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Basic MIP tasks: binary variables; logic constraints Description We wish to choose among items of different value and
weight those that result in the maximum total value for
a given weight limit.
Source Files By clicking on a file name, a preview is opened at the bottom of this page.
burglarl.mos (!******************************************************* * Mosel Example Problems * * ====================== * * * * file burglarl.mos * * ````````````````` * * Example for the use of the Mosel language * * (Burglar problem) * * -- Formulation of logical constraints -- * * * * (c) 2009 Fair Isaac Corporation * * author: S. Heipcke, June 2009 * *******************************************************!) model "Burglar (logic constraints)" uses "advmod" ! Use logic constraints package declarations Items={"camera", "necklace", "vase", "picture", "tv", "video", "chest", "brick"} ! Index set for items VALUE: array(Items) of real ! Value of items WEIGHT: array(Items) of real ! Weight of items WTMAX=102 ! Max weight allowed for haul x: array(Items) of mpvar ! 1 if we take item i; 0 otherwise end-declarations VALUE("camera") := 15; WEIGHT("camera") := 2 VALUE("necklace"):=100; WEIGHT("necklace"):= 20 VALUE("vase") := 90; WEIGHT("vase") := 20 VALUE("picture") := 60; WEIGHT("picture") := 30 VALUE("tv") := 40; WEIGHT("tv") := 40 VALUE("video") := 15; WEIGHT("video") := 30 VALUE("chest") := 10; WEIGHT("chest") := 60 VALUE("brick") := 1; WEIGHT("brick") := 10 MaxVal:= sum(i in Items) VALUE(i)*x(i) ! Objective: maximize total value ! Weight restriction WtMax:= sum(i in Items) WEIGHT(i)*x(i) <= WTMAX forall(i in Items) x(i) is_binary ! All x are 0/1 ! *** Logic constraint: ! *** Either take "vase" and "picture" or "tv" and "video" (but not both pairs). ! * Values within each pair are the same Log1:= x("vase") = x("picture") Log2:= x("tv") = x("video") ! * Choose exactly one pair (uncomment one of the 3 formulations A, B, or C) ! (A) MIP formulation ! Log3:= x("tv") = 1 - x("vase") ! (B) Logic constraint Log3:= xor(x("vase")+x("picture")>=2, x("tv")+x("video")>=2) ! (C) Alternative logic formulation (does not create additional binaries) (! Log3a:= indicator(1, x("vase"), x("tv")+x("video") <= 0) ! x("vase")=1 -> x("tv")+x("video")=0 Log3b:= indicator(-1, x("vase"), x("tv")+x("video") >= 2) ! x("vase")=0 -> x("tv")+x("video")=2 !) maximize(MaxVal) ! Solve the MIP-problem ! Print out the solution writeln("Solution:\n Objective: ", getobjval) forall(i in Items) writeln(" x(", i, "): ", x(i).sol) end-model | |||||||||||||||
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