 + 1X2+ 2X3+ 3X4 2350

2X1+ 4X2+ 1X3+ 2X4 2600 0.5X1+ 1X2+0.5X3+0.5X4 1200 X1 150 X2 100 X3 300 X4 400 X1, X2, X3, X4 0

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Production Scheduling ExampleMonth Mfg Cost Selling Price

JulyAugustSeptemberOctoberNovemberDecember

$ 60$ 60$ 50$ 60$ 70-

-$ 80$ 60$ 70$ 80$ 90

Production Lead time: 1 monthMaximum Sales for each month: 300 unitsMaximum Capacity of Warehouse: 100 units

Variables:X1, X2, X3, X4, X5, X6: number of units manufactured from July to Dec.Y1, Y2, Y3, Y4, Y5, Y6: number of units sold from July to Dec.

Objective Function:Max 80Y2+60Y3+70Y4+80Y5+90Y6-

(60X1+60X2+50X3+60X4+70X5)

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Production Schedule FormulationMax 80Y2+60Y3+70Y4+80Y5+90Y6-(60X1+60X2+50X3+60X4+70X5)s.t

I1 = X1

I2 = I1+X2-Y2

I3 = I2+X3-Y3 I4 = I3+X4-Y4

I5 = I4+X5-Y5 I6 = I5+X6-Y6

Inventory Constraints:

Inventory at end of this month = Inventory at end of prev. month +Current month’s production –This month’s Sales

Ii 100, for all iI6 = 0

Yi 300, for all i

Xi, Yi, Ii 0, for all i

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Diet Problem There are three grains for Caw; X, Y, and Z Four vitamins A, B, C, D in grain 1kg

Unit costs for grains; $0.02 (X), $0.04 (Y), $0.025 (Z)Minimum requirements per a caw:

over 64g (vitamin A), over 80g (vitamin B), over 16g (vitamin C), over 128g (vitamin D)

Grain Z can not buy no more than 80kgHow much grains should be bought to minimize the total cost?

Vitamin Grain X Grain Y Grain Z

ABCD

3 g/1kg2 g1 g6 g

2 g3 g0 g8 g

4 g1 g2 g4 g

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Diet Problem Formulation

Decision Variables:X1 = kg of grain X, X2 = kg of grain Y, X3 = kg of grain Z

Objective FunctionMinimize Z = 0.02X1+0.04X2+0.025X3

ConstraintsVitamin A constrains: 3X1 + 2X2 + 4X3 64

Vitamin B constrains: 2X1 + 3X2 + 1X3 80

Vitamin C constrains: 1X1 + 0X2 + 2X3 16

Vitamin D constrains: 6X1 + 8X2 + 4X3 128

Grain Z constraint: X3 80

Nonnegative Constraint: X1, X2, X3 0

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HW

Review All examples and solved problems by hands and with MS-Excel

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Good Bye!