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Fantasy OR: Sangraal (CP and MIP models)

Description
The Sangraal problem is an example of a mathematical problem embedded in a computer fantasy game. The description of the problem and the mathematical model introduced below draw on a publication by M.Chlond: M.J. Chlond, Fantasy OR, INFORMS Transactions on Education, Vol. 4, No. 3, 2004. http://ite.pubs.informs.org/Vol4No3/Chlond/

When the Sangraal (Holy Grail) is almost won the hero arrives at a castle where he finds 8 imprisoned knights. He is facing the task to bring the largest possible number of knights for the arrival of the Sangraal in twenty minutes' time. The time required for freeing a knight depends on his state of binding. A freed knight then needs a given amount of time to wash and recover himself physically. For every knight, the durations of freeing and preparing are given.

The problem of deciding in which order to free the knights is a standard scheduling problem, or to be more precise the problem of sequencing a set of disjunctive tasks. Typical objective functions in scheduling are to minimize the completion time of the last task (the so-called makespan) or the average completion time of all tasks. The objective to maximize the number of knights who are ready by a given time makes the problem slightly more challenging since we need to introduce additional variables for counting the knights who are ready on time.
• The MIP model (sangraal.mos) defines binary decision variables x(k,j) indicating whether knight k is the j'th knight to be freed. A formulation alternative uses indicator constraints (sangraalind.mos).
• The CP model (sangraal_ka.mos or sangraal2_ka.mos) uses the notion of tasks' with fixed durations and variable start times. These tasks need to be scheduled subject to precedence relations (freeing before preparing) and the disjunctive use of a resource (the hero's time).

Source Files
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sangraal.mos

(!******************************************************
Mosel Example Problems
======================

file sangraal.mos

Sangraal problem.

When the Sangraal (Holy Grail) is almost won the hero arrives
at a castle where he finds 8 imprisoned knights. He is facing
the task to bring the largest possible number of knights for
the arrival of the Sangraal in twenty minutes' time. The time
A freed knight then needs a given amount of time to wash and
recover himself physically.

Description and original model by M. Chlond:
https://doi.org/10.1287/ited.4.3.66

(c) 2008 Fair Isaac Corporation
author: S. Heipcke, 2005, rev. Mar. 2022
*******************************************************!)

model Sangraal
uses "mmxprs", "mmjobs"

forward procedure print_solution

declarations
KNIGHTS = {"Agravain", "Bors", "Caradoc", "Dagonet", "Ector", "Feirefiz",
"Gareth", "Harry"}
K = 8
POS = 1..K
FREE: array(KNIGHTS) of real   ! Time to free each knight
PREP: array(KNIGHTS) of real   ! Time to prepare each knight
x: array(KNIGHTS,POS) of mpvar ! x(k,j)=1 if knight k in position j,
! 0 otherwise
ontime: array(POS) of mpvar    ! ontime(j)=1 if position j finished within
! 20 minutes, 0 otherwise
ready: array(POS) of mpvar     ! Finish time for each position

SOLFOUND = 2                   ! Solution found event
pos: array(KNIGHTS) of integer ! Position of knight
end-declarations

FREE("Agravain"):=1;  PREP("Agravain"):=15
FREE("Bors")    :=1;  PREP("Bors")    := 5
FREE("Dagonet") :=2;  PREP("Dagonet") := 5
FREE("Ector")   :=3;  PREP("Ector")   :=10
FREE("Feirefiz"):=4;  PREP("Feirefiz"):=15
FREE("Gareth")  :=5;  PREP("Gareth")  :=10
FREE("Harry")   :=6;  PREP("Harry")   := 5

MAXT:= sum(k in KNIGHTS) FREE(k) + max(k in KNIGHTS) PREP(k)
MINT:= min(k in KNIGHTS) (PREP(k) + FREE(k))

forall(k in KNIGHTS, j in POS) x(k,j) is_binary
forall(j in POS) ontime(j) is_binary

! Maximize number of positions finished within 20 minutes
TotalFreed := sum(j in POS) ontime(j)

! Each knight in one position
forall(k in KNIGHTS) sum(j in POS) x(k,j) = 1

! Each position has one knight
forall(j in POS) sum(k in KNIGHTS) x(k,j) = 1

! Compute finish time for each position
forall(j in POS)
sum(k in KNIGHTS,l in 1..j-1) FREE(k)*x(k,l) +
sum(k in KNIGHTS) (FREE(k)+PREP(k))*x(k,j) = ready(j)

! ontime(j) = 1 if knight in position j is freed and prepared within 20 min.
forall(j in POS) do
end-do

! setparam("XPRS_VERBOSE", true)

! setparam("XPRS_HEUREMPHASIS", 0)
! setparam("XPRS_CUTSTRATEGY", 0)
! setparam("XPRS_PRESOLVE", 0)

setcallback(XPRS_CB_INTSOL,->print_solution)
maximize(TotalFreed)

!********************************************************************

procedure print_solution
obj:=getparam("XPRS_LPOBJVAL")
writeln("Number of knights freed on time: ", obj)