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Integer linear programming. When all of the decision variables in an LP must take only integer values, the resulting model is referred to as integer linear programming ( ILP ) model.
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Integer linear programming • When all of the decision variables in an LP must take only integer values, the resulting model is referred to as integer linear programming (ILP) model. • When the model requires both continuous and integer decision variables, it is called mixed integer linear programming (MILP) model. • When the LP model includes binary decision variables, it is called binary… (BILP) • Real-world problems often require integer variables. Many examples…assignment, scheduling (people, machines, airplanes, crews, classes, fabric-rolls, probation officers, locating ambulances, …)
Formulating IPL, MILP, BILP problems • Integer variables (a.k.a. general integer variables) without any other restrictions can take the values of 0, 1, 2, 3, … • Binary integer variables (a.k.a binary variables) can only be 0 or 1. • Typically, stating that some or all variables must be integer or binary is rather easy. MAX:350X1 + 300X2 S.T.: 1X1 + 1X2 <= 200 9X1 + 6X2 <= 1566 12X1 + 16X2 <= 2880 X1, X2>= 0 and must be integer But, solving an ILP can be very time consuming.
Solving ILPs • With ILPs there are no more corners to jump to, in search of the best (optimum) corner! • Rounding or truncating the optimal solution does not guarantee optimality and feasibility. • Branch & bound based algorithms are widely used. • Consider: Max 5x1 + 8x2 st x1 + x2 <= 6 5x1 + 9x2 <= 45 x1, x2 >= 0 and integer
o.f.v.= x1 = x2 = o.f.v.= x1 = x2 = o.f.v.= x1 = x2 = o.f.v.= x1 = x2 = o.f.v.= x1 = x2 = o.f.v.= x1 = x2 = o.f.v.= x1 = x2 =
B & B Example 7 X2 X1 <= 1 X1 >=2 6 5 X2 >=4 X2 >=5 4 3 X2 <=3 X2 <=4 2 1 0 0 1 2 3 4 5 6 7 8 9 10 X1 X1=2.25, X2=3.75 ofv=41.25
An Employee Scheduling Problem: Air-Express Day of Week Workers Needed Sunday 18 Monday 27 Tuesday 22 Wednesday 26 Thursday 25 Friday 21 Saturday 19 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 from one day to the next. 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:
Air-Express Cont’d Shift Days Off Wage 1 Sun & Mon $680 2 Mon & Tue $705 3 Tue & Wed $705 4 Wed & Thr $705 5 Thr & Fri $705 6 Fri & Sat $680 7 Sat & Sun $655 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 $ 655 per week. Because most workers prefer to have Saturday or Sunday off, the union has negotiated bonuses of $ 25 per day for its members who work on these days. The possible shifts and salaries for package handlers are: 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? Objective? Decision variables? Model, setup in Excel, solve via ASPE.
Formulation and Excel Setup Xi = the number of workers assigned to shift i, i=1 ,7 Min 680X1 +705X2 +705X3 +705X4 +705X5 +680X6 +655X7 ST X2+ X3+ X4+ X5+ X6 >= 18 } Sunday X3+ X4+ X5+ X6+ X7 >= 27 } Monday X1+ X4+ X5+ X6+ X7 >= 22 }Tuesday X1+ X2+ X5+ X6+ X7 >= 26 } Wednesday X1+ X2+ X3+ X6+ X7 >= 25 } Thursday X1+ X2+ X3+ X4+ X7 >= 21 } Friday X1+ X2+ X3+ X4+ X5 >= 19 } Saturday Xi >= 0 and integer for all i
Standard Solver Settings for ILP, BILP, MILP models • To declare some variables as “Integer” • Add a new constraint where • LHS is the variable range, e.g., $C$2:$I$2 • Choose “Int” (or “Bin” for binary) and • RHS will read integer (or binary) • Then select “Options” • Uncheck Ignore Integer Constraints • Set Integer Optimality (%) to zero
ASPE Settings for ILP, BILP, MILP models Assume decision variables in C2:I2 are integer then must add the same variable range here And to make sure that the solver to find the best integer solution Select the “Engine” tab Enter zero for “Integer Tolerance”
Model X1 = batches of CD players to produce X2 = “ “ tape decks “ X3 = “ “ stereo tuners “ MAX (75*150) X1 + (50*150) X2 + (40*150) X3 ST (3*150) X1 + (2*150) X2 + (1*150) X3 <= 400,000 50,000/150 <= X1 <= 150,000/150 50,000/150 <= X2 <= 100,000/150 50,000/150 <= X3 <= 90,000/150 Xi >= 0 and Integer for i=1,3
Binary integer variables • xi = 0 or 1 (binary) • Example 6.11 Capital Budgeting Problem Expected NPV Capital in ($000s) Required in Project (in $000s) Year 1 Year 2 Year 3 Year 4 Year 5 1 $141 $75 $25 $20 $15 $10 2 $187 $90 $35 $0 $0 $30 3 $121 $60 $15 $15 $15 $15 4 $83 $30 $20 $10 $5 $5 5 $265 $100 $25 $20 $20 $20 6 $127 $50 $20 $10 $30 $40 • The company currently has $250,000 available to invest in new • projects. It has budgeted $75,000 for continued support for • these projects in year 2 and $50,000 per year for years 3, 4, and 5.
Formulation • Let Xi = 1 if project i selected, 0 otherwise • max ROI as measured by NPV • Max 14X1 + 187X2 + … + 127X6 • ST • YEAR 1) 75X1 + 90X2 + … + 50X6 <= 250 • YEAR 2) 25X1 + … + 20X6 <= 75 • … • YEAR 5) 10X1 + … + 40X6 <= 50 • Note that for only BINARY decision variables we do not have to indicate that they are >= 0
Binary Variables & Logical Conditions • Binary variables are also useful in modeling a number of logical conditions. For example: • Of projects 1, 3 & 6, no more than one may be selected X1 + X3 + X6 <= 1 • Of projects 1, 3 & 6, exactly one must be selected X1 + X3 + X6 = 1 • Project 4 cannot be selected unless project 5 is also selected X4 – X5 <= 0 or X4 <= X5
Health Care Systems of Florida Health Care Systems of Florida ( HCSF) is planning to build a number of new emergency- care clinics in central Florida. HCSF management has divided a map of the area into seven regions. They want to locate the emergency centers so that all seven regions will be conveniently served by at least one facility. Five possible sites are available for constructing the new facilities. The regions that can be served conveniently by each site are indicated by X in the following table: Formulate as an ILP. Implement the formulation on Excel to solve using ASPE or solver.
Operation Prod. 1 Prod. 2 Prod. 3 Hours Available Machining 2 3 6 600 Grinding 6 3 4 300 Assembly 5 6 2 400 Unit Profit $48 $55 $50 Setup Cost $1000 $800 $900 The Fixed-Charge Problem • Remington Manufacturing Xi = the amount of product i to be produced, i = 1, 2, 3
Potential Pitfall • Do not use IF( ) functions to model the relationship between the Xi and Yi. • Suppose cell A5 represents X • Suppose cell A6 represents Y • You’ll want to let A6 = IF(A5>0,,0) • This will not work with Solver! • Treat the Yi just like any other variable. • Make them changing cells. • Use the linking constraints to enforce the proper relationship between the Xi and Yi.
Remington Manufacturing Formulation • Max 48x1 + 55x2 +50x3 –1000y1 – 800y2 – 900y3 • ST • 2x1 + 3x2 + 6x3 <= 600 • 6x1 + 3x2 + 4x3 <= 300 • 5x1 + 6x2 + 2x3 <= 400 • x1 – 999y1 <= 0 • x2 - 999y2 <=0 • x3 - 999y3 <= 0 • xi >= 0 (and integer, if desired/needed)
Suggested problems from chap 6 • Integer, mixed integer, binary models can be tricky – must practice more for these topics. • Suggested problems: • 6thed: # 7, 12, 13, 14, 15, 23, 25, 30
