(b) SCV of effective process times is zero since they are deterministic and have not random outages.
(c) The utilization is simply, r a
t e
.
r a
= 2 j/h = (1/30) j/min. Therefore u = 29/30 = 0.9667.
(d) There is no variability. The time in queue for the D/D/1 queue is always zero.
(e) The total cycle time at station 1 is the proces time, 29 minutes.
(f) The SCV of arrivals is also deterministic either by using equation (8.10) or by obvious observation.
(g) To get the utilization we must first compute t e
.
t e
= t
0
/ A therefore, t
, where e
=
A is the availabili ty, A =
26 / 0 .
909
=
28 .
6 and the
MTTF utilizatio
/( MTTF n will be
+ u
MTTR )
= r a t e
=
=
10 /( 10
+
1 )
( 1 / 30 )( 28 .
6 )
=
0
=
.
909
0 .
9533
(h) The SCV of process times is given by equation (8.6) c
2 e
= c
0
2 +
( 1
+ c r
2
) A ( 1
−
A ) m r t
0
=
0
+
( 1
+
1 )( 0 .
909 )( 1
−
0 .
909 )( 60 ) /( 26 )
=
0 .
3814
Because no information was given in the problem, we assumed the repair times were of moderate variability and therefore c r
2
= 1, a conservative assumption. Also, note that c
0
2
is zero because there is no variability.
(i) The cycle time at station 2 is given by equation (8.25) with SCV of arrivals being zero and SCV of process time that we computed in (h).
CT q
=
 c
2 a
+ c
2 e
2 
 u
1 − u t e
=
0 .
3814
2

  1
0 .
9533
− 0 .
9533


28 .
6
=
111 .
33 min
(j) Total cycle time will be cycle time in queue plus the effective process time,
111.33 min + 28.6 min = 139.93 min.
5. To get the number of hours per month first consider: 52 wks/yr/(12 mo/yr) = 4.333 wks/mo. We work 5 days per week, one shift, 7 (effective) hours per shift so we get: (4.333 wks/mo)(5 d/wk)(7h/d) = 151.667
h/mo.
(a) The demands per month and per hour are:
Part
B1
B2
B3
B4
B5
Total
D/mo D/hr
1800 11.86813
1000 6.593407
600 3.956044
350 2.307692
200 1.318681
3950 26.04396
Hence the r a
is 26.04 boxes per hour.
(b) If there were no setups the utilization would be r a t where t = 1 min = 1/60 hr. u
0
= 26.04/60 =0.4340
56

(c) If each batch were equal to 1000, then the run length would be 1000/60+4 = 20.67 hrs for each product.
The utilization would be u
CT
= i
5
∑
=
1 d i
1000
=
(
1000 / 60 + 4
)
=
1800 + 1000 + 600 + 350 + 200
( 1000 /
( 151 .
667 )( 1000 )
60 + 4 ) =
2
0 .
2  0
  1 −
.
5382
0 .
5382


20 .
667
+
20 .
667
=
14 .
45
+
20 .
667
=
35 .
12 hr
0 .
5382
(d) If we assume that all the batch sizes will be equal, the average CT will be
CT
=
2
0 .
2 
  1 u
− u
( k / 60
+
4
)
+ k / 60
+
4 where u = ( t + s / k ) i
5
∑
= 1 d i
= ( 26 .
04 )( 1 / 60 + 4 / k )
It turns out that the curve is very flat. Any batch size between 300 and 600 is pretty good. The best appears to be around 420 with a CT of 25.16 hours.
(e) Using batch sizes of 420, the number of times per month that we make each part is:
Part
B1
B2
B3
B4
B5 su/mo
4.286
2.381
1.429
0.833
0.476
6.
Demand is 160 parts per day. With a 16 hour day, this becomes 160/16 = 10 parts per hour.
(a) The maximum capacity is 125 parts divided by the heat treat time of 6 hours which is 20.833 per hour or 333.333 per 16 hour day.
(b) The WTBT would be (125-1)/(2*10pph) = 6.2 h. The utilization would be u = 10pph/20.833pph = 0.48.
The SCV of arrivals would be 1/125 = 0.008 and SCV of process time is (3h/6h)
2
= 0.25. Using the VUT equation we get
CT q
=
0 .
008
+
2
0 .
25  0
  1
−
.
48
0 .
48


6
=
0 .
714 h
Add these together with the process time and you get, 6.2h + 0.714h + 6h = 12.91h.
(c) The minimum batch size to meet demand is: k min
/ 6 h
>
10 pph or k min
>
60 therefore k min
=
61
57

(d) The average CT would be 56.95 h.
(e) Searching over the range reveals a minimum at batch size of 91 with a cycle time of 12.02 h.
7.
Note that TH will stay the same unless the arrival rate changes.
(a) Decrease arrival rate: WIP down, CT down, TH down
(b) Increase variability at one station: WIP up, CT up, TH same.
(c) Increase capacity at one station: WIP down, CT down, TH same
(d) Decrease capacity at all stations: WIP up, CT up, TH same
8.
The maximum the system can hold is b
=
7, 5 in the buffer and in each machine.
(a) For balanced lines, TH
Partial WIP
Total CT
=
= b / 2
40min
=
+
7 / 2 =
10 min
=
3
= b
.
5 ; b
+
1
50; r e
=
Using
Total
7
7
+
1 10
Little'
WIP
1 s
=
=
0 .
0875 jpm
Law for
TH * CT
Partial
=
CT = WIP/CT
0.0875jpm
* 50 min
=
=
3 .
5 / 0
4.375
.
0875 j
(b) TH
=
0.090
jpm; CT
=
60 min; WIP
=
5.4
j
= 40 min
(c) For unbalanced lines :
TH
=
1
1 −
− u b u b + 1 r a
=
1
−
( 12
1
−
( 12 /
/ 10 )
7
10 )
8
1
10
=
0 .
07828 jpm;
Partial WIP =
1 u
− u
−
( b
+
1 ) u b + 1
1
− u b + 1
(partial WIP is WIP at the second
=
1 .
2
1
−
1 .
2 station,
−
( 8
1
)(
− in the
1 .
2 )
1 .
2
8
8 buffer
= − 6 and
.
0 − at the
34
1
−
.
4
4 .
3 first
= 4 .
42 station wh en it is blocked)
Partial
CT
CT = Partial
=
56.5
min
+
10
WIP/CT min
=
=
66.5
4.42/0.078
min
28 = 56.5
WIP
=
(TH)(CT)
=
(0.07828)( 66.5)
=
5.2
(d) The calculations for TH and partial WIP are the same except that u is now 10/12 instead of 12/10.
TH stays the same. Partial WIP becomes 2.57. This leads to a partial CT = WIPP/TH = 32.9 min and a total CT of 32.9 + 10 = 42.9 min which produces a total WIP = TH*CT = 0.07828*43.9 = 3.51.
Notice that we now have less WIP and shorter CT than before (not surprising since the first machine is now the slower).
(e) We return to a balanced system with both machines taking 10 minutes, the buffer is 5, and the second machine SCV becomes ¼. Using equation (8.47) we can compute TH.
TH
≈ c
2 a
+
2 ( c
2 a c e
2
+
+ c
2 e
2 ( b
+ b
−
1 )
− 1 ) r e
=
1 + 0
2 ( 1 +
.
25
0 .
+
25
2
+
( 7
7
− 1 )
− 1 )
1
10
=
0 .
0914 jpm
58