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Readings. Readings. Chapter 7 Integer Linear Programming. Overview. Overview. Overview. Tool Summary. Tool Summary Use binary variables to indicate whether an activity, such as a production run, is undertaken.
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Readings • Readings • Chapter 7 • Integer Linear Programming
Overview • Overview
Tool Summary Tool Summary • Use binary variables to indicate whether an activity, such as a production run, is undertaken. • 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. • 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. • Write a conditional constraint: An inequality constraint so that one binary variable cannot equal unless certain other binary variables also equal 1. • Write a corequisite constraint: An equality constraint of binary variables, so are either both 0 or both 1.
Fixed Cost of Production • Fixed Cost of Production
Fixed Cost of Production 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 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 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.
Fixed Cost of Production 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 Publish finite math, business statistics, and financial accounting. Projected sales are 80,000.
Fixed Cost of Production Publish finite math, business statistics, and financial accounting. Projected sales are 80,000.
Fixed Cost of Production 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 Assignment • Fixed Cost of Assignment
Fixed Cost of Assignment 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 xij of job j is 5xij. On the one hand, if xij can take on any non-negative value between 0 and 1 (like xij 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 xij can take on only binary values, then cost is 0 if you are not assigned (xij = 0) and 5 if you are assigned the entire job (xij = 1), so 5 is the fixed cost of assignment (like the cost of commuting to a job across town).
Fixed Cost of Assignment 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 Garment12345 Wedding gown 19 23 20 21 18 Clown costume 11 14 X 12 10 Admiral's uniform 12 8 11 X 9 Bullfighter's outfit X 20 20 18 21
Fixed Cost of Assignment 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 • Answer: Define the decision variables = 1 if garment i is assigned to tailor j = 0 otherwise. • Find the number of decision variables = [(number of garments)x(number of tailors)] - (number of unacceptable assignments) = [4x5] - 3 = 17 • Find the number of constraints. 1 for each garment and each tailor = 9. xij
Fixed Cost of Assignment • Define the objective function Minimize total time spent making garments: Min 19x11 + 23x12 + 20x13 + 21x14 + 18x15 + 11x21 + 14x22 + 12x24 + 10x25 + 12x31 + 8x32 + 11x33 + 9x35 + 20x42 + 20x43 + 18x44 + 21x45
Fixed Cost of Assignment • Define the constraints of exactly one tailor per garment: 1) x11 + x12 + x13 + x14 + x15 = 1 2) x21 + x22 + x24 + x25 = 1 3) x31 + x32 + x33 + x35 = 1 4) x42 + x43 + x44 + x45 = 1
Fixed Cost of Assignment • Define the constraints of no more than one garment per tailor: 5) x11 + x21 + x31< 1 6) x12 + x22 + x32 + x42< 1 7) x13 + x33 + x43< 1 8) x14 + x24 + x44< 1 9) x15 + x25 + x35 + x45< 1
Fixed Cost of Assignment Minimum time: 55 hours Optimal assignments: Tailor Garment12345 Wedding gown 19 23 20 21 18 Clown costume 11 14 X 12 10 Admiral's uniform 12 8 11 X 9 Bullfighter's outfit X 20 20 18 21
Location Covering • Location Covering
Location Covering 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 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.
Location Covering 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 Answer: Min x1 + x2 + x3 + x4 + x5 + x6 s.t. x1 + x2 + x5 + x6> 2 x2 + x4 + x5> 2 x1 + x2 + x4 + x6> 2 x3 + x4 + x5> 2
Worker Covering • Worker Covering
Worker Covering 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 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.
Worker Covering 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 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
Transportation with New Origins • Transportation with New Origins
Transportation with New Origins 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 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:
Transportation with New Origins The company’s long-range planning group forecasts of the anticipated annual demand at the distribution centers are as follows:
Transportation with New Origins 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.
Transportation with New Origins 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 The objective is minimize total cost. From cost data shipping costs (in thousands of dollars) are 5X11 + 2X12 + 3X13+ 4X21 + 3X22 + 4X23 + 9X31 + 7X32 + 5X33 + 10X41 + 4X42 + 2X43 + 8X51 + 4X52 + 3X53
Transportation with New Origins From cost data plant construction costs (in thousands of dollars) are 175Y1 + 300Y2 + 375Y3 + 500Y4 Hence, the objective to minimize total costs is Min 5X11 + 2X12 + 3X13+ 4X21 + 3X22 + 4X23 + 9X31 + 7X32 + 5X33 + 10X41 + 4X42 + 2X43 + 8X51 + 4X52 + 3X53 + 175Y1 + 300Y2 + 375Y3 + 500Y4
Transportation with New Origins 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
Transportation with New Origins 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.
Transportation with New Origins 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.
Transportation with New Origins 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 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 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.