Operations Management Module B Linear Programming Power Point

Operations Management Module B – Linear Programming Power. Point presentation to accompany Heizer/Render Principles of Operations Management, 7 e Operations Management, 9 e © 2008 Prentice Hall, Inc. B–

Outline þ Requirements of a Linear Programming Problem þ Formulating Linear Programming Problems þ

Outline – Continued þ Graphical Solution to a Linear Programming Problem þ Graphical Representation of Constraints þ Iso-Profit Line Solution Method þ Corner-Point Solution Method © 2008 Prentice Hall, Inc. B– 3

Outline – Continued þ Sensitivity Analysis þ Sensitivity Report þ Changes in the Resources of the Right-Hand-Side Values þ Changes in the Objective Function Coefficient þ Solving Minimization Problems © 2008 Prentice Hall, Inc. B– 4

Outline – Continued þ Linear Programming Applications þ Production-Mix Example þ Diet Problem Example

Learning Objectives When you complete this module you should be able to: 1. Formulate linear programming models, including an objective function and constraints 2. Graphically solve an LP problem with the iso-profit line method 3. Graphically solve an LP problem with the corner-point method © 2008 Prentice Hall, Inc. B– 6

Learning Objectives When you complete this module you should be able to: 4. Interpret sensitivity analysis and shadow prices 5. Construct and solve a minimization problem 6. Formulate production-mix, diet, and labor scheduling problems © 2008 Prentice Hall, Inc. B– 7

Linear Programming þ A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources þ Will find the minimum or maximum value of the objective þ Guarantees the optimal solution to the model formulated © 2008 Prentice Hall, Inc. B– 8

LP Applications 1. Scheduling school buses to minimize total distance traveled 2. Allocating police patrol units to high crime areas in order to minimize response time to 911 calls 3. Scheduling tellers at banks so that needs are met during each hour of the day while minimizing the total cost of labor © 2008 Prentice Hall, Inc. B– 9

LP Applications 4. Selecting the product mix in a factory to make best use of machine- and labor -hours available while maximizing the firm’s profit 5. Picking blends of raw materials in feed mills to produce finished feed combinations at minimum costs 6. Determining the distribution system that will minimize total shipping cost © 2008 Prentice Hall, Inc. B – 10

LP Applications 7. Developing a production schedule that will satisfy future demands for a firm’s product and at the same time minimize total production and inventory costs 8. Allocating space for a tenant mix in a new shopping mall so as to maximize revenues to the leasing company © 2008 Prentice Hall, Inc. B – 11

Requirements of an LP Problem 1. LP problems seek to maximize or minimize some quantity (usually profit or cost) expressed as an objective function 2. The presence of restrictions, or constraints, limits the degree to which we can pursue our objective © 2008 Prentice Hall, Inc. B – 12

Requirements of an LP Problem 3. There must be alternative courses of action to choose from 4. The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities © 2008 Prentice Hall, Inc. B – 13

Formulating LP Problems The product-mix problem at Shader Electronics þ Two products 1. Shader X-pod, a portable music player 2. Shader Blue. Berry, an internetconnected color telephone þ Determine the mix of products that will produce the maximum profit © 2008 Prentice Hall, Inc. B – 14

Formulating LP Problems Hours Required to Produce 1 Unit Department Electronic Assembly Profit per unit X-pods (X 1 ) Blue. Berrys (X 2 ) Available Hours This Week 4 2 $7 3 1 $5 240 100 Table B. 1 Decision Variables: X 1 = number of X-pods to be produced X 2 = number of Blue. Berrys to be produced © 2008 Prentice Hall, Inc. B – 15

Formulating LP Problems Objective Function: Maximize Profit = $7 X 1 + $5 X 2 There are three types of constraints þ Upper limits where the amount used is ≤ the amount of a resource þ Lower limits where the amount used is ≥ the amount of the resource þ Equalities where the amount used is = the amount of the resource © 2008 Prentice Hall, Inc. B – 16

Formulating LP Problems First Constraint: Electronic is ≤ time used Electronic time available 4 X 1 + 3 X 2 ≤ 240 (hours of electronic time) Second Constraint: Assembly is ≤ time used Assembly time available 2 X 1 + 1 X 2 ≤ 100 (hours of assembly time) © 2008 Prentice Hall, Inc. B – 17

Graphical Solution þ Can be used when there are two decision variables 1. Plot the constraint equations at their limits by converting each equation to an equality 2. Identify the feasible solution space 3. Create an iso-profit line based on the objective function 4. Move this line outwards until the optimal point is identified © 2008 Prentice Hall, Inc. B – 18

Graphical Solution X 2 Number of Blue. Berrys 100 – 101 – 80 – – 60 – – 40 – – 20 – – Figure B. 3 Assembly (constraint B) |– 0 Electronics (constraint A) Feasible region | | 20 | | 40 | | 60 | | 80 | | 100 X 1 Number of X-pods © 2008 Prentice Hall, Inc. B – 19

Graphical Solution Iso-Profit Line Solution Method X 2 Number of Watch TVs Choose 100 a–possible value for the objective 101 –function 80 – Assembly (constraint B) $210 = 7 X 1 + 5 X 2 – 60 – Solve for –the axis intercepts of the function and plot line 40 the – – 20 – – Figure B. 3 |– 0 Electronics (constraint A) Feasible X 2 = 42 region | | 20 | | 40 X 1 = 30 | | 60 | | 80 | | 100 X 1 Number of X-pods © 2008 Prentice Hall, Inc. B – 20

Graphical Solution X 2 Number of Blue. Berrys 100 – 101 – 80 – – 60 – – 40 – $210 = $7 X 1 + $5 X 2 (0, 42) – (30, 0) 20 – – Figure B. 4 |– 0 | | 20 | | 40 | | 60 | | 80 | | 100 X 1 Number of X-pods © 2008 Prentice Hall, Inc. B – 21

Graphical Solution X 2 Number of Blue. Beryys 100 – $350 = $7 X 1 + $5 X 2 101 – 80 – $280 = $7 X 1 + $5 X 2 – 60 – $210 = $7 X 1 + $5 X 2 – 40 – – $420 = $7 X 1 + $5 X 2 20 – – Figure B. 5 |– 0 | | 20 | | 40 | | 60 | | 80 | | 100 X 1 Number of X-pods © 2008 Prentice Hall, Inc. B – 22

Graphical Solution X 2 Number of Blue. Berrys 100 – Maximum profit line 101 – 80 – – 60 – Optimal solution point (X 1 = 30, X 2 = 40) – 40 – – $410 = $7 X 1 + $5 X 2 20 – – Figure B. 6 |– 0 | | 20 | | 40 | | 60 | | 80 | | 100 X 1 Number of X-pods © 2008 Prentice Hall, Inc. B – 23

Corner-Point Method X 2 Number of Blue. Berrys 100 – 101 2 – 80 – – 60 – – 3 40 – – 20 – – Figure B. 7 1 |– 0 | | 20 | | 40 | 4 | 60 | | 80 | | 100 X 1 Number of X-pods © 2008 Prentice Hall, Inc. B – 24

Corner-Point Method þ The optimal value will always be at a corner point þ Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X 1 = 0, X 2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X 1 = 0, X 2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X 1 = 50, X 2 = 0) Profit $7(50) + $5(0) = $350 © 2008 Prentice Hall, Inc. B – 25

Corner-Point Method þ The optimal value will always be at a Solvepoint for the intersection of two constraints corner (electronics time) 1 + 3 X 2 ≤ 240 þ Find the 4 Xobjective function value at each (assembly time)with the corner 2 X point and choose the one 1 + 1 X 2 ≤ 100 highest profit 4 X 1 + 3 X 2 = 240 - 4 X 1 - 2 X 2 = -200 Point 1 : (X 1 = 0, X 2 = 0) + 1 X 2 = 40 4 X 1 + 3(40) = 240 4 X + 120 = 240 Profit 1$7(0) + $5(0) = $0 X 1 = 30 Point 2 : (X 1 = 0, X 2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X 1 = 50, X 2 = 0) Profit $7(50) + $5(0) = $350 © 2008 Prentice Hall, Inc. B – 26

Corner-Point Method þ The optimal value will always be at a corner point þ Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X 1 = 0, X 2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X 1 = 0, X 2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X 1 = 50, X 2 = 0) Profit $7(50) + $5(0) = $350 Point 3 : (X 1 = 30, X 2 = 40) Profit $7(30) + $5(40) = $410 © 2008 Prentice Hall, Inc. B – 27

Sensitivity Analysis þ How sensitive the results are to parameter changes þ þ Change in the value of coefficients Change in a right-hand-side value of a constraint þ Trial-and-error approach þ Analytic postoptimality method © 2008 Prentice Hall, Inc. B – 28

Sensitivity Report Program B. 1 © 2008 Prentice Hall, Inc. B – 29

Changes in Resources þ The right-hand-side values of constraint equations may change as resource availability changes þ 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-handside value of the constraint © 2008 Prentice Hall, Inc. B – 30

Changes in Resources þ Shadow prices are often explained as answering the question “How much would you pay for one additional unit of a resource? ” þ Shadow prices are only valid over a particular range of changes in right -hand-side values þ Sensitivity reports provide the upper and lower limits of this range © 2008 Prentice Hall, Inc. B – 31

Sensitivity Analysis X 2 – Changed assembly constraint from 2 X 1 + 1 X 2 = 100 to 2 X 1 + 1 X 2 = 110 100 – – 80 – 2 – Corner point 3 is still optimal, but values at this point are now X 1 = 45, X 2 = 20, with a profit = $415 60 – – 40 – – 20 – 1 © 2008 Prentice Hall, Inc. – |– 0 Electronics constraint is unchanged 3 | | 20 | | 4 60 | | 80 | | 100 X 1 Figure B. 8 (a) B – 32

Sensitivity Analysis X 2 – Changed assembly constraint from 2 X 1 + 1 X 2 = 100 to 2 X 1 + 1 X 2 = 90 100 – – 80 – 2 – Corner point 3 is still optimal, but values at this point are now X 1 = 15, X 2 = 60, with a profit = $405 60 – – 3 40 – – Electronics constraint is unchanged 20 – 1 © 2008 Prentice Hall, Inc. – |– 0 | | 20 | | | 40 4 | 60 | | 80 | | 100 X 1 Figure B. 8 (b) B – 33

Changes in the Objective Function þ A change in the coefficients in the objective function may cause a different corner point to become the optimal solution þ The sensitivity report shows how much objective function coefficients may change without changing the optimal solution point © 2008 Prentice Hall, Inc. B – 34

Solving Minimization Problems þ Formulated and solved in much the same way as maximization problems þ In the graphical approach an isocost line is used þ The objective is to move the isocost line inwards until it reaches the lowest corner point © 2008 Prentice Hall, Inc. B – 35

Minimization Example X 1 = number of tons of black-and-white picture chemical produced X 2 = number of tons of color picture chemical produced Minimize total cost = 2, 500 X 1 + 3, 000 X 2 Subject to: X 1 X 2 X 1 + X 2 X 1, X 2 © 2008 Prentice Hall, Inc. ≥ 30 tons of black-and-white chemical ≥ 20 tons of color chemical ≥ 60 tons total ≥ $0 nonnegativity requirements B – 36

Minimization Example Table B. 9 X 2 60 – X 1 + X 2

Minimization Example Total cost at a = 2, 500 X 1 + 3, 000 X 2 = 2, 500 (40) + 3, 000(20) = $160, 000 Total cost at b = 2, 500 X 1 + 3, 000 X 2 = 2, 500 (30) + 3, 000(30) = $165, 000 Lowest total cost is at point a © 2008 Prentice Hall, Inc. B – 38

LP Applications Production-Mix Example Department Product XJ 201 XM 897 TR 29 BR 788 Wiring Drilling. 5 1. 0 3 1 2 3 Assembly 2 4 1 2 Inspection. 5 1. 0. 5. 5 Unit Profit $ 9 $12 $15 $11 Department Capacity (in hours) Product Minimum Production Level Wiring Drilling Assembly Inspection 1, 500 2, 350 2, 600 1, 200 XJ 201 XM 897 TR 29 BR 788 150 100 300 400 © 2008 Prentice Hall, Inc. B – 39

LP Applications X 1 = number of units of XJ 201 produced X 2 = number of units of XM 897 produced X 3 = number of units of TR 29 produced X 4 = number of units of BR 788 produced Maximize profit = 9 X 1 + 12 X 2 + 15 X 3 + 11 X 4 subject to © 2008 Prentice Hall, Inc. . 5 X 1 + 1. 5 X 2 + 1. 5 X 3 + 3 X 1 + 1 X 2 + 2 X 3 + 2 X 1 + 4 X 2 + 1 X 3 +. 5 X 1 + 1 X 2 +. 5 X 3 + 1 X 4 3 X 4 2 X 4. 5 X 4 X 1 X 2 X 3 X 4 ≤ 1, 500 hours of wiring ≤ 2, 350 hours of drilling ≤ 2, 600 hours of assembly ≤ 1, 200 hours of inspection ≥ 150 units of XJ 201 ≥ 100 units of XM 897 ≥ 300 units of TR 29 ≥ 400 units of BR 788 B – 40

LP Applications Diet Problem Example Feed Product A B C D © 2008 Prentice

LP Applications X 1 = number of pounds of stock X purchased per cow each month X 2 = number of pounds of stock Y purchased per cow each month X 3 = number of pounds of stock Z purchased per cow each month Minimize cost =. 02 X 1 +. 04 X 2 +. 025 X 3 Ingredient A requirement: Ingredient B requirement: Ingredient C requirement: Ingredient D requirement: Stock Z limitation: 3 X 1 + 2 X 1 + 1 X 1 + 6 X 1 + 2 X 2 + 3 X 2 + 0 X 2 + 8 X 2 + 4 X 3 1 X 3 2 X 3 4 X 3 X 1, X 2, X 3 ≥ 64 ≥ 80 ≥ 16 ≥ 128 ≤ 80 ≥ 0 Cheapest solution is to purchase 40 pounds of grain X at a cost of $0. 80 per cow © 2008 Prentice Hall, Inc. B – 42

LP Applications Labor Scheduling Example Time Period Number of Tellers Required 9 AM - 10 AM - 11 AM - Noon - 1 PM 10 12 14 16 1 PM - 2 PM - 3 PM - 4 PM - 5 PM 18 17 15 10 F P 1 P 2 P 3 P 4 P 5 © 2008 Prentice Hall, Inc. = = = Full-time tellers Part-time tellers starting at 9 AM (leaving at 1 PM) Part-time tellers starting at 10 AM (leaving at 2 PM) Part-time tellers starting at 11 AM (leaving at 3 PM) Part-time tellers starting at noon (leaving at 4 PM) Part-time tellers starting at 1 PM (leaving at 5 PM) B – 43

LP Applications Minimize total daily manpower cost = $75 F + $24(P 1 + P 2 + P 3 + P 4 + P 5) F F 1/2 F F F + P 1 + P 2 + P 3 + P 4 + P 2 + P 3 + P 4 + P 5 ≥ 10 ≥ 12 ≥ 14 ≥ 16 ≥ 18 ≥ 17 ≥ 15 ≥ 10 ≤ 12 (9 AM - 10 AM needs) (10 AM - 11 AM needs) (11 AM - 11 AM needs) (noon - 1 PM needs) (1 PM - 2 PM needs) (2 PM - 3 PM needs) (3 PM - 7 PM needs) (4 PM - 5 PM needs) 4(P 1 + P 2 + P 3 + P 4 + P 5) ≤. 50(10 + 12 + 14 + 16 + 18 + 17 + 15 + 10) © 2008 Prentice Hall, Inc. B – 44

LP Applications Minimize total daily manpower cost = $75 F + $24(P 1 + P 2 + P 3 + P 4 + P 5) F F 1/2 F F F + P 1 + P 2 + P 3 + P 4 + P 2 + P 3 + P 4 + P 5 ≥ 10 ≥ 12 ≥ 14 ≥ 16 ≥ 18 ≥ 17 ≥ 15 ≥ 10 ≤ 12 (9 AM - 10 AM needs) (10 AM - 11 AM needs) (11 AM - 11 AM needs) (noon - 1 PM needs) (1 PM - 2 PM needs) (2 PM - 3 PM needs) (3 PM - 7 PM needs) (4 PM - 5 PM needs) 4(P 1 + P 2 + P 3 + P 4 + P 5) ≤. 50(112) F, P 1 , P 2 , P 3 , P 4 , P 5 ≥ 0 © 2008 Prentice Hall, Inc. B – 45

LP Applications Minimize total daily = $75 F + $24(P + P 2 + P 3 + P 4 + P 5) manpower cost There are two alternate optimal 1 solutions to this problem but both will cost per day F+P ≥ 10 ($1, 086 9 AM - 10 AM needs) 1 F + P 1 + P 2 1/2 F + P 1 + P 2 First + P 3 1/2 F + P 1 + Solution P 2 + P 3 + P 4 F +F P 2 +=P 310 + P 4 F P 1 +=P 30+ P 4 F + P 4 P 2 = 7 F P 3 = 2 F + P 5 ≥ 12 (10 AM - 11 AM needs) Second ≥ 14 (11 AM - 11 AM needs) Solution ≥ 16 (noon - 1 PM needs) ≥ 18 F (1=PM 10 - 2 PM needs) ≥ 1 P 7 (2=PM - 3 PM needs) 6 ≥ 151 (3 PM - 7 PM needs) P 2 (4=PM 1 - 5 PM needs) ≥ 10 P 3 = 2 ≤ 12 P 4 = 2 P 5 = 3 4(P 1 + P P 25 +=P 33+ P 4 + P 5) ≤. 50(112) F, P 1 , P 2 , P 3 , P 4 , P 5 ≥ 0 © 2008 Prentice Hall, Inc. B – 46

The Simplex Method þ Real world problems are too complex to be solved using the graphical method þ The simplex method is an algorithm for solving more complex problems þ Developed by George Dantzig in the late 1940 s þ Most computer-based LP packages use the simplex method © 2008 Prentice Hall, Inc. B – 47