This model illustrates the tradeoff between the cost of service people and the costs of machine downtime. In our model, we have ten machines that have a tendency to break down randomly. It costs $350/hour in lost production whenever a machine is broken. The question we need to answer is how many service people should we hire to minimize the total cost of down time and salaries for service people.

MODEL:

! Machine repair model;

SETS:

 NREP/1..5/: ! Consider 5 possible repair persons;

      NDOWN, ! Expected no. of down machines;

      CPERHR,! Expected cost/hour of down machines;

      TCOST; ! Total expected cost/hour;

ENDSETS

 

! The input data;

 NMACH = 10;! No. machines subject to breakdown;

 RTIME = 1; ! Average repair time;

 UPTIME = 5;! Mean time between failures;

 CR = 30;   ! Hourly cost of a repair person;

 CM = 350;  ! Hourly cost of a down machine

 

! The machine repairman queuing model;

! For each case of 1 - 5 service people calculate

  expected number of machines down, cost per hour

  of down machines, and total cost per hour of

  operations. @PFS calculates the Probability in

  a Finite Source, in this case expected number of

  machines under repair. ;

 @FOR( NREP( I):

  NDOWN( I) =

   @PFS( NMACH * RTIME / UPTIME, I, NMACH);

  CPERHR( I) = CM * NDOWN( I);

  TCOST( I) = CPERHR( I) + CR * I

 );

 

END

Model: MREPAR