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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.
burglari.mos
(!*******************************************************
* Mosel Example Problems *
* ====================== *
* *
* file burglari.mos *
* ````````````````` *
* Example for the use of the Mosel language *
* (Burglar problem) *
* *
* (c) 2008 Fair Isaac Corporation *
* author: S. Heipcke, 2001 *
*******************************************************!)
model "Burglar (index set)" ! Start a new model
uses "mmxprs" ! Load the optimizer library
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
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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