OPERATOR-MACHINE ASSIGNMENT PROBLEM
SFor K = 1 (1 machine assigned to a single operator), know:
SC (cycle time) = R + S (run time + setup time) {C in minutes}
SW (worker idleness per cycle) = R minutes/cycle
SM (machine interference idleness) = 0 minutes per cycle
S# cycle/hr = 60/C
S#unit/cycle = U; #unit/hour = Q = [60U]/C; #unit/shift = [hr/shift]*[60U]/C
SSystem cost per hour = Z = L + P (Labor rate + Production cost/hr per machine)
SMfg cost/unit = T[1] = Z/Q = [ (R+S)/(60U) ][L + P]
STo find optimal # machines to assign, use 3-step method:
SStep 1: Compute K # R/S + 1; that is, round down to nearest integer below; then K’ = K +
1.
SStep 2: Find costs for K and K’ = K+1 machines; one of these is the optimal number of
machines to assign:
ST[K] = [ (R+S)/60U][(L/K + P]
ST[K’] = [ (K’S)/60U][ (L/(K’) + P]
SStep 3: chose either K or K’ machines, whichever has lower per-unit mfg cost
SConduct follow-on analyses for optimal assignment:
SIf K is optimal:
SC = R+S
SM {per cycle} = Max{ [(K-1)S - R]; 0}
SW {per cycle} = Max{ [R - (K-1)S]; 0}
SOutput: Q = KU per cycle = [KU]*(60/C) per hour = [60KU/C]*(hr/shift) output per
shift
SCost/unit - T[K]
SIf K’ = (K+1) is optimal
SC = K’S
SM & W computed same as above; remember M*W = 0
SOutput: Q = K’U per cycle = [K’U]*(60/C) per hour = [60K’U/C]*(hr/shift) output per
shift
SCost/unit - T[K]
Sconduct sensitivity analyses on L, P
SIf K is optimal, set up relationship as T[K] < T[K’]
S[ (R+S)/(60U][(L/K) +P] < [ (K’S)/(60U)][ (L/K’) +P]
SPlug in values for R. S. U. K. K’, P; and solve for range for L
SPlug in values for R, S, U, K, K’, L; and solve for range for P
SNote that range for P should be in opposite direction as range for L { # versus $,
or vice versa}
Sif K’ optimal, set up relationship as T[K’] < T[K]
S[ (K’S)/(6-U)][L/K’ + P] < [ (R+S)/60U][ L/K + P]
SProceed with analyzing L and P as above.
SMaximum output per hour for operator and assigned machines is at K’ machines
