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Linear Programming Integer Linear Models. When Variables Have To Be Integers Example – one time production decisions –Fractional values make no sense.

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Presentation on theme: "Linear Programming Integer Linear Models. When Variables Have To Be Integers Example – one time production decisions –Fractional values make no sense."— Presentation transcript:

1 Linear Programming Integer Linear Models

2 When Variables Have To Be Integers Example – one time production decisions –Fractional values make no sense –But if ongoing process, fractional values could represent work in progress Example -- building houses or planes, or scheduling crews Binary variables Restricted to be 0 or 1 Example – Is a plant built?

3 Types of Integer Programs (ILP) All Integer Linear Programs (AILP) –All the decision variables are required to be integers Mixed Integer Linear Programs (MILP) –Only some of the variables are required to be integers Binary Integer Linear Programs (BILP) –Variables are restricted to be 0 or 1

4 Example Boxcar Burger will build restaurants in the suburbs and downtown Suburbs –Profit $12000/day –$2,000,000 investment –Requires 3 managers Downtown –Profit $20000/day –$6,000,000 investment –Requires 1 manager Constraints –$27,000,000 budget –At least 2 downtown restaurants –19 managers available

5 Decision Variables/Objective X 1 = Number of restaurants built in suburbs X 2 = Number of restaurants built downtown MAX Expected Daily Profit MAX 12X 1 + 20X 2 (in $1000’s)

6 Constraints Cannot invest more than $27,000,000 At least 2 downtown restaurants Number of managers used cannot exceed 19 Total Amount Invested Cannot Exceed 27 2X 1 + 6X 2 ≤ 27 # downtown restaurants Must be At least 2 X2X2 ≥ 2 # Managers used Cannot Exceed 19 3X 1 + 1X 2 ≤ 19 In $1,000,000’s

7 The Complete Model MAX12X 1 + 20X 2 (in $1000’s) s.t. 2X 1 + 6X 2  27 (Budget) X 2  2 (Downtown) 3X 1 + X 2  19 (Managers) Both X’s  0 Both X’s INTEGER!

8 The Linear Programming Feasible Region X2 6 5 4 3 2 1 0 12 X1 34 56 X 1, X 2 ≥ 0 2X 1 + 6X 2 ≤ 27 X 2 ≥ 2 3X 1 + 1X 2 ≤ 19 LP Feasible Region Max 12X 1 + 20X 2 12X 1 + 20X 2 Rounded down FEASIBLE Objective Value = 100 (5,2) LP Optimum (5 7/16, 2 11/16) Obj. Value = 119 Rounded up (6,3) Rounded off (5,3)

9 The Integer Programming Feasible Region X2 6 5 4 3 2 1 0 12 X1 34 56 X 1, X 2 ≥ 0 3X 1 + 1X 2 ≤ 19 2X 1 + 6X 2 ≤ 27 X 2 ≥ 2 3X 1 + 1X 2 ≤ 19 X 1, X 2 integer Max 12X 1 + 20X 2 12X 1 + 20X 2 ILP Optimum (4,3) OBJ. VALUE = 108

10 Why Not Round To Get the Optimal Integer Solution? Rounding may yield the optimal integer solution –None did in this example But it may yield an infeasible solution –Both (5,3) and (6,3) are infeasible solutions Or a feasible solution that is not optimal –(5,2) is feasible but not optimal –Many times a feasible rounded point gives a “good” solution (giving close to the optimal value of the objective function) -- BUT NOT ALWAYS

11 General Facts About Integer Models The solution time to solve integer models is longer than that of linear programs –Because many linear programs are solved en route to obtaining an optimal integer solution For maximization models, the optimal value of the objective function will be less (or at least not greater than) the value for the equivalent linear model –Because constraints have been added – the integer constraints There is no sensitivity analysis –Because the feasible region is not continuous

12 Solving ILP’s Using SOLVER The only change in SOLVER is to add the integer constraints –In the Add Constraints dialogue box, highlight the cells required to be integer and choose “int” from the pull down menu for the sign

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14 Optimal Build 4 Suburban Restaurants Build 3 Downtown Restaurants Average Daily Profit $108,000

15 Review When to use integer models Why rounding will not always work Solution time No sensitivity analysis Objective function value cannot improve SOLVER solution approach


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