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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
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 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. 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. June 2011 *******************************************************!) model Sangraal uses "mmxprs", "mmjobs" forward public 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("Caradoc") :=2; PREP("Caradoc") :=15 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 ready(j) >= 21-(21-MINT)*ontime(j) ready(j) <= MAXT-(MAXT-20)*ontime(j) 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) !******************************************************************** public procedure print_solution obj:=getparam("XPRS_LPOBJVAL") writeln("Number of knights freed on time: ", obj) writeln("Knight Position Ready <=20 min") forall(k in KNIGHTS) do pos(k):=round(getsol(sum(j in POS) j*x(k,j))) writeln(strfmt(k,-12), pos(k), " ", strfmt(getsol(ready(pos(k))),2), " ", if(getsol(ontime(pos(k)))=1,"yes","no")) end-do end-procedure end-model | |||||||||||||||||||

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