Introduction to Operations Management

1
Introduction to Operations Management

• 6S. Linear Programming
• Hansoo Kim (???)
• Dept. of Management Information Systems, YUST

2
OM Overview
X
Class Overview (Ch. 0)
Project Management (Ch. 17)
Operations, Productivity, and
Strategy (Ch. 1, 2)
Process Selection/Facility Layout LP (Ch.
6, 6S)
Strategic Capacity Planning (Ch. 5, 5S)
X
X
X
X
Mgmt of Quality/Six Sigma Quality (Ch. 9, 10)
Supply Chain Management
(Ch 11)
Location Planning and
Analysis (Ch. 8)
JIT Lean Mfg System
(Ch. 15)
Queueing/ Simulation (Ch. 18)
Demand MgmtForecasting (Ch 3)
Inventory Management (Ch.
12)
Aggregated Planning
(Ch. 13)
MRP ERP (Ch 14)
Term Project
3
Todays Outline

• What is LP?
• How to formulate (Model)?
• How to solve?
• Computing tools
• MS-Excel.Solver
• Lindo (or Lingo)
• How to apply?

4
What is LP (Linear Programming) ?

• Mathematical technique (Algorithm)
• Not computer programming
• Allocates limited resources to achieve an
  objective (??? ???? ?? ??? ??? ??? ????? ?? ??)
• Pioneered by George Dantzig in World War II
• Developed workable solution called Simplex
  Method in 1947

5
Example Problem (??)

• 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?

6
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
• X1?0, and X2?0 nonnegative constraints

7
Software for solving LP

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

8
Resource Constraints
9
Feasible Region
10
Objective Function
11
Extreme Points
12
Optimal Solution
X1 30 X2 40
13
Simplex Algorithm (Using Dictionary)

• Maximize 7X1 5X2
• 4X1 3X2 ? 240 electronics constraint
• 2X1 1X2 ? 100 assembly constraint
• X1?0, and X2?0 nonnegative constraints
• 4X13X2X3 240 gt X3 240 - 4X1 - 3X2
• 2X11X2X4 100 gt X4 100 - 2X1 - 1X2
• Z 7X15X2 gt Max Z 7X1 5X2
• X1, X2, X3, X4 ? 0

Basic variables
Nonbasic variables
14
Simplex Algorithm (Using Dictionary)

• X3 240 - 4X1 - 3X2
• X4 100 - 2X1 - 1X2
• Z 7X1 5X2
• (X10, X20, X3240, X4100, Z0)
• Deciding Entering Variable (???? ??),
• How much Z can increased, when X1 or X2 are
  increased? Find X such that Max7, 5 gt X1
• Hence Entering Variable is X1.
• Deciding Leaving Variable(???? ??),
• How much X1 can increased not to violate
  nonnegative constraint? Find X such that
  Min240/4, 100/2 gt X4
• X4 100 - 2X1 - 1X2 gt X1 50 X2/2 X4/2
  (1)
• Calculation,
• Enter (1) to Problem

15
Simplex Algorithm (Using Dictionary)

• X1 50 X2/2 X4/2
• X3 240 4(50 X2/2 X4/2 ) - 3X2
• Z 7(50 X2/2 X4/2 ) 5X2
• -----------------------------------------------
• X1 50 X2/2 X4/2
• X3 40 X2 2X4
• Z 350 3X2/2 7X4/2
• (X150, X20, X340, X40, Z350)
• -----------------------------------------------
• Deciding Entering Variable,
• Find X such that Max3/2, -7/2 gt X2
• Hence Entering Variable is X2.
• Deciding Leaving Variable,
• Find X such that Min50/(1/2), 40/(1) gt X3
• X3 40 X2 2X4 gt X2 40 X3 2X4 (1)
• Calculation,
• Enter (1) to Problem

16
Simplex Algorithm (Using Dictionary)

• X2 40 X3 2X4
• X1 50 ½(40-X32X4) X4/2
• Z 350 3/2(40-X32X4) -7X4/2
• --------------------------------------
• X2 40 X3 2X4
• X1 30 X3/2 - 3X4/2
• Z 410 3X3/2 - X4/2
• (X130, X240, X30, X40, Z410)
• -----------------------------------------
• Deciding Entering Variable,
• Find X such that Max-3/2, -1/2 lt 0 No more
  improvement!
• No possible Entering Variable The current
  solution is optimal!
• Z 410, X1 30, X2 40

17
Simplex Algorithm (Table Format)
Maximize 7X1 5X2 s.t. 4X1 3X2 ? 240 2X1
1X2 ? 100 X1?0, X2?0
Min -7X1 - 5X2 s.t. 4X1 3X2 X3
240 2X1 1X2 X4 100 X1?0,
X2?0, X3?0, X4?0
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 X3 240, X4 100 Z 0
18
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
Step3 Pivoting with X1
19
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 1 100
Z 1 0 3/2 0 -7/2 -350
X3 0 0 1 1 -2 40
X1 0 1 1/2 0 1/2 50
New Solution X1 50, X3 40 Z -350
Step1 Find Entering Variable among non-basic
variable Since Max 3/2, X2 is Entering
Variable Step2 Find Leaving Variable among basic
variable Since Min 40/140,50/(1/2)100, X3 is
Leaving Variable
20
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 50
Z 1 0 0 -3/2 -1/2 -410
X2 0 0 1 1 -2 40
X1 0 1 0 -1/2 3/2 30
New Solution X1 30, X2 40 Z -410
Step1 Find Entering Variable among non-basic
variable But, since all negative (-3/2, -1/2),
this solution is optimal
21
Solution Searching Path
22
Simplex Algorithm

• Step0Tablet Formulation
• Step1Find Entering Variable (Xk) among Nonbasic
  Variables
• Xk such that Max Zj-Cj gt 0
• If there is no candidate, the current is optimal
  solution
• Step2Find Leaving Variable among current Basic
  Variables
• Xr such that Min b-i/yik yik gt 0 Minimum
  Ratio Test
• If yik ? 0, Optimal Solution is unbounded
• Pivoting with Xk, and Repeat Step 1

23
Unbounded Case

• Max X1 3X2
• s.t. X1 - 2X2 ? 4
• -X1 X2 ? 3
• X1, X2 ? 0

24
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
25
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
26
Alternative Solution Case

• Min -2X1 - 4X2
• s.t. X1 2X2 X3 4
• -X1 X2 X4 3
• X1, X2, X3, X4 ? 0

X2
X1
27
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
28
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
29
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
30
Modeling Examples-Product Mix
Product Wiring Drilling Assembly Inspection Unit Profit
XJ201 XM897 TR29 BR788 0.5 1.5 1.5 1.0 3 1 2 3 2 4 1 2 0.5 1.0 0.5 0.5 9 12 15 11
Department Capacity (hr) Product MinimumProduction Level
Wiring Drilling Assembly Inspection 1,500 2,350 2,600 1,200 XJ201 XM897 TR29 BR788 150 100 300 400
31
Product Mix Formulation

• Max 9X112X215X311X4
• s.t. 0.5X11.5X21.5X3 1X4 ? 1500
• 3X1 1X2 2X3 3X4 ? 2350
• 2X1 4X2 1X3 2X4 ? 2600
• 0.5X1 1X20.5X30.5X4 ? 1200
• X1 ? 150
• X2 ? 100
• X3 ? 300
• X4 ? 400
• X1, X2, X3, X4
  ? 0

32
Production Scheduling Example
Month Mfg Cost Selling Price
July August September October November December 60 60 50 60 70 - - 80 60 70 80 90
Production Lead time 1 month Maximum Sales for
each month 300 units Maximum 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 80Y260Y370Y480Y5
90Y6- (60X160X250X360X470X5)
33
Production Schedule Formulation
Max 80Y260Y370Y480Y590Y6-(60X160X250X360X4
70X5)
s.t
I1 X1 I2 I1X2-Y2 I3 I2X3-Y3 I4
I3X4-Y4 I5 I4X5-Y5 I6 I5X6-Y6
Inventory Constraints Inventory at end of this
month Inventory at end of prev. month
Current months production This months Sales
Ii ? 100, for all i I6 0
Yi ? 300, for all i Xi, Yi, Ii ? 0, for all i
34
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 80kg
• How much grains should be bought to minimize the
  total cost?

Vitamin Grain X Grain Y Grain Z
A B C D 3 g/1kg 2 g 1 g 6 g 2 g 3 g 0 g 8 g 4 g 1 g 2 g 4 g
35
Diet Problem Formulation

• Decision VariablesX1 kg of grain X, X2 kg
  of grain Y, X3 kg of grain Z
• Objective Function
• Minimize Z 0.02X10.04X20.025X3
• Constraints
• Vitamin 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

36
HW

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

37
Good Bye!