1. Introduction to Operations Management
6S. Linear Programming
Hansoo Kim (金翰秀)
Dept. of Management Information Systems, YUST

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

3. Today’s 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 (목적을 추구하기 위해 제약된 자원을 어떻게 할당하는가 하는 문제)
Pioneered by George Dantzig in World War II
Developed workable solution called Simplex Method in 1947

5. 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?

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® (
WinQSB

8. Resource Constraints Electronics (Constraint A) Assembly20 40 60 80
100
120 10 30 50 70
Number of Walkmans (X1)
Number of Watch-TVs (X2)
Electronics
(Constraint A)
Assembly
(Constraint B)

9. Feasible Region Electronics (Constraint A) Assembly (Constraint B)20 40 60 80
100
120 10 30 50 70
Number of Walkmans (X1)
Number of Watch-TVs (X2)
Feasible
Region
Electronics
(Constraint A)
Assembly
(Constraint B)

10. Objective Function Electronics (Constraint A) Assembly (Constraint B)20 40 60 80
100
120 10 30 50 70
Number of Walkmans (X1)
Number of Watch-TVs (X2)
7*X1 + 5*X2 = 210
7*X1 + 5*X2 = 410
Electronics
(Constraint A)
Assembly
(Constraint B)
Iso-profit line

11. Extreme Points 20 40 60 80 100 120 10 30 50 70 Number of Walkmans (X1)20 40 60 80
100
120 10 30 50 70
Number of Walkmans (X1)
Number of Watch-TVs (X2)
Iso-profit line
Electronics
(Constraint A)
Assembly
(Constraint B)
Possible Corner Point Solution

12. Optimal Solution 20 40 60 80
100
120 10 30 50 70
Number of Walkmans (X1)
Number of Watch-TVs (X2)
Optimal solution
Iso-profit line
Electronics
(Constraint A)
Assembly
(Constraint B)
Possible Corner Point Solution
X1 = 30
X2 = 40

13. Simplex Algorithm (Using Dictionary)
Maximize 7X1 + 5X2
4X1 + 3X2  electronics constraint
2X1 + 1X2  assembly constraint
X10, and X2 nonnegative constraints
4X1+3X2+X3= 240 => X3 = X1 - 3X2
2X1+1X2+X4= 100 => X4 = X1 - 1X2
Z = 7X1+5X2 => Max Z = 7X1 + 5X2
X1, X2, X3, X4  0
Basic variables
Nonbasic variables

14. Simplex Algorithm (Using Dictionary)
X3 = X1 - 3X2
X4 = X1 - 1X2
Z = X1 + 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 = X1 - 1X2 = > 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 = (50 – X2/2 – X4/2 ) + 5X2
X1 = 50 – X2/2 – X4/2
X3 = 40 – X X4
Z = X2/2 – 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

16. Simplex Algorithm (Using Dictionary)
X2 = 40 – X3 + 2X4
X1 = 50 – ½(40-X3+2X4) – X4/2
Z = /2(40-X3+2X4) -7X4/2
X2 = 40 – X3 + 2X4
X1 = 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

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 + X = 240
2X1 + 1X X4 =100
X10, X20, X30, X40 Z X1 X2 X3 X4
RHS 1 7 5 4 3
240 2
100
Current Solution: X3 = 240, X4 = 100
Z = 0

18. Simplex Algorithm Z X1 X2 X3 X4 RHS 1 7 5 4 3 240 2 1004 3
240 2
100
Step1: Find Entering Variable among non-basic variable
Since Max {7,5}, X1 is Entering Variable
Step2: Find Leaving Variable among basic variable
Since Min {240/4=60,100/2=50}, X4 is Leaving Variable
Step3: Pivoting with X1

19. Pivoting with X1 Z X1 X2 X3 X4 RHS 1 7 5 4 3 240 2 100 Z 1 3/2 -7/24 3
240 2
100 Z 1
3/2
-7/2
-350 X3 1 -2 40 X1 1
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 {40/1=40,50/(1/2)=100}, X3 is Leaving Variable

20. Pivoting with X2 Z X1 X2 X3 X4 RHS 1 3/2 -7/2 -350 -2 40 1/2 50 Z 1
3/2
-7/2
-350 -2 40
1/2 50 Z 1
-3/2
-1/2
-410 X2 1 -2 40 X1 1
-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 Path20 40 60 80
100
120 10 30 50 70
Number of Walkmans (X1)
Number of Watch-TVs (X2)
Feasible
Region
Electronics
(Constraint A)
Assembly
(Constraint B)

22. 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

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 1 3 -2 4 -1 Z X1 X2 X3 X4 RHS 1 4 -3-2 4 -1 Z X1 X2 X3 X4
RHS 1 4 -3 -9 -1 2 10 3

25. Unbounded Case Z X1 X2 X3 X4 RHS 1 4 -3 -9 -1 2 10 3-3 -9 -1 2 10 3
Since all Yik  0, unbounded solution

26. Alternative Solution Case
Min -2X1 - 4X2
s.t. X1 + 2X2 + X3 = 4
-X1 + X X4 = 3
X1, X2, X3, X4  0 X2 X1

27. Alternative Case Z X1 X2 X3 X4 RHS 1 2 4 -1 Z X1 X2 X3 X4 RHS 1 6 -4 3-1 Z X1 X2 X3 X4
RHS 1 6 -4 3 -2 2 -1

28. Alternative Case Z X1 X2 X3 X4 RHS 1 6 -4 3 -2 2 -1 Z X1 X2 X3 X4 RHS-4 3 -2 2 -1 Z X1 X2 X3 X4
RHS 1 -2 -8
1/3
-2/3
2/3
5/3

29. Alternative Case Z X1 X2 X3 X4 RHS 1 -2 -8 1/3 -2/3 2/3 5/3 Z X1 X2 X3-2 -8
1/3
-2/3
2/3
5/3 Z X1 X2 X3 X4
RHS 1 -2 -8 4 3 5

30. Modeling Examples- Product Mix
Wiring
Drilling
Assembly
Inspection
Unit Profit
XJ201
XM897
TR29
BR788
0.5
1.5
1.0 3 1 2 4 $9
$12
$15
$11
Department
Capacity (hr)
Product
Minimum Production 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 9X1+12X2+15X3+11X4
s.t. 0.5X1+1.5X2+1.5X3+ 1X4  1500
3X X2+ 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

32. Production Scheduling Example
Month
Mfg Cost
Selling Price
July
August
September
October
November
December
$ 60
$ 50
$ 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 80Y2+60Y3+70Y4+80Y5+90Y6-
(60X1+60X2+50X3+60X4+70X5)

33. Production Schedule Formulation
Max 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 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
3 g
0 g
8 g
4 g

35. Diet Problem Formulation
Decision Variables: X1 = kg of grain X, X2 = kg of grain Y, X3 = kg of grain Z
Objective Function
Minimize Z = 0.02X1+0.04X X3
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!