Assignment 01 Operator-Machine Assignment Problem Dr. Stafford MSC 600 Spring 2005 Page 2
KEY: Operator-Machine Assignment Problem Assignment
MSC 600 Dr. Stafford Spring 2005
Question 1 Version A
Answer Item Computations for Answer
Part I 1 operator to 1 machine
C = R + S = 13 + 3 = 16 Minutes Q = 60KU/C = (60)(1)(3)/16 = 11.25 unit/hr
Z = L + KP = 14 + (1)(37) = $51 per hr
$4.533 1 T(K) = Z/Q = $51/11.25 = $4.533 per unit
48.75 2 Since K = 1, W = R = 13 min/cycle = 13(60/16) = 48.75 min/hr
90 3 Output/shift = (8 hr/shift) x (11.25 unit/hr) = 90 unit/shift
04
Since W > 0, M = 0 min/cycle
16 min 5 See computation above
Part II 1 operator to K machines R/S + 1 = 13/3 + 1 = 5.333
Thus K = 5 and K’ = 6
56
T(5) = [(R+S)/(60U)][(L/K) + P] = [16/{60(3)}][(14/5) + 37] = (.08888)(39.8) = $3.537
T(6) = [(K’S)/(60U)][(L/K’) + P] = [{(6)(3)}/{60(3)}][(14/6) + 37] = $3.933
Hence optimal number of machines is 5
$3.537 7 See item 6 above
16 min 8 C = R + S = 16 min
1 min 9 W = MAX{[R - (K-1)S];0} = Max{[13 - (4)(3)]; 0} = Max{1; 0} = 1 min per cycle
010
Since W > 0, M = 0 min/cycle, /hr, and /shift because WM = 0
2 Min 11 For K = 6, M = MAX{[(K-1)S - R]; 0} = Max{[(5)(3) - 13]; 0} = 2 min per cycle per machine
C = KS = (6)(3) = 18 min, and M/hr = (2)(60/18) = 6.67 min/hr per machine
26.25 min 12 W = MAX{[R - (K-1)S]; 0} = MAX{[13 - (2)(3)]; 0} = 7 min per cycle; and for K = 3, C = R+S = 16
min; so W per hr = (7)(60/16) = 26.25 min per hr
21 min 13 For K = 7, C = KS = (7)(3) = 21 min because (K-1)S = 6(3) = 16 > R = 13 min
Sensitivity Analysis
T[5] < T[6] T[5] < T[6]
[(R+S)/(60U)][L/K +P] < [(K’S/60U][L/K’ + P] [(R+S)/(60U)][L/K +P] < [(K’S/60U][L/K’ + P]
60U’s cancel 60U’s cancel
16[L/5 + 37] < 18[L/6 + 37] 16[14/5 +P] < 18[14/6 +P]
3.2L - 3L < 18(37) - 16(37) 16(14/5) - 18(14/6) < 18P - 16P
0.2L < 74, or (3.2 - 3)(14) < 2P
0 < L < $370 2P > 2.8, or 1.40 < P < infinity
