University:
California State University, Northridge
Course:
MGT 360 | Management and Organizational Behavior
Academic year:

2024
Views:

386

Pages:

46
Author:

Efrain Villa

Chapter 7 Integer Linear Programming     Types of Integer Linear Programming Models Graphical and Computer Solutions for an AllInteger Linear Program Applications Involving 0-1 Variables Modeling Flexibility Provided by 0-1 Variables Types of Integer Programming Models     An LP in which all the variables are restricted to be integers is called an all-integer linear program (ILP). The LP that results from dropping the integer requirements is called the LP Relaxation of the ILP. If only a subset of the variables are restricted to be integers, the problem is called a mixed-integer linear program (MILP). Binary variables are variables whose values are restricted to be 0 or 1. If all variables are restricted to be 0 or 1, the problem is called a 0-1 or binary integer linear program. 3 Example: All-Integer LP  Consider the following all-integer linear program: Max 3x1 + 2x2 s.t. 3x1 + x2 < 9 x1 + 3x2 < 7 -x1 + x2 < 1 x1, x2 > 0 and integer 4 Example: All-Integer LP  LP Relaxation Solving the problem as a linear program ignoring the integer constraints, the optimal solution to the linear program gives fractional values for both x1 and x2. The optimal solution to the linear program is: x1 = 2.5, x2 = 1.5, Max 3x1 + 2x2 = 10.5 If we round up the fractional solution (x1 = 2.5, x2 = 1.5) to the LP relaxation problem, we get x1 = 3 and x2 = 2. By checking the constraints, we see that this point lies outside the feasible region, making this solution infeasible. 5 Example: All-Integer LP  Rounding Down By rounding the optimal solution down to x1 = 2, x2 = 1, we see that this solution indeed is an integer solution within the feasible region, and substituting in the objective function, it gives 3x1 + 2x2 = 8. We have found a feasible all-integer solution, but have we found the OPTIMAL all-integer solution? --------------------The answer is NO! The optimal solution is x1 = 3 and x2 = 0 giving 3x1 + 2x2 = 9, as evidenced in the next slide. 6 Example: All-Integer LP  Complete Enumeration of Feasible ILP Solutions There are eight feasible integer solutions to this problem: x1 x2 3x1 + 2x2 1. 2. 3. 4. 5. 6. 7. 8. 0 1 2 3 0 1 2 1 0 0 0 0 1 1 1 2 0 3 6 9 2 5 8 7 optimal solution 7 Example: Capital Budgeting The Ice-Cold Refrigerator Company is considering investing in several projects that have varying capital requirements over the next four years. Faced with limited capital each year, management would like to select the most profitable projects. The estimated net present value for each project, the capital requirements, and the available capital over the fouryear period are shown on the next slide. 8 Example: Capital Budgeting  Problem Data 9 Example: Capital Budgeting  Decision Variables The four 0-1 decision variables are as follows: P = 1 if the plant expansion project is accepted; 0 if rejected W = 1 if the warehouse expansion project is accepted; 0 if rejected M = 1 if the new machinery project is accepted; 0 if rejected R = 1 if the new product research project is accepted; 0 if rejected 10 Example: Capital Budgeting  Problem Formulation Max 90P + 40W + 10M + 37R s.t. 15P + 10W + 10M + 15R < 40 20P + 15W + 10R < 50 20P + 20W + 10R < 40 15P + 5W + 4M + 10R < 35 P, W, M, R = 0, 1 (Yr. 1 capital avail.) (Yr. 2 capital avail. (Yr. 3 capital avail.) (Yr. 4 capital avail.) 11 Example: Capital Budgeting  Optimal Solution P = 1, W = 1, M = 1, R = 0. Total estimated net present value = $140,000. The company should fund the plant expansion, the warehouse expansion, and the new machinery projects. The new product research project should be put on hold unless additional capital funds become available. The company will have $5,000 remaining in year 1, $15,000 remaining in year 2, and $11,000 remaining in year 4. Additional capital funds of $10,000 in year 1 and $10,000 in year 3 will be needed to fund the new product research project. 12 Example: Fixed Cost Three raw materials are used to produce 3 products: a fuel additive, a solvent base, and a carpet cleaning fluid. The profit contributions are $40 per ton for the fuel additive, $30 per ton for the solvent base, and $50 per ton for the carpet cleaning fluid. Each ton of fuel additive is a blend of 0.4 tons of material 1 and 0.6 tons of material 3. Each ton of solvent base requires 0.5 tons of material 1, 0.2 tons of material 2, and 0.3 tons of material 3. Each ton of carpet cleaning fluid is a blend of 0.6 tons of material 1, 0.1 tons of material 2, and 0.3 tons of material 3. 13 Example: Fixed Cost RMC has 20 tons of material 1, 5 tons of material 2, and 21 tons of material 3, and is interested in determining the optimal production quantities for the upcoming planning period. There is a fixed cost for production setup of the products, as well as a maximum production quantity for each of the three products. Product Setup Cost Fuel additive $200 Solvent base $ 50 Cleaning fluid $400 Maximum Production 50 tons 25 tons 40 tons 14 Example: Fixed Cost  Decision Variables F = tons of fuel additive produced S = tons of solvent base produced C = tons of carpet cleaning fluid produced SF = 1 if the fuel additive is produced; 0 if not SS = 1 if the solvent base is produced; 0 if not SC = 1 if the cleaning fluid is produced; 0 if not 15 Example: Fixed Cost  Problem Formulation Max 40F + 30S + 50C – 200SF – 50SS – 400SC s.t. 0.4F + 0.5S + 0.6C < 20 0.2S + 0.1C < 5 0.6F + 0.3S + 0.3C < 21 F - 50SF < 0 S - 25SS < 0 C - 50SF < 0 F, S, C > 0; SF, SS, SC = 0, 1 (Mat’l. 1) (Mat’l. 2) (Mat’l. 3) (Max.F) (Max. S) (Max. C) 16 Example: Fixed Cost  Optimal Solution Produce 25 tons of fuel additive. Produce 20 tons of solvent base. Produce 0 tons of cleaning fluid. The value of the objective function after deducting the setup cost is $1350. The setup cost for the fuel additive and the solvent base is $200 + $50 = $250. The optimal solution shows SC = 0, which indicates that the more expensive $400 setup cost for the carpet cleaning fluid should be avoided. 17 Example: Distribution System Design The Martin-Beck Company operates a plant in St. Louis with an annual capacity of 30,000 units. Product is shipped to regional distribution centers located in Boston, Atlanta, and Houston. Because of an anticipated increase in demand, Martin-Beck plans to increase capacity by constructing a new plant in one or more of the following cities: Detroit, Toledo, Denver, or Kansas City. 18 Example: Distribution System Design The estimated annual fixed cost and the annual capacity for the four proposed plants are as follows: Proposed Plant Annual Fixed Cost Annual Capacity Detroit $175,000 10,000 Toledo $300,000 20,000 Denver $375,000 30,000 Kansas City $500,000 40,000 19 Example: Distribution System Design The company’s long-range planning group developed forecasts of the anticipated annual demand at the distribution centers as follows: Distribution Center Annual Demand Boston 30,000 Atlanta 20,000 Houston 20,000 20 Example: Distribution System Design The shipping cost per unit from each plant to each distribution center is shown below. 21 Example: Distribution System Design  Decision Variables y1 = 1 if a plant is constructed in Detroit; 0 if not y2 = 1 if a plant is constructed in Toledo; 0 if not y3 = 1 if a plant is constructed in Denver; 0 if not y4 = 1 if a plant is constructed in Kansas City; 0 if not xij = the units shipped (in 1000s) from plant i to distribution center j , with i = 1, 2, 3, 4, 5 and j = 1, 2, 3 22 Example: Distribution System Design  Problem Formulation 23 Example: Distribution System Design  Optimal Solution Construct a plant in Kansas City (y4 = 1). 20,000 units will be shipped from Kansas City to Atlanta (x42 = 20), 20,000 units will be shipped from Kansas City to Houston (x43 = 20), and 30,000 units will be shipped from St. Louis to Boston (x51 = 30). The total cost of this solution including the fixed cost of $500,000 for the plant in Kansas City is $860,000. 24 Example: Bank Location The long-range planning department for the Ohio Trust Company is considering expanding its operation into a 20-county region in northeastern Ohio. Ohio Trust does not have, at this time, a principal place of business in any of the 20 counties. According to the banking laws in Ohio, if a bank establishes a principal place of business (PPB) in any county, branch banks can be established in that county and in any adjacent county. To establish a new PPB, Ohio Trust must either obtain approval for a new bank from the state’s superintendent of banks or purchase an existing bank. 25 Example: Bank Location The 20 counties in the region and adjacent counties are listed on the next slide. For example, Ashtabula County is adjacent to Lake, Geauga, and Trumbull counties; Lake County is adjacent to Ashtabula, Cuyahoga, and Geauga counties; and so on. As an initial step in its planning, Ohio Trust would like to determine the minimum number of PPBs necessary to do business throughout the 20-county region. A 0-1 integer programming model can be used to solve this location problem for Ohio Trust. 26 Example: Bank Location 27 Example: Bank Location  Decision Variables xi = 1 if a PBB is established in county i; 0 otherwise  Problem Formulation 28 Example: Bank Location  Optimal Solution For this 20-variable, 20-constraint problem: Establish PPBs in Ashland, Stark, and Geauga counties. (With PPBs in these three counties, Ohio Trust can place branch banks in all 20 counties.) All other decision variables have an optimal value of zero, indicating that a PPB should not be placed in these counties. 29 Example: Product Design & Market Share Market Pulse Research has conducted a study for Lucas Furniture on some designs for a new commercial office desk. Three attributes were found to be most influential in determining which desk is most desirable: number of file drawers, the presence or absence of pullout writing boards, and simulated wood or solid color finish. Listed on the next slide are the part-worths for each level of each attribute provided by a sample of 7 potential Lucas customers. 30 Example: Product Design & Market Share  Part-Worths File Drawer Pullout Writing Boards Finish Consumer 0 1 2 Present Absent Simulated Wood Solid Color 1 5 26 20 18 11 17 10 2 18 11 5 12 16 15 26 3 4 16 22 7 13 11 19 4 12 8 4 18 9 22 14 5 19 9 3 4 14 30 19 6 6 15 21 8 17 20 11 7 9 6 3 13 5 16 28 31 Example: Product Design & Market Share Suppose the overall utility (sum of part-worths) of the current favorite commercial office desk is 50 for each customer. What is the product design that will maximize the share of choices for the seven sample participants? Formulate and solve this 0 – 1 integer programming problem. 32 Example: Product Design & Market Share  Decision Variables There are 7 lij decision variables, one for each level of attribute. lij = 1 if Lucas chooses level i for attribute j; 0 otherwise. There are 7 Yk decision variables, one for each consumer in the sample. Yk = 1 if consumer k chooses the Lucas brand; 0 otherwise 33 Example: Product Design & Market Share  Objective Function Maximize the number of consumers preferring the Lucas brand desk. Max Y1 + Y2 + Y3 + Y4 + Y5 + Y6 + Y7 34 Example: Product Design & Market Share  Constraints There is one constraint for each consumer in the sample. 5l11 + 26l21 + 20l31 + 18l12 + 11l22 + 17l13 + 10l23 – 50Y1 > 1 18l11 + 11l21 + 5l31 + 12l12 + 16l22 + 15l13 + 26l23 – 50Y2 > 1 4l11 + 16l21 + 22l31 + 7l12 + 13l22 + 11l13 + 19l23 – 50Y3 > 1 12l11 + 8l21 + 4l31 + 18l12 + 9l22 + 22l13 + 14l23 – 50Y4 > 1 19l11 + 9l21 + 3l31 + 4l12 + 14l22 + 30l13 + 19l23 – 50Y5 > 1 6l11 + 15l21 + 21l31 + 8l12 + 17l22 + 20l13 + 11l23 – 50Y6 > 1 9l11 + 6l21 + 3l31 + 13l12 + 5l22 + 16l13 + 28l23 – 50Y7 > 1 35 Example: Product Design & Market Share  Constraints There is one constraint for each attribute. l11 + l21 + l31 = 1 l12 + l22 = 1 l13 + l23 = 1 36 Example: Product Design & Market Share  Optimal Solution Lucas should choose these product features: 1 file drawer (l21 = 1) No pullout writing boards (l22 = 1) Simulated wood finish (l13 = 1) Three sample participants would choose the Lucas design: Participant 1 (Y1 = 1) Participant 5 (Y5 = 1) Participant 6 (Y6 = 1) 37 Modeling Flexibility Provided by 0-1 Variables  When xi and xj represent binary variables designating whether projects i and j have been completed, the following special constraints may be formulated: • • • • At most k out of n projects will be completed: Σxj < k j Project j is conditional on project i: xj - xi < 0 Project i is a corequisite for project j: xj - xi = 0 Projects i and j are mutually exclusive: xi + xj < 1 38 Example: Metropolitan Microwaves Metropolitan Microwaves, Inc. is planning to expand its sales operation by offering other electronic appliances. The company has identified seven new product lines it can carry. Relevant information about each line follows on the next slide. 39 Example: Metropolitan Microwaves Product Line 1. 2. 3. 4. 5. 6. 7. TV/VCRs TVs Projection TVs VCRs DVD Players Video Games Home Computers Initial Floor Space Exp. Rate Invest. (Sq.Ft.) of Return $ 6,000 12,000 20,000 14,000 15,000 2,000 32,000 125 150 200 40 40 20 100 8.1% 9.0 11.0 10.2 10.5 14.1 13.2 40 Example: Metropolitan Microwaves Metropolitan has decided that they should not stock projection TVs unless they stock either TV/VCRs or TVs. Also, they will not stock both VCRs and DVD players, and they will stock video games if they stock TVs. Finally, the company wishes to introduce at least three new product lines. If the company has $45,000 to invest and 420 sq. ft. of floor space available, formulate an integer linear program for Metropolitan to maximize its overall expected return. 41 Example: Metropolitan Microwaves  Define the Decision Variables xj = 1 if product line j is introduced; = 0 otherwise. where: Product line 1 = TV/VCRs Product line 2 = TVs Product line 3 = Projection TVs Product line 4 = VCRs Product line 5 = DVD Players Product line 6 = Video Games Product line 7 = Home Computers 42 Example: Metropolitan Microwaves  Define the Decision Variables xj = 1 if product line j is introduced; = 0 otherwise.  Define the Objective Function Maximize total expected return: Max .081(6000)x1 + .09(12000)x2 + .11(20000)x3 + .102(14000)x4 + .105(15000)x5 + .141(2000)x6 + .132(32000)x7 43 Example: Metropolitan Microwaves  Define the Constraints 1) Money: 6x1 + 12x2 + 20x3 + 14x4 + 15x5 + 2x6 + 32x7 < 45 2) Space: 125x1 +150x2 +200x3 +40x4 +40x5 +20x6 +100x7 < 420 3) Stock projection TVs only if stock TV/VCRs or TVs: x1 + x2 > x3 or x1 + x2 - x3 > 0 44 Example: Metropolitan Microwaves  Define the Constraints (continued) 4) Do not stock both VCRs and DVD players: x4 + x5 < 1 5) Stock video games if they stock TV's: x2 - x6 > 0 6) Introduce at least 3 new lines: x1 + x2 + x3 + x4 + x5 + x6 + x7 > 3 7) Variables are 0 or 1: xj = 0 or 1 for j = 1, , , 7 45 Example: Metropolitan Microwaves  Optimal Solution Introduce: TV/VCRs, Projection TVs, and DVD Players Do Not Introduce: TVs, VCRs, Video Games, and Home Computers Total Expected Return: $4,261 46 Cautionary Note About Sensitivity Analysis    Sensitivity analysis often is more crucial for ILP problems than for LP problems. A small change in a constraint coefficient can cause a relatively large change in the optimal solution. Recommendation: Resolve the ILP problem several times with slight variations in the coefficients before choosing the “best” solution for implementation. 47