KEYS: In-Class Practice Problems MSC 385 Dr. Stafford Revised: 08 July 2005 Page 1
KEYS TO IN-CLASS AND PRACTICE PROBLEMS, MSC 385
Operator-Machine Problem 1
R 14 min.
S 3 min. {1 to unload, 2 to reload}
U 2 parts/mach/cycle
L $20 per hr.
P $40 per hr, per machine
Optimal Solution:
K # R/S + 1 = 14/3 + 1 = 4.67 + 1 = 5.67; therefore: K = 5 machines and K’ = K+1 = 6 machines
TK RS
U
L
KP
[] ()() [/ ][] .
=+
⎡
⎣
⎢⎤
⎦
⎥+
⎡
⎣
⎢⎤
⎦
⎥=+⎡
⎣
⎢⎤
⎦
⎥+
⎡
⎣
⎢⎤
⎦
⎥==
60
14 3
60 2
20
540 17 120 44 62333
TK KS
U
L
KP
['] '
'
()()
()() [/ ][. ] .
=⎡
⎣
⎢⎤
⎦
⎥+
⎡
⎣
⎢⎤
⎦
⎥=⎡
⎣
⎢⎤
⎦
⎥+
⎡
⎣
⎢⎤
⎦
⎥==
60
63
60 2
20
640 18 120 43 333 650
Therefore:
1K* = 5 machines
2 T[K*] = $6.233 per unit
3 C{K*} = 14 + 3 = 17 minutes
4 W = Max{[R - (K-1)S]; 0} = Max{[14 - (4)(3)]; 0} = Max{2, 0] = 2 min/cycle
5 Since W > 0, and we know MW = 0, M = 0 min/cycle = 0 min/hr
6 For K = 7, M = Max{[(K-1)S - R]; 0} = Max{[(6)(3) -14];0} = Max{4; 0} = 4 min/cycle per machine
7 For K = 4, W = Max{[R - (K-1)S]; 0} = Max{[14 -(3)(3)]; 0} = 5 min/cycle
W{hr} = [W{cycle}][cycle/hr] = [5][60/17] = 17.647 min/hr
{since K = 4 # K = 5 from above, C = R+S = 17 min}
8Since K = 7 $ K’ = 6 from above, C = KS = (7)(3) = 21 minutes
