Mathematics for Management Science Notes 0 prepared by Professor Jenny Baglivo Jenny A. Baglivo 00. All rights reserved. Integer Linear Programming (ILP) When the values of the decision variables in a linear programming problem are restricted to integers, we say that we have an integer linear programming problem or an ILP problem. When there are a large number of feasible solutions, we usually solve the LP relaxation problem (that is, the problem you get by ignoring the restriction to integer decision variables) and round the values at the end. When there are (relatively) few feasible solutions, it is best to solve the problem by trying all possibilities. Example ILP problem: Maximize OBJECTIVE = x + x Subject to: Constraint : x + x. Constraint :. x + x. Non-negativity: x 0, x 0 Integer: x, x integer () The feasible set for the ILP problem is the collection of points highlighted below. The feasible set for the LP relaxation problem is the gray area plus the boundary lines. The dashed line is x + x =.0. page of
() The solution to the ILP problem is x= and x=. The maximum value of the objective function is. Feasible Point OBJECTIVE Feasible Point OBJECTIVE (0,0) 0 (,0) (0,) (,) (0,) (,) Max (,0) (,0) (,) (,) (,) () The solution to the LP relaxation problem is x=/ and x=/. The maximum value of the objective function is.0. Corner Point OBJECTIVE value (0,0) 0 (0,/). (/,/).0 Max (/,0) () If you round the solution to the LP relaxation problem, then you get an infeasible solution (that is, one that does not satisfy the constraints). Bounding Principles: () The maximum value for an ILP problem is always less than or equal to the maximum value for the LP relaxation: MAX for ILP MAX for LP relaxation () The minimum value for an ILP problem is always greater than or equal to the minimum for the LP relaxation: MIN for ILP MIN for LP relaxation page of
Implementing ILP problems in Solver: Step : Add constraints for each integer decision variable: () Put the address of the decision variable in the lefthand side, () Select int where you would normally select a relation (,, =), () The word integer should appear on the righthand side. If nothing appears, then you should type =integer on the righthand side. Step : Make appropriate selections in the options window : () Assume Linear Model () Assume Non-negative () 0% Tolerance () 00 Iterations (or more, if needed). Comments: If one or more variables are integer, then Solver switches from the usual linear programming algorithm to one that searches the integer points. It uses a technique called branch-and-bound so that it doesn t have to look at every single integer point before finding the optimal solution. This algorithm does not produce sensitivity reports. The tolerance default value is %. If the % value stays in effect, then Solver will return a solution whose objective function value is within % of the best possible. For maximization problems, the stopping criterion is 0. * Maximum for LP relaxation Current objective function value. For minimization problems, the stopping criterion is Current objective function value.0 * Minimum for LP relaxation. page of
Simple example solution sheet: 0 0 A B C D E Simple Example Variable Variable Bound Constraint. Constraint.. Objective M O D E L Decision Variables Variable Variable #Units: M a x i m i z e Objective: Constraints L H S R H S Constraint <=. Constraint <=. Maximize: B By Changing: B:C Subject to: B:B <= C:D B:C = integer Options: Assume Linear Model Assume Non-negative 0% Tolerance 00 Iterations page of
Exercise : Air-Express is an express shipping service that guarantees overnight delivery of packages anywhere in the continental United States. The company has various operations centers, called hubs, at airports in major cities across the country. Packages are received at hubs from other locations and then shipped to intermediate hubs or to their final destinations. The manager of the Air-Express hub in Baltimore, Maryland, is concerned about labor costs at the hub and is interested in determining the most effective way to schedule workers. The hub operates seven days a week, and the number of packages it handles each day varies. Using historical data on the average number of packages received each day, the manager estimates the number of workers needed to handle the packages as: Day Sun Mon Tue Wed Thu Fri Sat # Required The package handlers working for Air-Express are unionized and are guaranteed a five-day work week with two consecutive days off. The base wage for the handlers is $ per week. Because most workers prefer to have Saturday or Sunday off, the union has negotiated bonuses of $ per day for its members who work on these days. The possible shifts and salaries for package handlers are: Shift Number Days Off Wages Sun and Mon Mon and Tue Tue and Wed Wed and Thu Thu and Fri Fri and Sat Sat and Sun The manager wants to keep the total wage expense for the hub as low as possible. With this in mind, how many package handlers should be assigned to each shift if the manager wants to have a sufficient number of workers available each day? Summarize the problem page of
Define the decision variables precisely Completely specify the ILP model page of
Clearly state the optimal solution Exercise solution sheet: 0 0 Air A B C D E F G H I J Express Shift Shift Shift Shift Shift Shift Shift Req: Sunday 0 0 Monday 0 0 Tuesday 0 0 Wednesday 0 0 Thursday 0 0 Friday 0 0 Saturday 0 0 Wages/wk: 0 0 0 0 0 0 M O D E L Decision Variables: Shift Shift Shift Shift Shift Shift Shift #Workers:.E- 0 M i n i m i z e Total Wages: 0 Subject to L H S R H S Sunday >= Monday >= Tuesday >= Wednesday >= Thursday >= Friday >= Saturday >= Days On =, Days Off = 0 Solver: Minimize B By Changing B:H Subject to B:B >= D:D B:H = integer Options: Assume Linear Model Assume Non-negative 0% Tolerance 00 Iterations Min page of
Formulas sheet: 0 0 Air A B C D Express Days On =, Days Off = 0 Shift Shift Shift Sunday 0 Monday 0 0 Tuesday 0 0 Wednesday 0 Thursday Friday Saturday Wages/wk: =+*(B+B) =+*(C+C) =+*(D+D) M O D E L Decision Variables: Shift Shift Shift #Workers: 0 0 0 M i n i m i z e Total Wages: =SUMPRODUCT(B:H,B:H) Subject to L H S Sunday =SUMPRODUCT(B:H,$B$:$H$) >= =J Monday =SUMPRODUCT(B:H,$B$:$H$) >= =J Tuesday =SUMPRODUCT(B:H,$B$:$H$) >= =J Wednesday =SUMPRODUCT(B:H,$B$:$H$) >= =J Thursday =SUMPRODUCT(B:H,$B$:$H$) >= =J Friday =SUMPRODUCT(B:H,$B$:$H$) >= =J Saturday =SUMPRODUCT(B:H,$B$:$H$) >= =J R H S 0 0 E F G H I J Min Shift Shift Shift Shift Req: 0 0 0 0 0 0 0 0 =+*(E+E) =+*(F+F) =+*(G+G) =+*(H+H) Shift Shift Shift Shift 0 0 0 0 page of
Programming with binary (0 ) variables To restrict an integer variable to the values 0 and only, add a constraint as follows:. Put the address of the decision variable in the lefthand side,. Select bin where you would normally select a relation (,, = ),. The word binary should appear on the righthand side. If nothing appears, then you should type =binary on the righthand side. The next exercise is an example of a capital budgeting problem, where the decision variable equals when a project is selected and equals 0 otherwise. Exercise : In his position as vice president of research and development (R&D) for CRT Technologies, Mark Schwartz is responsible for evaluating and choosing which R&D projects to support. The company received R&D proposals from its scientists and engineers and identified six projects as being consistent with the company s mission. However, the company does not have the funds available to undertake all six projects. Mark must determine which of the projects to select. The funding requirements for each project are summarized below, along with the expected net present value (NPV) the company expects each project to generate. Expected Capital (Thous $$) Required in: NPV Project (Thous $$) Year Year Year Year Year 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The company currently has $0,000 available to invest in new projects. It has budgeted $,000 for continued support for these projects in year and $0,000 per year in years,, and. Which projects should CRT support in order to maximize total expected NPV? Summarize the problem page of
Define the decision variables precisely Completely specify the ILP model Clearly state the optimal solution page of
Exercise solutions and formulas sheets: 0 0 A B C D E F G H CRT Tech. Thousands of Dollars: Max Project Project Project Project Project Project Avlbl: Year 0 0 0 0 0 0 Year 0 0 Year 0 0 0 0 Year 0 0 0 0 Year 0 0 0 0 NPV: M O D E L Decision Variables: Project Project Project Project Project Project Select? 0 0 0 (=Yes,0=No) M a x i m i z e NPV : Subject to: L H S R H S Year 0 <= 0 Year 0 <= Year 0 <= 0 Year 0 <= 0 Year <= 0 Maximize B By Changing B:G Subject to: B:B <= D:D B:G = binary Assume Linear Model Assume Non-negative 0% Tolerance 00 Iterations 0 A B C D E F G H CRT Tech. Thousands of Dollars: Max Project Project Project Project Project Project Avlbl: Year 0 0 0 0 0 0 Year 0 0 Year 0 0 0 0 Year 0 0 0 0 Year 0 0 0 0 NPV: M O D E L Decision Variables: Project Project Project Project Project Project Select? 0 0 0 0 0 0 (=Yes,0=No) M a x i m i z e NPV : =SUMPRODUCT(B:G,B:G) Subject to: L H S R H S Year =SUMPRODUCT(B:G,$B$:$G$) <= =H Year =SUMPRODUCT(B:G,$B$:$G$) <= =H Year =SUMPRODUCT(B:G,$B$:$G$) <= =H Year =SUMPRODUCT(B:G,$B$:$G$) <= =H Year =SUMPRODUCT(B:G,$B$:$G$) <= =H page of
The next exercise is an example of a fixed cost problem. It is assumed that the cost of production includes a setup cost (which is a fixed cost) and a variable cost (which is directly related to the quantity produced). The setup cost is only incurred if you choose to produce the product. Products are indexed using i=,, etc. For product i, let. X i equal the number of items produced,. Y i equal when the fixed cost is incurred and 0 otherwise, and. M i equal the maximum number that could be produced under given constraints. A constraint of the form X i M i * Y i assures that Y i = when and only when X i is positive. Exercise : Remington Manufacturing is planning its next production cycle. The company can produce three products, each of which must undergo machining, grinding, and assembly operations. The table below summarizes the hours of machining, grinding and assembling required for each unit, and the total hours of capacity available in the next production cycle. Hours Required by: Operation Product Product Product Hours Available Machining 00 Grinding 00 Assembly 00 Products,, and have unit profits of $, $, and $0, respectively. There are also production line setup costs associated with producing each product. These costs are $,000 for Product, $00 for Product, and $00 for Product. The marketing department believes it can sell all the products produced. Therefore, the management of Remington wants to determine the most profitable mix of products to produce. What are the values of M, M, and M? page of
Define the decision variables precisely Completely specify the ILP model page of
Clearly state the optimal solution Exercise solution sheet: 0 0 A B C D E F G Remington Mfg Hours: Max Product Product Product Avlbl: Machining 00 Grinding 00 Assembly 00 Unit Profit ($): 0 Setup charge ($): 00 00 00 Max #Units: 0. 0 M O D E L Decision Variables: Product Product Product #Units: 0 Setup? 0 (=Yes,0=No) Product Profits: 0 Fixed Charges: 00 Max Profit: 0 Subject to: L H S R H S Machining <= 00 Grinding <= 00 Assembly <= 00 Product Max 0 <= 0 Product Max <=. Product Max <= 0 Maximize B By Changing B:D Subject to: B:B <= D:D B:D = integer B:D = binary Assume Linear Model Assume Non-negative 0% Tolerance 00 Iterations page of
Formulas sheet: 0 0 A B C D E Remington Mfg Hours: Max Product Product Product Avlbl: Machining 00 Grinding 00 Assembly 00 Unit Profit ($): 0 Setup charge ($): 00 00 00 Max #Units: =MIN(E/B,E/B,E/B) =MIN(E/C,E/C,E/C) =MIN(E/D,E/D,E/D) M O D E L Decision Variables: Product Product Product #Units: 0 0 0 Setup? 0 0 0 (=Yes,0=No) Product Profits: =SUMPRODUCT(B:D,B:D) Fixed Charges: =SUMPRODUCT(B:D,B:D) Max Profit: =B-B Subject to: L H S R H S Machining =SUMPRODUCT(B:D,$B$:$D$) <= =E Grinding =SUMPRODUCT(B:D,$B$:$D$) <= =E Assembly =SUMPRODUCT(B:D,$B$:$D$) <= =E Product Max =B <= =B*B Product Max =C <= =C*C Product Max =D <= =D*D page of
Exercise : In the Remington Manufacturing problem, suppose that management does not want to produce a product unless it produces at least 0 units of that product. The sheet below gives the new optimal solution. 0 0 A B C D E Remington II Hours: Max Product Product Product Avlbl: Machining 00 Grinding 00 Assembly 00 Unit Profit ($): 0 Setup charge ($): 00 00 00 Max #Units: 0. 0 M O D E L Decision Variables: Product Product Product #Units: 0 0 Setup? 0 (=Yes,0=No) Product Profits: 0 Fixed Charges: 00 Max Profit: 0 Subject to: L H S R H S Machining <= 00 Grinding <= 00 Assembly <= 00 Product Max 0 <= 0 Product Max <=. Product Max 0 <= 0 Product Min 0 >= 0 Product Min >= 0 Product Min 0 >= 0 How does the original model change? page of
Exercise : The Martin-Beck Company operates a plant in St. Louis which has an annual production capacity of 0,000 units. The final product is shipped to regional distribution centers located in Boston, Atlanta, and Houston. Because of an anticipated increase in demand, Martin-Beck plans to increase capacity by constructing a new plant in one or more of the following cities: Detroit, Toledo, Denver, Kansas City. The estimated annual fixed cost and the annual capacity for the four proposed plants are as follows: Proposed Plant Annual Fixed Cost ($) Annual Capacity (units) Detroit,00,000 Toledo 0,000 0,000 Denver,000 0,000 Kansas City 00,000,000 The company's long-range planning group has developed the following forecasts of the anticipated annual demand at the distribution centers: Distribution Center Annual Demand (units) Boston 0,000 Atlanta 0,000 Houston 0,000 The shipping cost ($) per unit from each (proposed) plant to each distribution center is shown in the following table: to Boston to Atlanta to Houston From Detroit From Toledo From Denver From Kansas City From St. Louis Finally, Martin-Beck would like to locate a plant in either Detroit or Toledo, but not both. Martin-Beck would like to determine where to locate the new plant(s) and how much should be shipped from each plant to each distribution center in order to minimize the total annual fixed costs and transportation costs. Notation: Source nodes: Detroit, Toledo, Denver, Kansas City, St Louis Destination nodes: Boston, Atlanta, Houston page of
Define the decision variables precisely Completely specify the ILP model page of
Clearly state the optimal solution Exercise solution sheet: 0 0 A B C D E F G H I J M a r t i n - B e c k Annual Unit Shipping Costs: Production Fixed to Boston to Atlanta to Houston Capacity: Cost: from Detroit 000 00 from Toledo 0000 0000 from Denver 0000 000 from Kansas City 000 00000 from St. Louis 0000 M O D E L Demand: 0000 0000 0000 Transp. Costs: 0000 Fixed Costs: 00 Minimize Cost 00 Decision Variables: Setup? to Boston to Atlanta to Houston (=Yes,0=No) L H S R H S from Detroit 000 0 0 000 <= 000 from Toledo 0 0 0.E- 0 <=.E- from Denver.0E- 0 0 0.0E- <= 0 from Kansas City 0 000 0000 000 <= 000 from St. Louis 000 000 0 0000 <= 0000 L H S 0000 0000 0000 = = = R H S 0000 0000 0000 L H S R H S Detroit/Toledo = Minimize B By Changing B:D, F:F Subject to: H:H <= J:J B:D = B:D B = D B:D = integer F:F = binary Assume: linear model, non-negative 0% Tolerance 00 Iterations page of
Formula sheet: 0 0 A B C D E M a r t i n - B e c k Unit Shipping Costs: to Boston to Atlanta to Houston from Detroit from Toledo from Denver from Kansas City from St. Louis M O D E L Demand: 0000 0000 0000 Transp. Costs: =SUMPRODUCT(B:D,B:D) Fixed Costs: =SUMPRODUCT(G:G,F:F) Minimize Cost =B+B Decision Variables: to Boston to Atlanta to Houston from Detroit 0 0 0 from Toledo 0 0 0 from Denver 0 0 0 from Kansas City 0 0 0 from St. Louis 0 0 0 L H S =SUM(B:B) =SUM(C:C) =SUM(D:D) = = = R H S =B =C =D L H S Detroit/Toledo =F+F = R H S 0 0 F G H I J Annual Production Fixed Capacity: Cost: =.*000 =.*000 =.*0000 =.*00000 =0000 =000 =0.*0000 =0.*00000 0000 Setup? (=Yes,0=No) L H S R H S 0 =SUM(B:D) <= =F*F 0 =SUM(B:D) <= =F*F 0 =SUM(B:D) <= =F*F 0 =SUM(B:D) <= =F*F =SUM(B:D) <= =F page 0 of
Some modelling tips for binary variables: Suppose that X i = when a project is chosen and 0 otherwise, for i=,,..., n. () To choose exactly k of the projects, add the constraint X + X +... X n = k. Replace with when you want k or more projects chosen. Replace with when you want k or fewer projects chosen. () To choose project i only when project j is chosen, add the constraint X i = X j. () The conditional constraint has the following effect: X i X j If X i equals, then it forces X j to equal. If X j equals 0, then it forces X i to equal 0. page of
Exercise : Office Warehouse (OW) has been downsizing its operations. It is in the process of moving to a much smaller location and reducing the number of different computer products it carries. Coming under scrutiny are ten products OW has carried for the past year. For each of these products, OW has estimated the floor space required for effective display, the capital required to restock if the product line is retained, and the short-term loss that OW will incur if the corresponding product is eliminated (through liquidation sales, etc.) Cost to Cost to Floor Product Product Manufactured Liquidate Restock Space Number Line by (thous $) (thous $) (sqr.ft.) Notebook Toshiba 0 Notebook Compaq 0 PC Compaq 0 00 PC Packard Bell 00 Macintosh Apple 0 Monitor Packard Bell Monitor Sony 0 Printer Apple 0 Printer HP 0 Printer Epson ("Notebook" refers to a notebook computer.) Office Warehouse wishes to minimize the loss due to the liquidation of product lines subject to the following conditions: () At least four product lines will be eliminated. () The remaining products will occupy no more than 00 sqr. ft. of floor space. () At most $,000 is to be spent on restocking the product lines. () If one product from a particular manufacturer is eliminated, then all products from that manufacturer will be eliminated. () At least two of the five computer models (Toshiba NB, Compaq NB, Compaq PC, Packard Bell PC, Apple Macintosh) will continue to be carried by Office Warehouse. () If the Toshiba notebook computer is to be retained, then the Epson line of printers will also be retained. page of
Define the decision variables precisely Completely specify the ILP model page of
Clearly state the optimal solution Exercise solution sheet: 0 0 A B C D E F G H I J K Office Whse Product Numbers: Line: NB NB PC PC Mac Monitor Monitor Printer Printer Printer Mfg: Toshiba Compaq Compaq PB Apple PB Sony Apple HP Epson Liquidation Cost (thous $): 0 Restock Cost (thous $): 0 Floor space (sqr feet): 0 0 00 00 0 0 0 Max Space Min # Max Cost Used by Computer for Max # Remaining Models Restocking Remaining: (sqr ft): Remaining: (thous $): 00 M O D E L Decision Variables: Keep? 0 0 0 0 0 (=Yes,0=No) M i n i m i z e L i q u i d a t i o n : Subject to: L H S R H S #Remaining: <= Space: 0 <= 00 Restocking <= Compaq 0 = 0 PB 0 = 0 Apple = Computers >= If #, then # <= Minimize B By Changing B:K Subject to: B:B <= D:D B:B = D:D B >= D B <= D B:K = binary Assume: linear model, non-negative 0% Tolerance 00 Iterations page of
0 0 A B C D Office Whse Product Numbers: Line: NB NB PC Mfg: Toshiba Compaq Compaq Liquidation Cost (thous $): 0 Restock Cost (thous $): Floor space (sqr feet): 0 0 00 Max Space Used by Max # Remaining Remaining: (sqr ft): 00 M O D E L Decision Variables: Keep? 0 0 (=Yes,0=No) M i n i m i z e L i q u i d a t i o n : =SUM(B:K)-SUMPRODUCT(B:K,B:K) Subject to: L H S R H S #Remaining: =SUM(B:K) <= =B0 Space: =SUMPRODUCT(B:K,B:K) <= =D0 Restocking =SUMPRODUCT(B:K,B:K) <= =H0 Compaq =C = =D PB =E = =G Apple =F = =I Computers =SUM(B:F) >= =F0 If #, then # =B <= =K 0 E F G H I J K PC Mac Monitor Monitor Printer Printer Printer PB Apple PB Sony Apple HP Epson 0 00 0 0 0 Min # Max Cost Computer for Models Restocking Remaining: (thous $): 0 0 0 page of