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Timetabling and personnel planning Description
Further explanation of this example:
'Applications of optimization with Xpress-MP', Chapter 14: Timetabling and personnel planning
Source Files By clicking on a file name, a preview is opened at the bottom of this page.
Data Files
i1assign.mos (!****************************************************** Mosel Example Problems ====================== file i1assign.mos ````````````````` Assigning workers to machines Given a set of operators and a set of machines, an operator must be assigned to each of the six machines in a workshop. The productivity (pieces per hour) of each operator and each machine is given. What operator should be assigned to each machine to maximize the total productivity if machines are running in parallel? What should the assignment be if machines are working in series? For the parallel-machines problem version, a simple IP model is implemented. For the machines-working-in-series, since total productivity is defined by the bottleneck (operator-machine assignment with lowest productivity), a new variable is introduced to define a minmax problem. A procedure to obtain a heuristic solution is also implemented. (c) 2008-2022 Fair Isaac Corporation author: S. Heipcke, Mar. 2002, re. Mar. 2022 *******************************************************!) model "I-1 Personnel assignment" uses "mmxprs", "mmsystem" forward procedure parallelheur forward procedure printsol(txt1,txt2:text) declarations PERS = 1..6 ! Personnel MACH = 1..6 ! Machines OUTP: array(PERS,MACH) of integer ! Productivity end-declarations initializations from 'i1assign.dat' OUTP end-initializations ! **** Heuristic solution for parallel assignment **** parallelheur ! **** Exact solution for parallel assignment **** declarations assign: array(PERS,MACH) of mpvar ! 1 if person assigned to machine, ! 0 otherwise end-declarations ! Objective: total productivity TotalProd:= sum(p in PERS, m in MACH) OUTP(p,m)*assign(p,m) ! One machine per person forall(p in PERS) sum(m in MACH) assign(p,m) = 1 ! One person per machine forall(m in MACH) sum(p in PERS) assign(p,m) = 1 ! Solve the problem maximize(TotalProd) printsol("Exact solution (parallel assignment)", "Total") ! **** Exact solution for serial machines **** declarations pmin: mpvar ! Minimum productivity end-declarations ! Calculate minimum productivity forall(p in PERS) sum(m in MACH) OUTP(p,m)*assign(p,m) >= pmin forall(p in PERS, m in MACH) assign(p,m) is_binary ! Solve the problem maximize(pmin) printsol("Exact solution (serial machines)", "Minimum") !----------------------------------------------------------------- ! Heuristic solution for parallel assignment procedure parallelheur declarations ALLP, ALLM: set of integer ! Copies of sets PERS and MACH HProd: integer ! Total productivity value pmax,omax,mmax: integer end-declarations writeln("Heuristic solution:") ! Copy the sets of workers and machines forall(p in PERS) ALLP+={p} forall(m in MACH) ALLM+={m} ! Assign workers to machines as long as there are unassigned persons while (ALLP<>{}) do pmax:=0; mmax:=0; omax:=0 ! Find the highest productivity among the remaining workers and machines forall(p in ALLP, m in ALLM) if OUTP(p,m) > omax then omax:=OUTP(p,m) pmax:=p; mmax:=m end-if HProd+=omax ! Add to total productivity ALLP-={pmax}; ALLM-={mmax} ! Remove person and machine from sets writeln(" ", pmax, " operates machine ", mmax, " (", omax, ")") end-do writeln(" Total productivity: ", HProd) end-procedure !----------------------------------------------------------------- ! Solution printing procedure printsol(txt1,txt2:text) writeln(txt1, ":") forall(p in PERS) do mp:=round(getsol(sum(m in MACH) m*assign(p,m))) writeln(" ", p, " operates machine ", mp, " (", OUTP(p,mp), ")") end-do writeln(" ", txt2, " productivity: ", getobjval) end-procedure end-model | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

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