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

   file sangraalind.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.

   Model formulation using indicator constraints.

   Description and original model by M. Chlond:
   https://doi.org/10.1287/ited.4.3.66
    
   (c) 2010 Fair Isaac Corporation
       author: S. Heipcke, Nov. 2010, 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

  pos: array(KNIGHTS) of integer ! Position of knight
 end-declarations
 
 FREE :: (["Agravain", "Bors", "Caradoc", "Dagonet", "Ector", "Feirefiz", 
           "Gareth", "Harry"])[1, 1, 2,2, 3, 4, 5,6]
 PREP :: (["Agravain", "Bors", "Caradoc", "Dagonet", "Ector", "Feirefiz", 
           "Gareth", "Harry"])[15,5,15,5,10,15,10,5]

 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)

 ! if ontime(j) = 1, then knight in position j is freed and prepared within 
 ! 20 min.  [ indicator constraint:  ontime(j)=1 -> ready(j)<=20 ]
 forall(j in POS)
  indicator(1, ontime(j), ready(j)<=20)
 
 setparam("XPRS_VERBOSE", true)   

 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