```! Single Machine Job Selection and Sequencing. Given a Set of jobs, each with an earliest begin time, latest finish time, processing time, and value if processed, and a machine that can process only one job at a time, and changeover times between jobs, Decide which jobs to accept, and the sequence in which to process accepted jobs; ! Keywords: Changeover times, Due dates, Job selection, Machine sequencing, Scheduling, Sequence dependent setups, Sequencing, Setup time; SETS: !Parameters: r( k) = earliest allowed start time of job k, d( k) = latest allowed finish time of job k, p( k) = processing time of job k, cot( i, k) = changeover time from job i to k, v( k) = value of job k, Variables: y( i, k) = 1 if both jobs i and k are accepted and job i immediately precedes job k, else 0, s( k) = start time of job k, drop( k) = 1 if we do not do job k, y0( k) = 1 if job k is first, yL( k) = 1 if job k is last; job: r, p, d, v, s, y0, yL, drop; jxj( job, job): COT, y; ENDSETS DATA: ! 1 2 3 4 5 6 7 8 9 10 11 12 ! Ready times; r = 0 0 2 2 3 5 5 6 12 20 20 21; ! The processing times; p = 18 24 4 8 13 1 12 2 3 6 5 5; ! Due date of each job; d = 22 25 8 11 29 10 25 14 16 29 32 31; ! Value of processing the job; v = 2 6 4 2 3 5 7 4 4 3 1 6; ! Change over time matrix; COT= 0 1 3 2 1 2 5 3 2 1 1 1 2 0 2 3 5 1 2 2 0 3 2 1 3 4 0 3 4 2 1 1 2 1 4 1 1 1 3 0 4 1 2 1 1 3 3 1 2 3 2 3 0 3 5 3 2 2 2 1 4 1 2 1 4 0 2 2 4 3 2 4 5 0 2 3 4 4 0 3 4 3 0 4 1 2 2 3 1 1 2 0 4 4 2 4 2 0 2 2 4 2 3 3 0 3 1 4 2 1 2 3 4 1 2 3 4 0 2 4 2 0 2 3 4 0 2 3 4 3 0 4 4 1 1 1 1 1 3 3 3 3 4 0 ; ENDDATA SUBMODEL JOBROUTE: ! Maximize the value of accepted jobs; MAX = OBJ ; OBJ = @SUM( job( j): v( j)*(1- DROP( j))); ! For each JOB k...; @FOR( JOB( k): ! A job cannot follow itself...; y( k, k) = 0; ! it must be either the first job, or preceded by some other job i, or not done; [NTR] y0( k) + @SUM( JOB( i)| i #NE# k : y( i, k)) + DROP( k) = 1; ! it must be either the last job, or followed by some other job j, or not done; [XIT] yL( k) + @SUM( JOB( j)| j #NE# k : y( k, j)) + DROP( k) = 1; ); ! There can be only one job that is...; @SUM( JOB( k): y0( k)) <= 1; ! ... first; @SUM( JOB( k): y0( k)) = @SUM( JOB( k): yL( k)); ! ... last; ! Make the y's binary, 0 or 1; @FOR( jxj( j, k): @BIN( y( j, k))); ! Compute time at which job k starts; @FOR( JOB( k): @FOR( JOB( i) | i #NE# k: ! Time of start of job k if preceding job was i; [RTM] s( k) >= s( i) + ( p( i) + COT( i, k))* y( i, k) + ( r( k) - d( i) + p( i))*(1- y( i, k)); ! otherwise; ); s( k) >= r( k); ! Not early; s( k) + p( k) <= d( k); ! Not late. Assume r( k) + p( k) <= d( k); ); ENDSUBMODEL CALC: @SET( 'TERSEO', 2); ! Make default output terse; @SET( 'IPTOLR', .01); ! Relative optimality tolerance; @SET( 'TIM2RL', 30); ! Time to turn on IPTOLR in seconds; @SOLVE( JOBROUTE); ! Display a little report; @WRITE(' Which jobs should we do, in what order to Max Profit,',@NEWLINE(1), ' taking into account ready, changeover(COT), processing, & due times?', @NEWLINE(1)); @WRITE(@NEWLINE(1),' Various times', @NEWLINE(1)); @WRITE(' Job Ready Processing Due Value COT Start Finish ',@NEWLINE(1)); @FOR( job( i): @WRITE( @FORMAT( i,"3.0f"),' ', @FORMAT(r( i),"3.0f") ,' ', @FORMAT(p( i),"3.0f"), ' ', @FORMAT(d( i) ,"3.0f"),' ', @FORMAT(v( i),"3.0f")); @IFC( drop( i) #LT# 0.5: cotij = @SUM( job( k): cot( k, i)*y( k, i)); @WRITE(' ',@FORMAT( cotij, "3.0f"), ' ', @FORMAT( s( i),"3.0f"), ' ', @FORMAT( s( i)+ p( i),"3.0f"),@NEWLINE(1)); @ELSE @WRITE(' -- -- --',@NEWLINE(1)) ); ); @WRITE(@NEWLINE(1),' Total value of accepted jobs= ',OBJ, @NEWLINE(1)); ENDCALC ```