(!*******************************************************
* Mosel Example Problems *
* ====================== *
* *
* file burglar.mos *
* ```````````````` *
* Example for the use of the Mosel language *
* (Burglar problem) *
* *
* (c) 2008 Fair Isaac Corporation *
* author: S. Heipcke, 2001 *
*******************************************************!)
model Burglar ! Start a new model
uses "mmxprs" ! Load the optimizer library
declarations
Items=1..8 ! Index range 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
! Item: 1 2 3 4 5 6 7 8
VALUE :: [15,100, 90, 60, 40, 15, 10, 1]
WEIGHT:: [ 2, 20, 20, 30, 40, 30, 60, 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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(!*******************************************************
* 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
|
(!*******************************************************
* 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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