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BA 452 Lesson B.4 Binary Fixed Costs 11ReadingsReadings Chapter 7 Integer Linear Programming

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BA 452 Lesson B.4 Binary Fixed Costs 22OverviewOverview

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33Overview Fixed Costs of Production are those production costs that are present whenever production is positive. The simplest model of fixed costs in a linear program restricts decision variables to binary (0 or 1). Fixed Costs of Assignment are modeled like Fixed Costs of Production. The simplest model of fixed costs restricts decision variables (the fraction of work completed) to be binary. Location Covering Problems are a special kind of Linear Programming problem when outputs are fixed because the firm has only established customers, who require coverage by service centers. Worker Covering Problems are like Location Covering Problems, where customers require scheduling workers so that at each time period is covered by a prescribed minimal number of workers. Transportation Problems with New Origins are Transportation Problems extended so that new origins may be added, at a fixed cost. They choose the best plant locations and how much to ship. Transshipment Problems with New Nodes are Transshipment Problems extended so that new transshipment nodes may be added, at a fixed cost. They choose the best transshipment locations.

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BA 452 Lesson B.4 Binary Fixed Costs 44 Tool Summary n Use binary variables to indicate whether an activity, such as a production run, is undertaken. n Write a multiple-choice constraint: The sum of two or more binary variables equals 1, so any feasible solution choose one variable to equal 1. n Write a mutually-exclusive constraint: The sum of two or more binary variables is at most 1, so any feasible solution chooses at most one variable to equal 1. All variables could equal 0. n Write a conditional constraint: An inequality constraint so that one binary variable cannot equal unless certain other binary variables also equal 1. n Write a corequisite constraint: An equality constraint of binary variables, so are either both 0 or both 1. Tool Summary

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BA 452 Lesson B.4 Binary Fixed Costs 55 Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 66 Overview Fixed Costs of Production are those production costs that are present whenever production is positive. The simplest way to model fixed costs in a linear program is to restrict decision variables to be binary (0 or 1). For example, suppose the cost of producing quantity x is 5x. On the one hand, if x can take on any non-negative value (like x is the pounds of hamburger produced), then 5 is the constant unit cost of production. On the other hand, if x can take on only binary values (like x is the number of new books adopted), then cost is 0 if there is no production and 5 if there is positive production, so 5 is the fixed cost of production. Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 77 Question: W. W. Norton & Company, the oldest and largest publishing company wholly owned by its employees, must decide which new textbooks to adopt next year. The books considered are described along with their expected three-year sales: Fixed Cost of Production Book Projected Sales Business Calculus 20 Finite Math 30 General Statistics 15 Mathematical Statistics 10 Business Statistics 25 Finance 18 Financial Accounting 25 Managerial Accounting 50 English Literature 20 German 30

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BA 452 Lesson B.4 Binary Fixed Costs 88 Three individuals in the company can be assigned to these projects, all of whom have varying amounts of time available. John has 60 days, Susan has 52 days, and Monica has 43 days. The days required by each person to complete each project are showing in the following table. For example, if the business calculus book is published, it will require 30 days of John’s time and 40 days of Susan’s time. Book Projected SalesJohnSusanMonica Business Calculus X Finite Math X General Statistics 1524X30 Mathematical Statistics 1020X24 Business Statistics 2510X16 Finance 18XX14 Financial Accounting 25X2426 Managerial Accounting 50X2830 English Literature German 30x5036 Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 99 Norton will not publish more than two statistics books or more than one accounting book in a single year. In addition, one of the math books (business calculus or finite math) must be published, but not both. Which books should be published, and what are the projected sales? If Monica has 1 more day available, which books should be published, and what are the projected sales? Comment. Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1010 iBook 1Business Calculus 2Finite Math 3General Statistics 4Mathematical Statistics 5Business Statistics 6Finance 7Financial Accounting 8Managerial Accounting 9English Literature 10German Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1111 Max 20x1+30x2+15x3+10x4+25x5+18x6+25x7+50x8+20x9+30x10 s.t. 30x1+16x2+24x3+20x4+10x5+40x9 60 John 40x1+24x2+24x7+28x8+34x9+50x10 52 Susan 30x3+24x4+16x5+14x6+26x7+30x8+30x9+36x10 43 Monica x3x3+x4x4+x5x5 2 No. of Stat Books x7x7+x8x8 1 Account Book x1x1+x2x2= 1 Math Book iBook Projected SalesJohnSusanMonica 1Business Calculus X 2Finite Math X 3General Statistics 1524X30 4Mathematical Statistics 1020X24 5Business Statistics 2510X16 6Finance 18XX14 7Financial Accounting 25X2426 8Managerial Accounting 50X2830 9English Literature German 30x5036 Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1212 Max 20x1+30x2+15x3+10x4+25x5+18x6+25x7+50x8+20x9+30x10 s.t. 30x1+16x2+24x3+20x4+10x5+40x9 60 John 40x1+24x2+24x7+28x8+34x9+50x10 52 Susan 30x3+24x4+16x5+14x6+26x7+30x8+30x9+36x10 43 Monica x3x3+x4x4+x5x5 2 No. of Stat Books x7x7+x8x8 1 Account Book x1x1+x2x2= 1 Math Book Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1313 Publish finite math, business statistics, and financial accounting. Projected sales are 80,000. iBook 1Business Calculus 2Finite Math 3General Statistics 4Mathematical Statistics 5Business Statistics 6Finance 7Financial Accounting 8Managerial Accounting 9English Literature 10German Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1414 Publish finite math, business statistics, and financial accounting. Projected sales are 80,000. iBook 1Business Calculus 2Finite Math 3General Statistics 4Mathematical Statistics 5Business Statistics 6Finance 7Financial Accounting 8Managerial Accounting 9English Literature 10German Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1515 If Monica is available 1 more day (44 days total), optimal projected sales are now 98,000. A big (discrete) gain for a small increase in a constraint. Fixed Cost of Production

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BA 452 Lesson B.4 Binary Fixed Costs 1616 Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 1717 Overview Fixed Costs of Assignment are modeled like Fixed Costs of Production. The simplest model of fixed costs restricts decision variables to be binary. For example, suppose the cost of worker i performing the fraction x ij of job j is 5x ij. On the one hand, if x ij can take on any non-negative value between 0 and 1 (like x ij is the fraction of the job performed), then 5 is the constant unit cost of assignment (like the time cost of working). On the other hand, if x ij can take on only binary values, then cost is 0 if you are not assigned (x ij = 0) and 5 if you are assigned the entire job (x ij = 1), so 5 is the fixed cost of assignment (like the cost of commuting to a job across town). Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 1818 Question: Tina's Tailoring has five idle tailors and four custom garments to make. The estimated time (in hours) it would take each tailor to make each garment is shown in the next slide. (An 'X' in the table indicates an unacceptable tailor-garment assignment.) Tailor Tailor Garment Garment Wedding gown Wedding gown Clown costume X Clown costume X Admiral's uniform X 9 Admiral's uniform X 9 Bullfighter's outfit X Bullfighter's outfit X Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 1919 Formulate an integer program for determining the tailor- garment assignments that minimize the total estimated time spent making the four garments. No tailor is to be assigned more than one garment, and each garment is to be worked on by only one tailor. This particular problem can be formulated as either a binary program or as an integer program. Any feasible solution to the latter program is binary (0-1). Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 2020 Answer: Define the decision variables = 1 if garment i is assigned to tailor j = 1 if garment i is assigned to tailor j = 0 otherwise. = 0 otherwise. n Find the number of decision variables = [(number of garments)x(number of tailors)] = [(number of garments)x(number of tailors)] - (number of unacceptable assignments) - (number of unacceptable assignments) = [4x5] - 3 = 17 = [4x5] - 3 = 17 n Find the number of constraints. 1 for each garment and each tailor = 9. x ij Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 2121 n Define the objective function Minimize total time spent making garments: Minimize total time spent making garments: Min 19x x x x x x 21 Min 19x x x x x x x x x x x x x x x x x x x x x x x x x x x x 45 Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 2222 n Define the constraints of exactly one tailor per garment: 1) x 11 + x 12 + x 13 + x 14 + x 15 = 1 1) x 11 + x 12 + x 13 + x 14 + x 15 = 1 2) x 21 + x 22 + x 24 + x 25 = 1 2) x 21 + x 22 + x 24 + x 25 = 1 3) x 31 + x 32 + x 33 + x 35 = 1 3) x 31 + x 32 + x 33 + x 35 = 1 4) x 42 + x 43 + x 44 + x 45 = 1 4) x 42 + x 43 + x 44 + x 45 = 1 Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 2323 n Define the constraints of no more than one garment per tailor: 5) x 11 + x 21 + x 31 < 1 6) x 12 + x 22 + x 32 + x 42 < 1 7) x 13 + x 33 + x 43 < 1 8) x 14 + x 24 + x 44 < 1 9) x 15 + x 25 + x 35 + x 45 < 1 Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 2424 Tailor Tailor Garment Garment Wedding gown Wedding gown Clown costume X Clown costume X Admiral's uniform X 9 Admiral's uniform X 9 Bullfighter's outfit X Bullfighter's outfit X Minimum time: 55 hours Optimal assignments: Fixed Cost of Assignment

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BA 452 Lesson B.4 Binary Fixed Costs 2525 Location Covering

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BA 452 Lesson B.4 Binary Fixed Costs 2626 Overview Location Covering Problems are a special kind of Linear Programming problem when outputs are fixed because the firm has only established customers. Commitments to established customers require building service centers so that each customer area is covered by a prescribed minimal number of service centers. The objective is to minimize the cost of building service centers. If each center is equally costly, the objective reduces to minimizing the number of service centers. The simplest way to model the all-or-nothing decision to build or not build a serve center is to restrict the fraction of the center built to be a binary (0 or 1) decision variable. Location Covering

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BA 452 Lesson B.4 Binary Fixed Costs 2727 Question: UPS is drawing up new zones for the location of drop boxes for customers. The city has been divided into the four zones shown below. You have targeted six possible locations for drop boxes (numbered 1 through 6). The list of which drop boxes could be reached easily from each zone is listed below. Zone Can be Served by Drop Box Locations: Downtown Financial 1, 2, 5, 6 Downtown Legal 2, 4, 5 Retail South 1, 2, 4, 6 Retail North 3, 4, 5 Location Covering

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BA 452 Lesson B.4 Binary Fixed Costs 2828 Formulate and solve a model to provide the fewest drop- box locations yet make sure that each zone is covered by at least two boxes. Location Covering

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BA 452 Lesson B.4 Binary Fixed Costs 2929 Answer: Min x 1 + x 2 + x 3 + x 4 + x 5 + x 6 s.t.x 1 + x 2 + x 5 + x 6 > 2 x 2 + x 4 + x 5 > 2 x 1 + x 2 + x 4 + x 6 > 2 x 3 + x 4 + x 5 > 2 x 3 + x 4 + x 5 > 2 Location Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3030 Worker Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3131 Overview Worker Covering Problems are like Location Covering Problems. Outputs are fixed because the firm has only established customers. Commitments to established customers require scheduling workers so that at each time period customer needs are covered by a prescribed minimal number of workers. The objective is to minimize the cost of scheduling workers. If each worker is equally costly, the objective reduces to minimizing the number of workers scheduled. The simplest way to model the all-or- nothing decision to schedule a worker is to restrict the number of workers scheduled at each time period to be an integer decision variable. Worker Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3232 Question: Amazon.com is open 24 hours a day. The number of phone operators need in each four hour period of a day is listed below. Period Operators Needed 10 p.m. to 2 a.m. 8 2 a.m. to 6 a.m. 2 a.m. to 6 a.m.4 6 a.m. to 10 a.m. 6 a.m. to 10 a.m.7 10 a.m. to 2 p.m p.m. to 6 p.m p.m. to 10 p.m. 15 Worker Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3333 Suppose operators work for eight consecutive hours. Formulate and solve the company’s problem of determining how many operators should be scheduled to begin working in each period in order to minimize the number of cashiers needed? (Hint: Workers can work from 6 p.m. to 2 a.m.) Worker Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3434 Answer: Define the decision variables TNP = the number of operators who begin working at 10 p.m. TWA = the number of operators who begin working at 2 a.m. SXA = the number of operators who begin working at 6 a.m. TNA = the number of operators who begin working at 10 a.m. TWP = the number of operators who begin working at 2 p.m. SXP = the number of operators who begin working at 6 p.m. Min TNP + TWA + SXA + TNA + TWP + SXP s.t.TNP + TWA > 4 TWA + SXA > 7 SXA + TNA > 12 TNA + TWP > 10 TWP + SXP > 15 SXP + TNP > 8, all variables > 0 Worker Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3535 Worker Covering

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BA 452 Lesson B.4 Binary Fixed Costs 3636 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 3737 Transportation Problems with New Origins are Transportation Problems extended so that new origins may be added, at a fixed cost. Also called Distribution System Design Problems, they choose the best plant locations and determine how much to ship from each plant. The simplest way to model those fixed costs in a linear program is with binary (0 or 1) variables. For example, consider a Transportation Problem extended by allowing the possibility of developing a new origin. Suppose the fixed development cost is 5 and the new origin’s supply capacity is 7. That is a linear program with a supply of 7Y at the new origin and added cost term of 5Y, where Y = 0 indicates the new origin is not developed and Y = 1 indicates the new origin is developed. Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 3838 Question: Linksys operates a plant to produce its wireless routers in St. Louis with an annual capacity of 30,000 units. Product is shipped to regional distribution centers in Boston, Atlanta, and Houston. Because of an anticipated increase in demand, Linksys plans to increase capacity by constructing a new plant in one or more of the following cities: Detroit, Toledo, Denver, or Kansas City. The estimated annual fixed cost and annual capacity for the four proposed plants are as follows: Proposed Plant Annual Fixed Cost Annual Capacity Detroit$175,00010,000 Toledo$300,00020,000 Denver$375,00030,000 Kansas City $500,00040,000 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 3939 The company’s long-range planning group forecasts of the anticipated annual demand at the distribution centers are as follows: Distribution Center Annual Demand Boston30,000 Atlanta20,000 Houston20,000 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4040 The shipping cost per unit from each plant to each distribution center is as follows: Formulate and solve the problem of minimizing the cost of meeting all demands. Plant\DistributionBostonAtlantaHouston Detroit$5$2$3 Toledo$4$3$4 Denver$9$7$5 Kansas City $10$4$2 St. Louis $8$4$3 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4141 Answer: Define binary variables for plant construction, Y1 = 1 if a plan is constructed in Detroit; 0, if not Y2 = 1 if a plan is constructed in Toledo; 0, if not Y3 = 1 if a plan is constructed in Denver; 0, if not Y4 = 1 if a plan is constructed in Kansas City; 0, if not Define shipment variables just as in transportation problems, Xij = the units shipped (in thousands) from plant i (i = 1, 2, 3, 4, 5) to distribution center j (j = 1, 2, 3) each year. Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4242 The objective is minimize total cost. From cost data shipping costs (in thousands of dollars) are 5X11 + 2X12 + 3X13+ 4X21 + 3X22 + 4X23 + 9X31 + 7X32 + 5X X41 + 4X42 + 2X43 + 8X51 + 4X52 + 3X53 Plant\DistributionBostonAtlantaHouston Detroit$5$2$3 Toledo$4$3$4 Denver$9$7$5 Kansas City $10$4$2 St. Louis $8$4$3 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4343 From cost data plant construction costs (in thousands of dollars) are 175Y Y Y Y4 Hence, the objective to minimize total costs is Min 5X11 + 2X12 + 3X13+ 4X21 + 3X22 + 4X23 + 9X31 + 7X32 + 5X X41 + 4X42 + 2X43 + 8X51 + 4X52 + 3X Y Y Y Y4 Proposed Plant Annual Fixed Cost Annual Capacity Detroit$175,00010,000 Toledo$300,00020,000 Denver$375,00030,000 Kansas City $500,00040,000 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4444 From capacity data Detroit capacity constraint is X11 + X12 + X13 < 10Y1 Toledo capacity constraint is X21 + X22 + X23 < 20Y2 Denver capacity constraint is X31 + X32 + X33 < 30Y3 Kansas City capacity constraint is X41 + X42 + X43 < 40Y4 And St. Louis capacity constraint is X51 + X52 + X53 < 30 Proposed Plant Annual Fixed Cost Annual Capacity Detroit$175,00010,000 Toledo$300,00020,000 Denver$375,00030,000 Kansas City $500,00040,000 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4545 From demand data Boston demand constraint is X11 + X21 + X31 + X41 + X51 = 30 Atlanta demand constraint is X12 + X22 + X32 + X42 + X52 = 20 Houston demand constraint is X13 + X23 + X33 + X43 + X53 = 20 Non-negativity constraints complete the linear programming formulation. Distribution Center Annual Demand Boston30,000 Atlanta20,000 Houston20,000 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4646 From demand data Boston demand constraint is X11 + X21 + X31 + X41 + X51 = 30 Atlanta demand constraint is X12 + X22 + X32 + X42 + X52 = 20 Houston demand constraint is X13 + X23 + X33 + X43 + X53 = 20 Non-negativity constraints complete the linear programming formulation. Distribution Center Annual Demand Boston30,000 Atlanta20,000 Houston20,000 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4747 The Management Scientist solves this mixed integer linear program of 4 binary variables Yj and 15 continuous variables Xij, and 8 constraints. Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4848 Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 4949 The Management Scientist solves this mixed integer linear program of 4 binary variables Yj and 15 continuous variables Xij, and 8 constraints. Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 5050 All variables at the optimum are zero except: X42 = 20, X43 = 20, X51 = 30, and Y4 = 1. So, the Kansas City plant should be built; 20,000 units should be shipped from Kansas City to Atlanta; 20,000 units should be shipped from Kansas City to Houston; and 30,000 units should be shipped from St. Louis to Boston. Transportation with New Origins

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BA 452 Lesson B.4 Binary Fixed Costs 5151 Transshipment with New Nodes

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BA 452 Lesson B.4 Binary Fixed Costs 5252 Overview Transshipment Problems with New Transshipment Nodes are Transshipment Problems extended so that new transshipment nodes may be added, at a fixed cost. They choose the best transshipment locations and determine how much to ship through each location. The simplest way to model those fixed costs in a linear program is with binary (0 or 1) variables. Transshipment with New Nodes

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BA 452 Lesson B.4 Binary Fixed Costs 5353 Question: Zeron Industries supplies three firms (Zrox, Hewes, Rockrite) with customized shelving for its offices. Zeron orders shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently, weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockrite. Both Arnold and Supershelf can supply up to 75 units to its customers. Zeron currently ships from its Northside facilities, but it can develop Southside facilities for a fixed cost of 8. Unit costs from the manufacturers to the suppliers are: Zeron N Zeron S Zeron N Zeron S Arnold 5 8 Arnold 5 8 Supershelf 7 4 Supershelf 7 4 The costs to install the shelving at the various locations are: The costs to install the shelving at the various locations are: Zrox Hewes Rockrite Zrox Hewes Rockrite Zeron N Zeron N Zeron S Zeron S Transshipment with New Nodes

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BA 452 Lesson B.4 Binary Fixed Costs 5454 ARNOLD WASH BURN HEWES Arnold SuperShelf Hewes Zrox ZeronN ZeronS Rock-Rite Transshipment with New Nodes

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BA 452 Lesson B.4 Binary Fixed Costs 5555 n Define decision variables: x ij = amount shipped from manufacturer i to supplier j x jk = amount shipped from supplier j to customer k x jk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) where i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) y = 1 if Zeron S is developed, y = 0 if Zeron S is not developed y = 1 if Zeron S is developed, y = 0 if Zeron S is not developed Transshipment with New Nodes

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BA 452 Lesson B.4 Binary Fixed Costs 5656 n Define objective function: Minimize total shipping costs plus setup costs. Min 5x x x x x x x x x x y Min 5x x x x x x x x x x y n Constrain amount out of Arnold: x 13 + x 14 < 75 n Constrain amount out of Supershelf: x 23 + x 24 < 75 n Constrain amount through Zeron N: x 13 + x 23 - x 35 - x 36 - x 37 = 0 n Constrain amount through Zeron S: x 14 + x 24 - x 45 - x 46 - x 47 = 0 n Constrain amount into Zrox: x 35 + x 45 = 50 n Constrain amount into Hewes: x 36 + x 46 = 60 n Constrain amount into Rockrite: x 37 + x 47 = 40 n Setup indicator for Zeron S: x 45 + x 46 + x 47 < 150y (The first 4 constraints imply x 45 + x 46 + x 47 < 150, so the setup indicator constraint “x 45 + x 46 + x 47 < 150y” means, at an optimum, y = 1 if any material x 45 or x 46 or x 47 transships through Zeron S, and y = 0 if no material x 45 or x 46 or x 47 transships through Zeron S.) Transshipment with New Nodes

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BA 452 Lesson B.4 Binary Fixed Costs 5757 BA 452 Quantitative Analysis End of Lesson B.4

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