# WOOD 492 MODELLING FOR DECISION SUPPORT Lecture 17 Integer Programming.

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WOOD 492 MODELLING FOR DECISION SUPPORT Lecture 17 Integer Programming

Integer Programming (IP) For discrete inputs/outputs –Number of workers required in a factory For binary variables (yes/no decisions) –Building a facility For logical conditions (if {x} then {y}) –If sawing pattern 1 is selected, then sawing pattern 2 can not be selected Oct 17, 20122Wood 492 - Saba Vahid

Oct 17, 2012 A facility location problem –A manufacturing company has \$10 million available capital and would like to invest this in building new factories and warehouses in Los Angeles and San Francisco in order to maximize Net Present Value (NPV). –Decision Variables: which facilities to build? (either build or dont build, binary) –Constraints: –Factories can be built in L.A. or S.F, or both –Only one warehouse can be located and should be in a city where a new factory is being built Example 9: Integer Formulation Example Yes/No VariableNPVCapital required Factory in L.A.X1\$9 m\$6 m Factory in S.FX2\$5 m\$3 m Warehouse in L.A.X3\$6 m\$5 m Warehouse in S.F.X4\$4 m\$2 m 3Wood 492 - Saba Vahid

Oct 17, 2012 This is a Binary Integer Program (BIP) Obj Z=9 X1 +5 X2 +6 X3 +4 X4 (Total NPV) Subject to: 6 X1+3 X2+ 5 X3+ 2 X4 10 (Total Capital) X3+ X4 1 (One warehouse) -X1 + X3 0 (warehouse and factory - X2 + X4 0 in the same city) Xi 1 (Upper bound) Xi 0 (Lower Bound) LP Matrix 4Wood 492 - Saba Vahid

Oct 17, 2012 Indicators are binary variables used for modelling problems with: –Fixed costs –Either/OR constraints –If/Then constraints –… Using indicator variables 5Wood 492 - Saba Vahid

Oct 17, 2012 Fixed-Charge Problems When taking up an activity has a one-time fixed cost –production line set-up cost –Road establishment cost Example: –A cabinet manufacturer can outsource the cabinet doors at \$20/door or make them in the factory with \$500 line set-up cost and then \$10/door production cost. 6Wood 492 - Saba Vahid

Oct 17, 2012 Objective is to minimize costs Decision variables are – X1 number of outsourced doors – X2 number of manufactured doors Total cost for outsourced doors: 20 X1 Total cost for the manufactured doors: 0 if X2=0 500 + 10 X2 if X2>0 Fixed-Charge Problems 7Wood 492 - Saba Vahid

Oct 17, 2012 How can we force the \$500 costs when X2 is greater than 0? We define an indicator variable Y2 : –if X2=0 then Y2=0 –otherwise Y2=1 –Now, rewrite the formula for production cost: Manufacturing cost: 500 Y2 + 10 X2 Objective: Min Z= 20 X1 + 500 Y2 + 10 X2 Fixed-Charge Problems 8Wood 492 - Saba Vahid

Oct 17, 2012 Now, how to ensure that Y2 value changes according to X2 value? Big M method: –Choose a big number M (relative to problem parameters, bigger than any possible value of X2) –Add the following to the existing constraints X2 <= M.Y2 Y2 is binary –How does this constraint work? If X2>0, then Y2 has to be 1, so that X2<=M If X2=0, then Y2 can be either 0 or 1, however since 500Y2 is a term in the objective function, the solution algorithm always picks Y2=0, because it results in the lower objective function value Fixed-Charge Problems 9Wood 492 - Saba Vahid

Oct 17, 2012 Either/OR Constraints When only one of two constraints must hold –E.g. there are two suppliers for cabinet doors –Each supplier has a certain number of doors available for shipping –Only one of the suppliers can be selected Variables: number of doors purchased (X) Supply constraints (depending on which supplier is elected): –Either X <= 100 –OR X <= 150 10Wood 492 - Saba Vahid

Oct 17, 2012 Either/OR Constraints Define an indicator variable Y: –If first supplier is selected, Y=1 and first constraint is activated –otherwise Y=0 and second constraint is activated Use the Big M Method –Constraint 1: X <= 100 + M. (1-Y) –Constraint 2: X <= 150 + M. Y How does this formulation work? –If supplier 1 is selected, then Y=1, so the first constraint will be X<=100 and the second constraint will be X<=150+M which in effect eliminates the second constraint (since M is a really big number) –If supplier 2 is selected, then Y=0 and the first constraint is eliminated –The solution algorithm picks the Y value that results in the best value of the objective function 11Wood 492 - Saba Vahid

Lab 6 preview Same problem as in Lab 3 Adding extra constraints: –Building the roads to each cut block (if we harvest, we have to build the road first) –Selecting only one sawing pattern for all log sizes –Selecting the optimal number of shifts to run the sawmill (1,2, or 3) Oct 17, 201212Wood 492 - Saba Vahid LP matrix

Next Class Branch and bound Oct 17, 201213Wood 492 - Saba Vahid

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