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

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

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

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

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?

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

Software for solving LP MS-Excel Solver Lindo® (www.lindo.com) WinQSB

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

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)

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

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

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

Simplex Algorithm (Using Dictionary) Maximize 7X1 + 5X2 4X1 + 3X2  240 electronics constraint 2X1 + 1X2  100 assembly constraint X10, and X20 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

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

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 (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

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

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 1 7 5 4 3 240 2 100 Current Solution: X3 = 240, X4 = 100 Z = 0

Simplex Algorithm Z X1 X2 X3 X4 RHS 1 7 5 4 3 240 2 100 4 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

Pivoting with X1 Z X1 X2 X3 X4 RHS 1 7 5 4 3 240 2 100 Z 1 3/2 -7/2 4 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

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

Solution Searching Path 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)

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

Unbounded Case Max X1 + 3X2 s.t. X1 - 2X2  4 -X1 + X2  3 X1, X2  0

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

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

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

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

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

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

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

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

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)

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

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

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+0.04X2+0.025X3 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

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

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