Ch13 - NLP,DP,GP

1Nonlinear Models: Dynamic, Goal and Nonlinear Programming Chapter 13 2 13.1 Introduction to Nonlinear Programming  Most real-life situations are more accurately depicted by nonlinear models than by linear models.  Nonlinear models formulated in this chapter fall into three broad categories:  Dynamic programming models (DP).  Goal programming models (GP).  General nonlinear programming models (NLP). 3 13.2 Dynamic Programming  Dynamic programming problems can be thought of as multistage problems, in which decisions are made “in sequence.”  Known applications of dynamic programming in business:  Resource allocation,  Equipment replacement,  Production and inventory control,  Reliability. 4 13.2 Dynamic Programming  At stage n a system is found to be in a certain state.  A decision made at stage n takes the system to stage n+1, and leaves it in a new state for a stage- state related cost/profit.  The decision maker’s challenge is to find a set of optimal decisions for the entire process. 5 THE U.S. DEPARTMENT OF LABOR 6  The U.S. Department of Labor has made up to $5 million available to the city of Houston for job creation.  Requests for funding are to be prepared by four departments.  The Department of Labor would like to allocate funds to maximize the number of jobs created. THE U.S. DEPARTMENT OF LABOR 7  Data Department Cost $ millions Jobs Created Housinng 4 225 Employment 1 45 2 125 3 190 Highway 50 jobs per $1 million funding Law Enforcement 2 75 3 155 4 220 Estimated new jobs created for proposal THE U.S. DEPARTMENT OF LABOR 8 SOLUTION  The U.S. Department of Labor wants to:  Maximize the total number of new jobs.  Limit funding to $5 million.  Fund at most one proposal from each department 9  Notation D j = the amount allocated to department j, where j is: 1 - housing, 2 - Employment, 3 - Highway, 4 - Law Enforcement. R(D j ) = the number of new jobs created by funding department j with $D j million.  The model Max R 1 (D 1 ) + R 2 (D 2 ) + R 3 (D 3 ) + R 4 (D 4 ) ST D 1 + D 2 + D 3 + D 4 <= 5 D 1 , D 2 , D 3 , D 4 >= 0 Non-linear functions SOLUTION 10  Define F j (X j ) as the maximum number of new jobs created by departments (stages) j, j+1,…, 4, given that there is $X j million in funding available for departments j (state) through 4. The Backward Dynamic Programming 11  Stage 4: The Law Enforcement,(LED)  Start with the last stage j=4 (Law Enforcement department, LED). Allocate to this department funds that maximize the number of new jobs created  Obviously, the optimal solution for the last department is to use all the amount available at this stage).  The optimal solution for the last stage is called “The boundary condition The Backward Dynamic Programming 12  Stage 4: The Law Enforcement table Available Optimal Funding Maximum jobs Funding For Stage 4 F4(X4) 0 0 0 1 0 0 2 2 75 3 3 155 4 4 220 5 5 220 States Cost Jobs Law Enforcement 2 75 3 155 4 220 Recall SOLUTION 13  Stage 3: The Highway Department, (HD)  At this stage we consider the funding of both the Highway department and the Law Enforcement department.  For a given amount of funds available for funding of these two departments, the decision regarding the funds allocated to the HD affects the funds available for the LED). SOLUTION 14 Available Possible Remaining Max. New Jobs Optimal F3(X3) Funding Funding Funds for When Allocating for Stage 3,4 for Stage 3 Stage 4 D3 to Stage 3 Optimal D3 (X3) (D3) (X3-D3) R3(D3)+F4(X3-D3) 0 0 0 0 + 0 F3(0) = 0; D3(0) = 0 1 0 1 0+0 1 0 50+0 = 50 F3(1) = 50; D3 = 1 2 0 2 0+75 = 75 1 1 50+0 = 50 2 0 100+0 F3(3) = 100; D3 = 2 3 0 3 0+155 = 155 1 2 50+75 = 125 2 1 100+0 = 100 3 0 150+0 = 150 F3(3) = 155; D3 = 0 Stage 3: The Highway Department table SOLUTION 15 Available Possible Remaining Max. New Jobs Optimal F3(X3) Funding Funding Funds for When Allocating for Stage 3,4 for Stage 3 Stage 4 D3 to Stage 3 Optimal D3 (X3) (D3) (X3-D3) R3(D3)+F4(X3-D3) 4 0 4 0+220 = 220 1 3 50+155 = 205 2 2 100+75 = 175 3 1 150+ 0 = 150 4 0 200+ 0 = 220 F3(4) = 220; D3 = 0 5 0 5 0+220 = 220 1 4 50+220 = 270 2 3 100+155 = 255 3 2 150+75 = 225 4 1 200+0 = 200 5 0 250+0 = 250 F3(5) = 270; D3 = 1 SOLUTION Stage 3: The Highway Department table 16  Stage 2: The Employment Department, (ED)  At this stage we consider the funding of both the Employment Department and the previous departments (HD and the LED)  For a given state (the amount of funds available for funding of these three departments), the decision regarding the funds allocated to the ED affects the funds available for the HD and LED. (the state at the stage j=3) SOLUTION 17 Funding Funding Funds for When Allocating for Stage 2,3,4 for Stage 2 Stage 3,4 D2 to Stage 2 Optimal D2 (X2) (D2) (X2-D2) R2(D2)+F3(X2-X2) 0 0 0 0 + 0 F2(0) = 0; D2 = 0 1 0 1 0+50 = 50 1 0 45+0 = 45 F2(1) = 50; D2 = 0 2 0 2 0+100 = 100 1 1 45+50 = 95 2 0 125+0 = 125 F2(2) = 125; D2 = 2 3 0 3 0+155 = 155 1 2 45+100 = 145 2 1 125+50 = 175 3 0 190+0 = 190 F2(3) = 190; D2 = 3 4 0 4 0+220 = 220 1 3 45+155 = 195 2 2 125+100 = 225 3 1 190+50 = 240 F2(4) = 240; D2 = 3 5 0 5 0+270 = 270 1 4 45+220 = 265 2 3 125+155 = 280 3 2 190+100 = 290 F2(5) = 290; D2 = 3 SOLUTION 18  Stage 1: The Housing Department  At this stage we consider the funding for the Housing Department and all previous departments.  Note that at stage 1 we know there is exactly $5 million to allocate (X 1 = 5). SOLUTION 19  Stage 1: The Housing Department submitted only one proposal of $4 million, thus (D 1 = 0, 4). Available Possible Remaining Max. New Jobs Optimal F1(X1) Funding Funding Funds for When Allocating for Stage 1,2,3,4 for Stage 1 Stage 2,3,4 D3 to Stage 3 Optimal D1 (X1) (D1) (X1-D1) R1(D1)+F2(X1-D1) 5 0 5 0+290 = 290 4 1 225+50 = 275 F1(5) = 290; D1 = 0 Do not fund Fund the project SOLUTION 20 Stage State 0 1 2 3 4 5 1 D1 = 0 2 D2 = 0 D2 = 0 D2 = 2 D2 = 3 D2 = 3 D2 = 3 3 D3 = 0 D3 = 1 D3 = 2 D3 = 0 D3 = 0 D3 = 1 4 D4 = 0 D4 = 1 D4 = 2 D4 = 3 D4 = 4 D4 = 5 In summary, the optimal funds allocation to maximize the number of jobs created is: Housing = $0 Employment = $3 million Highways = $2 million Law Enforcement = $0 Maximum number of jobs created = 290 SOLUTION 21 Notation Description Example Stage Variable j A decision point The four departments State Variable Xj The amount of resource Funds left to allocate left to be allocated to departments Decision Variable Dj A possible decision Funds that could have at stage j been give to department j Stage Return Values Rj(Dj) The stage return for Number of jobs created making decision Dj at department j if funding Dj Optimal Value Fj(Xj) The best cumulative Maximum number of new return for stages j and jobs created for dept j,…,4. remaining stages at state Xj. Boundary Condition F(XN) A set of optimal values The new jobs created from for the last stage (N) funding Dept 4 with $X4 million. Optimal Solution F1(T) The best cumulative The maximum total number Value return of new jobs given $Tmillion to allocate to all departments Components of Dynamic Programming 22  Bellman‟s principle of optimality.  From a given state at a given stage, the optimal solution for the remainder of the process is independent of any previous decisions made to that point.  Dynamic programming is a recursive process.  The following recursive relationship describes the process for the U.S. Department of Labor problem.  Define F j (X j ) as the maximum number of new jobs created by departments (stages) j, j+1,…, 4, given that there is $X j million in funding available for departments j (state) through 4. F j (X j ) = Max {(R j (X j ) + F j+1 (X j - D j )} all feasible X j Dynamic Recursive Relationship 23 The form of the recursion relation differs from problem to problem, but the general idea is the same: Do the best you can for the remaining stages with the remaining resources available. Dynamic Recursive Relationship 24 13.3 Computational Properties of Dynamic Programming  Dynamic programming versus total enumeration Dynamic programming approach eliminates from consideration combinations that have no chance of being optimal.  Tricks for reducing computations further We can reduce computational effort by… • “smart” definition of state variable(s); • noticing the feasible range of state variables. 25  Networks can be used to model multistage decision problems that are solved by dynamic programming approach.  Each node represents a state variable value at each stage.  Each arc represents a possible decision at the particular state.  The stage return is the value assigned to each arc.  The objective is to find the longest (shortest) path. Dynamic programs as networks 26 Network Presentation- The US Department of Labor 27  Dynamic programming problems are typically solved using computer programs.  Since, dynamic problems differ from one another, there is no universal code to solve them.  Some problems do fit a particular type, and are presented in the next section. Computers and dynamic programming 28 13.4 Dynamic Programming - Examples  In our examples we focus on identifying the following elements:  The stage variable.  The state variable.  The decision variable.  The stage return or cost function(s).  The optimal value function.  The boundary conditions.  The optimal solution value (stopping rule).  The recurrence relation. 29 KNAPSACK PROBLEMS  This problem belongs to the family of resource allocation problems.  A hiker carries a knapsack with limited space.  Each item (j) has a volume (W i ) and a value V i.  Limited quantity of each item (U i ) may be available.  There is a minimum requirement (L i ) for the amount of each item that is to be taken.  Fill the knapsack with items such that the total value of items selected is maximized. 30 KNAPSACK PROBLEMS – The Integer Linear Model . eger int is D ) j item each for ( U D ) j item each for ( L D C D W ST D V Maximize i i i i i i i i i i s > s ¿ ¿ where D i is the units of item j put in the knapsack. 31  The NWC firm‟s representatives need to deliver boxes of items the company wants to present in a conference.  The company‟s small jet will be used to transport the team. It can carry a maximum of 127 pounds of cargo for the show.  Quantities of four different items must be carried : 4Advertising buttons 4 Laser printers 4Advertisig brochures 4 Notebook computers NORTHWESTERN COMPUTER 32  Data  At least one box of brochures, two printers, and two notebook computers must be taken.  Availability, weight, and profit per item are: Weight Profit Item Available (lbs.) Potential ($) Boxes of Buttons 4 4 600 Boxes of Brochures 5 9 1,100 Laser Printers 4 21 4,000 Notebook Computers 5 12 1,500 How many of each item should be loaded on the small jet to maximize the total profit potential of the trip? NORTHWESTERN COMPUTER 33  Stage variable= the item j under consideration. j=1 buttons; j=2 brochures; j=3 printers; j=4 computers  State variable X j = The amount of weight left for transporting item j, j+1, …,notebook computers.  Stage return function = the added value obtained by taking D j additional type j items.  Optimal value function F j (X j ) = the maximum total profit potential of transporting item j through 4 (notebook computers) if X j pounds remain to take these items. Northwestern – Definitions 34 Let j = 4 represent notebook computer. F 4 (X 4 ) = 0 for all values of X 4 = 0,1,2,…,11; D 4 = 0 F 4 (X 4 ) = 1,500 for all values of X 4 = 12, 13,…,23; D 4 = 1 F 4 (X 4 ) = 3,000 for all values of X 4 = 24, 25, …,35; D 4 = 2 F 4 (X 4 ) = 4,500 for all values of X 4 = 36, 37, …,48; D 4 = 3 Free space = 127 - 4 lbs of buttons box - 7 lbs of brochure box - 2(21) lbs of printer boxes - 2(12) lbs of computer boxes = 48 lbs. Northwestern - Boundary Conditions 35 F j (X j ) = Maximum{V j D j + F j+1 (X j - W j D j ) D j s the number of remaining items of type j W j D j sX j Recursive Calculations See a demonstration of several recursive calculations next 36 Recursive Calculations 37 RELIABILITY PROBLEMS  Product reliability and safety are significant concern for consumers and manufacturers.  Products must be designed with an acceptable probability of operating satisfactorily.  Dynamic programming can be used to determine minimum cost product design that ensures  minimum required level of reliability, or  highest possible degree of reliability. 38 PLAYCO TOYS, INC.  Playtco Toys manufactures the Watkins Glen remote-control race car.  Manufacturing cost reduction must come from three electronic components ordered from a supplier at the following cost per unit:  Power unit $12  Sound unit $ 5  Remote control unit $ 8 $25 To remain competitive the total cost per unit of $25 must be reduced to $18. 39  Other suppliers offer reduced costs for these components, but the reliability level of their components has been proven to be lower.  Playco needs to determine which manufacturer component to use in order to get the highest degree of reliability without exceeding the $18 limit. Power Sound Remote Manufacturer Unit Cost Reliability Unit Cost Reliability Unit Cost Reliability DSL, Inc. 12 0.998 5 0.995 5 0.998 Karrow Industries 10 0.994 4 0.987 8 0.999 ELO Elaectronic 5 0.975 2 0.980 4 0.995 PLAYCO TOYS, INC. 40 PLAYCO TOYS - SOLUTION  Define: P j (D j ) = the probability that component j is operational if it is purchased for $D j .  The reliability of the product is the probability that all three units operate properly (the product is functional).  Assuming independence between the components, the reliability is calculated by : P 1 (D 1 ) P 2 (D 2 ) P 3 (D 3 ) - - 41  Stage variable j = the electrical component.  State variable X j = the amount of money left to purchase component j, and subsequent components.  Decision variable D j = the amount spent for component j.  Optimal value function F j (X j ) = the highest probability that all the components j through 3 will function if $X j is spent to purchase these components. PLAYCO TOYS – Definitions 42  Playco must spend at least $4 on remote control (stage 3). F 3 (X 3 ) = 0 for X 3 = 0, 1, 2, 3 D 3 = 0 F 3 (X 3 ) = 0.995 for X 3 = 4 D 3 = 4 F 3 (X 3 ) = 0.998 for X 3 = 5, 6, 7 D 3 = 5 F 3 (X 3 ) = 0.999 for X 3 = 8, 9,…, 18 D 3 = 8 Boundary Conditions – Stage 3: Remote 43 F j (X j ) = Max {P j (D j ) F j+1 (X j - D j )} Over all values of D j The Recursion – Stage 2: Sound unit  At least $6 is required to purchase both a sound unit and a remote control unit. At most $13 is required to purchase the most reliable sound and remote control units.  Thus, the feasible range for X 2 = 3, 4, …,13. - 44 Amount left Reliability of for allocation to Allocation to Components 2 and 3 Optimal F2(X2) Stage 2 and 3 (X2) Stage 2 (D2) P(D2)F3(X2 - D2) Optimal D2 0 - 5 0 ********** 0 6 2 .980(.995) 0.975100 7 2 .980(.998) 0.978040 8 2 .980(.998) 0.97804 4 .987(.995) 0.982065 9 2 .980(.998) 0.978040 4 .987(.998) 0.985026 5 .995(.995) 0.990025 10 2 .980(.999) 0.979020 4 .987(.998) 0.985026 5 .995(.998) 0.993010 Sound Remote Unit Cost Reliability Unit Cost Reliability 5 0.995 5 0.998 4 0.987 8 0.999 2 0.980 4 0.995 Recursive Calculations Stage 2: Sound unit 45  We only need to calculate F 1 (18) Reliability of Components Optimal Value Allocation to 1 through 3 and Decision Stage 1 (D1) P(D1)F(X1-D1) F1(X1) 5 .975(.994005) 0.969155 10 .994(.982065) 0.976173 12 .998(.975100) 0.973150 Power Unit Cost Reliability 12 0.998 10 0.994 5 0.975 Recursive Calculations – Stage 1: Power Supply 46 Amount left for allocation to Allocation to Stage 2 and 3 (X2) Stage 2 (D2) 0 - 5 0 6 2 7 2 8 2 4 Allocation to Stage 1 (D1) 5 10 12 Amount left for allocation for stage 3 Allocation 4 4 Purchase power supply from Karrow Industries Purchase sound unit from Karrow Industries Purchase remote unit from ELO Electronic PLAYCO TOYS – Solution summary 47 PRODUCTION AND INVENTORY PROBLEMS  During the course of a certain production run certain parameters may vary (prices of raw material, storage space available, etc.)  A dynamic programming modeling approach can be found very useful in these situations. 48 CJM INDUSTRIES  CJM builds classic “replicars” using original car bodies from the 1920‟s, „30‟s, „40‟s, and „50‟s.  CJM is planning a 15 cars production run of 1957 Chrysler convertibles for the months of May, June,…, September. 49  Data  Maximum production level is 3 for July, and 4 for any other month.  Up to 2 vehicles may be stored and displayed at a cost of $2,500 per month ($3,000 for May).  Fixed costs (due to utilities, insurance, etc.) are incurred only when cars are been actually produced. Month Fixed Production Cars to be j Costs Sj ($) Costs Pj ($) Delivered Cj May 2000 21,000 3 June 3000 16,000 2 July 4000 9,000 1 August 3000 13,000 5 September 2000 23,000 4 CJM INDUSTRIES 50 Develop a production schedule that minimizes total cost. CJM INDUSTRIES – the Objective 51  Stage variable j: The current month.  State variable X j : The number of cars in inventory at the beginning of month j.  Decision variable D j : The production quantity for month j. CJM - Definitions 52 • Production costs PC j (D j ) in month j are proportional to the number of cars produced PC j (D j ) = P j D j • Holding (storage) costs HC j (D j ) in month j are paid for cars that are left unsold at the end of this month. For month j =1: HC 1 (D 1 ) = 3000(X 1 + D 1 - C 1 )= 3000D 1 – 9000 For months j = 2, 3, 4, 5: HC j (D j ) =2500(X j +D j - C j )  Stage cost function: • Fixed costs FC j (D j ) are incurred in month j if there is a car production. That is FC j (D j ) = S j if D j > 0 FC j (0) = 0 if D j = 0 CJM - Definitions Given: X 1 = 0 and C 1 = 3 53  The optimal value function F j (X j ) in month j is the minimum total cost incurred from month j through month 5 (September), given X j cars are inventoried at the beginning of month j.  Boundary conditions F 5 (X 5 ): F 5 (0) = 2,000 + 23,000(4) = $94,000; D 5 = 4 F 5 (1) = 2,000 + 23,000(3) = $71,000; D 5 = 3 F 5 (2) = 2,000 + 23,000(2) = $48,000; D 5 = 2  Optimal solution F 1 (0) is the minimum cost from month 1 (May) through month 5 (September) given that there is no initial inventory. CJM - Definitions 54 F j (X j ) = Min{FC j (D j ) + PC j (D j ) + HC j (D j ) + F j+1 (X j +D j - C j )}, D j is feasible only if it satisfies the following conditions: D j + X j > C j ; D 3 s 3 for July; D j s 4 for j = 1, 2, 4, 5; D j + X j - C j s 2; D j > 0 over all feasible D j . The Recursion 55 Recursive Calculations – Stage 4: August X 4 Possible D 4 X 4 + D 4 – C 4 FC 4 PC 4 HC 4 F 5 (X 4 +D 4 –C 4 ) Total Optimal Production Units Stored Cost Value 0 Infeasible Infeasible Infeasible 1 4 0 3 52 0 94 149.0 F 4 =149 D 4 =4 2 4 1 3 52 2.5 71 128.5 F 4 =128.5 3 0 3 39 0 94 136.0 D 4 =4 56 X 3 Possible D 3 X 3 + D 3 – C 3 FC 3 PC 3 HC 3 F 4 (X 3 +D 3 –C 3 ) Total Optimal Production Units Stored Cost Value 0 2 1 4 18 2.5 149 173.5 F 3 =164.5 3 2 4 27 5.0 128.5 164.5 D 3 =3 1 1 1 4 9 2.5 149 164.5 F 3 =155.5 2 2 4 18 5 128.5 155.5 D 3 =2 2 0 1 0 0 2.5 149 151.5 F 3 =146.5 1 2 4 9 5 128.5 146.5 D 3 =1 Recursive Calculations – Stage 3: July 57 13.5 Goal Programming  In real life decision situations, virtually all problems have several objectives.  When objectives are conflicting, the optimal decision is not obvious.  Goal programming is an approach that seeks to simultaneously take into account several objectives. 58  Goals are prioritized in some sense, and their level of aspiration is stated.  An optimal solution is attained when all the goals are reached as close as possible to their aspiration level, while satisfying a set of constraints.  There are two types of goal programming models:  Nonpreemtive goal programming - no goal is pre-determined to dominate any other goal.  Preemtive goal programming - goals are assigned different priority levels. Level 1 goal dominates level 2 goal, and so on. 13.5 Goal Programming 59  A company is considering three forms of advertising.  Goals  Goal 1: Spend no more $25,000 on advertising.  Goal 2: Reach at least 30,000 new potential customers.  Goal 3: Run at least 10 television spots. NONPREEMTIVE GOAL PROGRAMMING An Advertisement Example Cost per Ad Customers Television 3000 1000 Radio 800 500 Newspaper 250 200 60  If these were constraints rather than goals we would have: 3000X 1 + 800X 2 + 250X 3 s25,000 1000X 1 + 500X 2 + 200X 3 > 30,000 X 1 > 10  No feasible solution exists that satisfies all the constraints.  When these constraints are simply goals they are to be reached as close as possible. An Advertisement Example 61  Detrimental variables Ui = the amount by which the left hand side falls short of (under) its right had side value. Ei = the amount by which the left side exceeds its right had side value.  The goal equations 3000X 1 + 800X 2 + 250X 3 + U 1 – E 1 = 25,000 1000X 1 + 500X 2 + 200X 3 + U 2 – E 2 = 30,000 X 1 + U 3 – E 3 = 10 An Advertisement Example 62  The objective is to minimize the penalty of not meeting the goals, represented by the detrimental variables E1, U2, U3. An Advertisement Example s25,000 > 30,000 > 10 63  The penalties are estimated to be as follows:  Each extra dollar spent on advertisement above $25,000 cost the company $1.  There is a loss of $5 to the company for each customer not being reached, below the goal of 30,000.  Each television spot below 10 is worth 100 times each dollar over budget. An Advertisement Example 64  It is assumed that no advantage is gained by overachieving a goal. Minimize 1E 1 + 5U 2 + 100U 3 s.t. 3000X 1 + 800X 2 + 250X 3 + U 1 – E 1 = 25,000 1000X 1 + 500X 2 + 200X 3 + U 2 – E 2 = 30,000 X 1 + U 3 – E 3 = 10 All variables are non-negative. An Advertisement Example – The goal programming model 65  The NECC is planning next month production of its two bicycles B2 and S10.  Data  Both models use the same seats and tires.  2000 seats are available; 2400 tires are available.  1000 gear assembly are available (used only in the S10 model).  Production time per unit: 2 hours for B2; 3 hours for S10.  Profit: $40 for each B2; 10$ for each S10. PREEMTIVE GOAL PROGRAMMING - New England Cycle Company 66  Priority 1: Fulfill a contract for 400 B2 bicycles to be delivered next month. NECC – Prioritized Goals Priority 4:  At least 200 tires left over at the end of the month.  At least 100 gear assemblies left over at the end of the month.  Priority 2: Produce at least 1000 total bicycles during the month.  Priority 3:  Achieve at least $100,000 profit for the month.  Use no more than 1600 labor-hours during the month. 67 Management wants to determine the production schedule that best meets its prioritized schedule. New England Cycle Company Example 68 NECC - SOLUTION  Decision variables X 1 = The number of B2s to be produced next month X 2 = The number of S10s to be produced next month  Functional / nonnegativity constraints 0 X , X Tires 2400 X 2 X 2 assemblies Gear 1000 X Seats 2000 X X 2 2 1 2 1 2 2 1 > s + s s + 69  Goal constraints Priority 1 (goal 1): Production of at least 400 B2s X 1 + U 1 - E 1 = 400 Priority 2 (goal 2): Production of at least 1000 total cycles X 1 + X 2 + U 2 - E 2 = 1000 Priority 3 (goal 3) Profit of at least $100,000 .04X 1 + .10X 2 + U 3 - E 3 = 100 (in $1000) Priority 3 (goal 4) Use a maximum of 1600 labor hours 2X 1 + 3X 2 + U 4 - E 4 = 1600 Priority 4 (goal 5) At least 200 leftover tires 2X 1 + 2X 2 + U 5 - E 5 = 2200 Priority 4 (goal 6) At least 100 leftover gear assembly X 2 + U 6 - E 6 = 900 NECC - SOLUTION 70  Priority level objectives Priority 1: Underachieving a production of 400 B2s: Minimize U 1 Priority 2: Underachieving a total production of 1000: Minimize U 2 NECC - SOLUTION 71  Priority level objectives  Priority 3: Underachieving a $100,000 profit Using more than 1600 labor-hours Minimize 30U 3 + E 4 NECC - SOLUTION Each $1,000 short of the $100,000 goal is considered 30 times as important as utilizing an extra labor-hour. 72  Priority level objectives Priority 4: Using more than 2200 tires Using more than 900 gear assemblies Minimize E 5 + 2E 6 NECC - SOLUTION Each leftover gear assembly is deemed twice as important as leftover tire. 73  Solve the linear goal programming for priority 1 objective, under the set of regular constraints and goal constraint as shown below Minimize U 1 ST NECC - The solution procedure X 1 + U 1 - E 1 = 400 Tires 2400 X 2 X 2 Gear 1000 X Seats 2000 X X 2 2 1 2 2 1 s + s s + X 1 = 400, thus U 1 = 0, and priority 1 goal is fully achieved. 74  Solve the linear goal programming for priority 2 level objective, under the set of original constraints plus the constraint X 1 > 400 (maintain the level of achievement of the priority 1 goal). Minimize U 2 ST NECC - The solution procedure Every point that satisfies X 1 + X 2 > 1000 yields U 2 = 0, and therefore, priority 2 goal is fully achieved. Tires 2400 X 2 X 2 Gear 1000 X Seats 2000 X X 2 2 1 2 2 1 s + s s + X 1 > 400 X 1 + X 2 + U 2 - E 2 = 1000 75  Solve the linear goal programming for priority 3 level objective, under the set of original constraints plus the constraint X 1 > 400 (maintain the level of achievement of the priority 1 goal), plus the constraint X 1 + X 2 > 1000 (maintain the level of achievement of the priority 2 goal). • Every point in the range X 1 = 400 and 600 s X 2 s 800 is optimal for this model; 30U 3 + E 4 = 1720 is the level of achievement for the priority 3 goal. NECC - The solution procedure 76  Solve the linear goal programming for priority 4 level objective, and notice that after the previous step the feasible region is reduced to a segment of a straight line between the points (400,600) and (400,800). • X1 =400; 600 s X2 s 700 and E5 + 2E6 = 0 NECC - The solution procedure 77  In summary NECC should produce  400 B2 model  Between 600 and 700 S10 model NECC - Solution Summary 78 13.6 Nonlinear Programming  A nonlinear programming problem (NLP) is one in which the objective function, F, and or one or more constraint functions, G i , possess nonlinear terms.  There is no a universal algorithm that can find the optimal solution to every NLP.  One class of NLPs “Convex Programming Problems” can be solved by algorithms that are guaranteed to converge to the optimal solution. 79  The objective is to maximize a concave function or to minimize a convex function.  The set of constraints form a convex set. Properties of Convex Programming Problems 80  A smooth function (no sharp points, no discontinuities)  One global maximum (minimum).  A line drawn between any two points on the curve of the function will lie below (above) the curve or on the curve. A One Variable Concave (Convex) Function X A Concave function A convex function X 81 An illustration of a two variable convex function 82 If a straight line that joins any two points in the set lies within the set, then the set is called a convex set. Convex set Non-convex set Convex Sets 83  In a NLP model all the constraints are of the “less than or equal to” form G i (X) s B.  If all the functions G i are convex, the set of constraints forms a convex set. NLP and Convex Sets 84 13.7 Unconstrained Nonlinear Programming  One-variable unconstrained problems are demonstrated by the Toshi Camera problem.  The inverse relationships between demand for an item and its value (price) are utilized in this problem. 85 TOSHI CAMERA  Toshi camera of Japan has just developed a new product, the Zoomcam.  It is believed that demand for the initial product will be linearly related to the price. Price Estimated P ($) Demand (X) 100 350,000 150 300,000 200 250,000 250 200,000 300 150,000 350 100,000 86  Unit production cost is estimated to be $50.  What is the production quantity that maximizes the total profit from the initial production run? SOLUTION Total profit = Revenue - Production cost F(X) = PX - 50X TOSHI CAMERA 87 From the Price / Demand table it can be verified that P = 450 - .001X The Profit function becomes F(X) = (450 - .001X)X = 400X - .001X 2 This is a concave function. 400,000 0 TOSHI CAMERA 88  To obtain an optimal solution (maximum profit), two conditions must be satisfied:  A necessary condition dF/dX = 0  A sufficient condition d 2 F/dX 2 < 0.  The necessary condition is satisfied at: dF/dX = 400 - 2(.001)X = 0; X = 200,000.  The sufficient condition is satisfied since d 2 F/dX 2 = -.002.  The optimal solution:  Produce 200,000 cameras.  The profit is F(200,000) = $40,000,000. TOSHI CAMERA 89 If a function is known to be concave or convex at all points, the following condition is both a necessary and sufficient condition for optimality: The point X* gives the maximum value for a concave function, or the minimum value for a convex function, F(X), if at X* dF/dX = 0 Optimal solutions for concave/convex functions with one variable 90  Determining whether or not a multivariate function is concave or convex requires analysis of the second derivatives of the function.  A point X* is optimal for a concave (convex) function if all its partial derivatives are equal to zero at X*.  For example, in the three variable case: c c c c c c F X F X F X 1 1 2 2 3 3 0 0 0 = = = ; ; ; Optimal solutions for concave/convex functions with more than one variable 91 13.8 Constrained Nonlinear Programming Problems – one variable  The feasible region for a one variable problem is a segment on a straight line (X > a or X s b).  When the objective function is nonlinear the optimal solution must not be at an extreme point. 92 TOSHI CAMERA - revisited  Toshi Camera needs to determine the optimal production level from among the following three alternatives: 150,000 s X s 300,000 50,000 s X s 175,000 150,000 s X s 350,000 93 400 0 X > 250,000 X s 350,000 X* = 250,000 400 0 400 0 Maximize F(X) = 400X - .001X 2 X > 150,000 X s 300,000 150 X > 50,000 X s 175,000 50 175 X*=200,000 X* = 175,000 The objective function does not change: TOSHI CAMERA – solution 94 Constrained Nonlinear Programming Problems – m variables m n 2 1 2 n 2 1 1 n 2 1 n 2 1 B ) X ..., , X , Gm(X . . . B ) X ..., , X , G2(X B ) X ..., , X , G1(X ST ) X ..., , X , X ( F Maximize s s s Let us define Y 1 , Y 2 , …,Y m as the instantaneous improvement in the value of F for one unit increase in B 1 , B 2 , …B m respectively. 95  This is a set of “necessary conditions” for optimality of most nonlinear problems.  If the problem is convex, the K-T conditions are also sufficient for a point X* to be optimal. | . | \ | c c + + | . | \ | c c + | . | \ | c c = c c | . | \ | c c + + | | . | \ | c c + | . | \ | c c = c c = = = > m m 2 1 1 m m 2 1 1 1 m m 2 2 1 1 m 2 1 X Gm Y ... X 2 G 2 Y X 1 G Y Xn F . . . X Gm Y ... X 2 G Y X 1 G Y X F . 4 0 S Y ..., , 0 S Y , 0 S Y . 3 0 Y ..., , Y , Y . 2 . feasible is * X . 1 2 S1, S2, …,Sm are defined as the slack variables in each constraint. Kuhn-Tucker optimality conditions 96 PBI INDUSTRIES  PBI wants to determine an optimal production schedule for its two CD players during the month of April.  Data  Unit production cost for the portable CD player = $50.  Unit production cost for the deluxe table player = $90.  There is additional “intermix” cost of $0.01(the number of portable CD‟s)(the number of deluxe CD‟s). 97  Forecasts indicate that unit selling price for each CD player is related to the number of units sold as follows:  Portable CD player unit price = 150 - .01X1  Deluxe CD player unit price = 350 - .02X2 PBI INDUSTRIES 98  Resource usage  Each portable CD player uses 1 unit of a particular electrical component, and .1 labor hour.  Each deluxe CD player uses 2 units of the electrical component, and .3 labor hour.  Resource availabilty  10,000 units of the electrical component units;  1,500 labor hours. PBI INDUSTRIES 99 PBI INDUSTRIES – SOLUTION  Decision variables X1 - the number of portable CD players to produce X2 - the number of deluxe CD players to produce  The model Max (150-. X X X X X ST X X X 01X1)X1+(350-.02X2)X2 - 50X1- 90X2-.01X1X2 = -.01X1 .1X1 +.3X2 1,500 - X1 2 ÷ ÷ + + + s s s ÷ s . . , 02 2 01 1 2 100 1 260 2 1 2 2 10 000 0 2 0 2 Production cannot be negative Resource constraints 100  For a point X1, X2 to be optimal, the K-T conditions require that: Y1S1 = 0; Y2S2 = 0; Y3S3 = 0; Y4S4 = 0, and ) 1 ( Y ) 0 ( Y ) 3 (. Y ) 2 ( Y 260 2 X 04 . X 01 . or X G Y X G Y X G Y X G Y X F ) 0 ( Y ) 1 ( Y ) 1 (. Y ) 1 ( Y 100 X 01 . X 02 . or X G 4 Y X G Y X G Y X G Y X F 4 3 2 1 1 2 4 4 2 3 3 2 2 2 2 1 1 2 4 3 2 1 2 1 1 4 1 3 3 1 2 2 1 1 1 1 ÷ + + + = + ÷ ÷ | . | \ | c c + | . | \ | c c + | . | \ | c c + | . | \ | c c = c c + ÷ + + = + ÷ ÷ | . | \ | c c + | . | \ | c c + | . | \ | c c + | . | \ | c c = c c PBI INDUSTRIES – SOLUTION 101  Finding an optimal production plan.  Assume X1>0 and X2>0. • The assumption implies S3>0 and S4>0. • Thus, Y3 = 0 and Y4 = 0.  Add the assumption that S1 = 0 and S2 = 0. • From the first two constraints we have X1 = 0 and X2 = 5000. X1= 0 A contradiction X1>0 AS A RESULT THE SECOND ASSUMPTION CANNOT BE TRUE PBI INDUSTRIES – SOLUTION 102  Change the second assumption. Assume that S1 = 0 and S2 > 0. • As before, from the first assumption S3 > 0 and S4 > 0. • Thus, Y3 = 0 and Y4 = 0. • From the second assumption Y2 = 0. • Substituting the values of all the Ys in the partial derivative equations we get the following two equations: -.02X1 - .01X2 + 100 = Y1 -.01X1 - .04X2 + 260 = 2Y1 • Also, since S1 = 0 (by the second assumption) X1 + 2X2 = 10,000 • Solving the set of three equations in three unknowns we get: PBI INDUSTRIES – SOLUTION 103 X1 = 1000, X2 = 4500, Y1 = 35  This solution is a feasible point (check the constraints). • X1 and X2 are positive. • 1X1 + 2X2 <= 10,000 [1000 + 2(4500) = 10,000] • .1X1 + .3X2 <= 1,500 [.1(1000) + .3(4500) = 1450]  This problem represents a convex program since • It can be shown that the objective function F is concave. • All the constraints are linear, thus, form a convex set. The K-T conditions yielded an optimal solution PBI INDUSTRIES – SOLUTION 104 Portable Deluxe Total Profit April Production 1000 4500 810000 Portable Deluxe Used Available Electrical Components 1 2 10000 10000 Labor Hours 0.1 0.3 1450 1500 PBI INDUSTRIES PBI INDUSTRIES – Excel SOLUTION 105 Copyright © 2002 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that named in Section 117 of the United States Copyright Act without the express written consent of the copyright owner is unlawful. Requests for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Adopters of the textbook are granted permission to make back-up copies for their own use only, to make copies for distribution to students of the course the textbook is used in, and to modify this material to best suit their instructional needs. Under no circumstances can copies be made for resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein.