ISyE 6201: Manufacturing Systems
Instructor : Spyros Reveliotis
Spring 2006
Solutions for Homework #6
ISYE 6201 Spring 2006
Homework 6 Solution
A. Questions
Chapter 7
Question 5
n
By the definition of the critical WIP, W0  rbT0  rb  t i , where rb denotes the bottleneck
i 1
rate and ti denotes the processing time at the i-th station of the line. Also, by the
definition of the bottleneck rate rb,
(i)
rb = Nb/tb, where Nb is the number of machines at a bottleneck workstation
and tb is the processing time at that station;
(ii)
rb  ri = Ni/ti,  rbti  Ni, i=1,…,n, where Ni and ti are respectively the
number of machines and the processing time at station i.
n
n
i 1
i 1
Hence, W0   rb t i   N i
More intuitively, W0 denotes the minimum WIP that can ensure 100% utilization of the
system bottlenecks, in a deterministic operational setting. Keeping a WIP which is larger
than the total number of the machines in the system, will only increase the waiting time
of the jobs, without enhancing the utilization of the system bottlenecks.
n
Finally, notice that W0   N i only when rbti = Ni, i=1,…,n, i.e., only when the line is
i 1
balanced.
Chapter 8
Question 1
The coefficient of variation represents relative variability, i.e., it characterizes the st. dev.
of the considered random variable as a percentage of its mean value. For instance, the
standard deviation, alone, is not very meaningful when considering process times that can
range from fractions of seconds to several hours.
Question 6
For the M/M/1 case, the number of customers is adequate to fully define the state of the
system because process and inter-arrival times are memoryless. Thus, knowledge of how
long a station has been working on a particular job or how much time has passed since
the last arrival, are irrelevant. This is not the case for the G/G/1 system. In a G/G/1
queuing station, the time that a particular job has spent in processing and the time since
the last arrival are important pieces of information for analyzing the future evolution of
the system, and therefore, they should be part of the state definition.
Chapter 9
Question 3
Both variability reduction and capacity expansion are mechanisms that can be employed
in order to reduce the job cycle time, in particular, the experienced waiting time. The
difference is that capacity expansion is a rather “brute force” approach, since it relies on
buying more machines, while variability reduction involves understanding and managing
the process better (with all the potential benefits from such an exercise).
2
ISYE 6201 Spring 2006
Homework 6 Solution
Question 5
While the instance of “worst case” performance materializes under the assumption of
deterministic processing times, it does involve variability when considering the interarrival times experienced by the various single jobs in the batch. More specifically, at any
given station i, the inter-arrival time between the last job of batch j and the first job of
batch j+1 is equal to Wti-1, while the inter-arrival times between the remaining jobs in
batch j are equal to zero.
Question 6
In this case, the variability will prevent the effective synchronization of the plant
operation with the upstream and downstream processes. More specifically, although there
will be maximum throughput with a minimum WIP of 4 jobs, even with variable process
times, there is still the issue of raw material and finished goods inventories. If customers
showed up just when we completed the production of one product, there would be no
finished goods inventory. Of course, this is hardly likely and therefore we would either
have finished goods inventory or customers would have to wait. In either case, there
would be a buffer. We would need similar coordination for raw materials or else we
would need to keep a raw material inventory buffer or else periodically starve the system
for WIP.
B. Problems
Chapter 7
Problem 1
a. rb = 0.5 job/hr and T0 = 5 hr
b. Adding another machine at station two makes the parameters
rb = 1 job/hr and T0 = 5 hr
so the bottleneck rate increases and the raw process time stays the same. This
will decrease cycle time and increase throughput when WIP is greater than the
critical WIP.
c. Speeding up station two makes the parameters
rb = 1 job/hr and T0 = 4 hr
and so it increases the bottleneck rate and decreases the raw process time. In
this case, throughput will go up and cycle time will decrease regardless of the
WIP level.
d. There is no change in the parameters from (a). Since there is no variability,
there will be no change in performance.
e. This will make the parameters
rb = 0.5 job/hr and T0 = 4.5 hr
and it will reduce the cycle time and increase the throughput and, but only when
WIP is below the critical WIP. So, in this (deterministic) case, it is true that
saving an hour at a non-bottleneck does not help much.
3
ISYE 6201 Spring 2006
Homework 6 Solution
Problem 2
When all jobs are processed before moving, we have the worst case performance with
cycle time given by CT = wT0 and TH=1/T0.
a. In this new setting, the base parameters rb and T0 will be the same with those in
Problem 1, for all four cases (a-d). However, the impact of the suggested
changes on the system performance can be different, as discussed below.
b. Since T0 remains the same as in (a), there is no change from (a) regarding cycle
time and throughput. This is in contrast with Problem 1.
c. Speeding up station 2 reduces the raw process time and therefore reduces cycle
time and increases throughput for all WIP levels.
d. If all the jobs are worked on by only one machine at station 1 (as assumed in the
worst-case scenario, since the batch moves as a single block), there is absolutely
no change in performance.
e. If station one is speeded up this will reduce the raw process time as in Problem
1 and so will improve performance for all WIP levels.
Problem 3
The practical worst case assumes exponential process times and a balanced line using
only a single machine at each station.
a. Similar with Problem 2, the base parameters for the system do not change, but
the observed performance sometimes does.
b. Throughput will increase and cycle times will decrease for all WIP levels.
c. Speeding up station 2 reduces the raw process time and therefore reduces cycle
time and increases throughput for all WIP levels.
d. Adding a second machine at a non-bottleneck will improve throughput and
reduce cycle times for all WIP levels above WIP = 1.
e. Speeding up station one will improve performance for all WIP levels.
Problem 6
rb = 2000 per day = 125 per hour, and T0 = 0.5 hr, W0 = 62.5 cases, TH = 1700
cases/day = 106.25 per hour, CT = 3.5 hours.
a. Average WIP level = TH*CT = 106.25 cases per hour * 3.5 hours = 372 cases
372
 125 cases/hour = 107.27
b. At w = 372, we have THPWC(372) =
62.5  372  1
372  1
 3.468 hours, so we are roughly operating at the
and CTPWC = 0.5 
125
PWC level.
c. Throughput would increase (or at least not decrease), because bottleneck would
be blocked/starved less. Unbalancing the PWC line causes it to perform better.
d. Throughput would increase (or at least not decrease), because bottleneck would
be blocked/starved less. Replacing single machine stations with parallel
4
ISYE 6201 Spring 2006
Homework 6 Solution
machine stations in the PWC causes it to perform better. This is an example of
performance improvement through flexibility enhancement.
e. Moving cases in batches would further inflate cycle time by adding “wait for
batch” time, and increasing the variability experienced in the flow of the line.
Chapter 8
Problem 1
a. The mean is 5 and the variance is 0. The coefficient of variation is also zero.
These process times could be from a highly automated machine dedicated to
one product type.
b. The mean is 5, the standard deviation is 0.115 and the CV is 0.023. These
process times might be from a machine that has some slight variability in
process times.
c. The mean of these is 11.7 and the standard deviation is 14.22 so the CV is 1.22.
The times appear to be from a highly regular machine that is subject to random
outages.
d. The mean is 2. If this pattern repeats itself over the long run, the standard
deviation will be 4 (otherwise, for these 10 observations it is 4.2). The CV will
be 2.0. The pattern suggests a machine that processes a batch of 5 items before
moving any of the parts.
Problem 3
a. The natural CV is 1.5/2 = 0.75.
b. The mean would be (60)(2) = 120 min. The variance would be (60)(1.5)2 = 135.
The CV will be CV = √135/120 = 0.0968
c. The availability will be A = mf /(mf + mr) = 60/(60+2) = 0.9677. The effective
mean will be te = t0/A = 120/0.9677 = 124.
The effective SCV will be
ce2 = c02 + 2(1-A)Amr/t0 = 0.009375 + 2(1-0.9677)(0.9677)(120min)/120 =
0.07181. So, ce = 0.268.
Problem 5
In the tables below TH is the throughput that is given, te, is the mean effective process
time, ce2 is the effective SCV of the process times, u is the utilization given by (TH)(t e),
CTq is the expected time in queue given by
(1  ce2 ) u
CTq 
t e for the single machine case and
2 (1  u)
(1  ce2 ) u 2 ( m 1) 1
t e for the case with m machines.
2
m(1  u )
Finally, CT is the sum of CTq and te
CTq 
5
ISYE 6201 Spring 2006
Homework 6 Solution
a. In the first case, even though B has greater capacity, A has shorter cycle time
since its SCV is much smaller.
Machine
TH
te
2
ce
m
u
CTq
CT
A
B
0.92
1
0.92
0.85
0.25
1
0.92
7.1875
8.1875
4
1
0.782
7.6227
8.4727
b. Doubling the arrival rate (TH) and the number of tools makes station A have a
longer average cycle time than B.
Machine
TH
te
2
ce
m
u
CTq
CT
A
B
1.84
1
1.84
0.85
0.25
2
0.92
3.4616
4.4616
4
2
0.782
3.4125
4.2625
c. Note the large increase in cycle time with the modest increase in throughput as
compare to (a).
Machine
TH
te
2
ce
m
u
CTq
CT
A
B
0.95
1
0.95
0.85
0.25
1
0.95
11.8750
12.8750
4
1
0.8075
8.9140
9.7640
d. We now consider Machine A only.
i) First we increase TH by 1% from 0.5. The increase in cycle time is less than
one percent.
ii) Next we increase TH by 1% from 0.95. The increase in cycle time is almost
23%.
(i)
Machine
TH
te
2
ce
m
u
CTq
CT
% Increase
A
0.5
1
0.25
1
0.5
0.6250
1.6250
(ii)
A
0.505
1
0.25
1
0.505
0.6376
1.6376
0.7770
A
0.95
1
0.25
1
0.95
11.8750
12.8750
A
0.9595
1
0.25
1
0.9595
14.8071
15.8071
22.7736
6
ISYE 6201 Spring 2006
Homework 6 Solution
Problem 9
The same computations as in problem 5 are displayed for each option below. It is clear
that Option a is the best since it not only has the lowest cost but it results in less than 2
days average cycle time.
Machine
TH
te
ce2
m
u
CTq
CT
Cost
Current
10
1
Option a
10
1
5
11
0.9091
2.0689
3.0689
$
1
11
0.9091
0.6896
1.6896
8,000 $
Option b
10
1
Option c
10
0.96
5
12
0.8333
0.7104
1.7104
10,000 $
5
11
0.8727
1.2099
2.1699
8,500
Chapter 9
Problem 10
a. TH decreases. CT decreases.
b. TH increases. CT decreases.
c. TH increases. CT decreases.
d. TH decreases. CT increases.
e. TH decreases. CT decreases.
f. TH increases. CT decreases. Reducing the buffer size will reduce the WIP that
keeps the machine utilizations and the TH high. However, if the process times
variability is reduced at the same time, such protection may not be necessary and
the TH will still increase.
Chapter 10
Problem 1
~  3 u  3 0.9  27
a. w
1 u
1  0.9
w
27
rb 
0.5  0.466
b. TH 
w  W0  1
27  3  1
It can be seen that the CONWIP line achieves a higher throughput for this same WIP
level. This effect results from the fact that the push system may release work into the
line when the queue is very long, causing congestion that inflates the cycle time and (by
Little’s law) reduces the throughput.
7
ISYE 6201 Spring 2006
Homework 6 Solution
Problem 5
a. (i)
j
1
2
3
4
te(j)
2.5
2
2
2
ce(j)
1
1
1
1
w CT1(w) WIP1(w) CT2(w) WIP2(w) CT3(w) WIP3(w) CT4(w) WIP4(w)
0
0
0
0
0
1
2.500
0.294
2.000
0.235
2.000
0.235
2.000
0.235
2
3.235
0.608
2.471
0.464
2.471
0.464
2.471
0.464
3
4.019
0.942
2.928
0.686
2.928
0.686
2.928
0.686
4
4.854
1.297
3.372
0.901
3.372
0.901
3.372
0.901
5
5.743
1.674
3.802
1.109
3.802
1.109
3.802
1.109
6
6.686
2.075
4.217
1.308
4.217
1.308
4.217
1.308
7
7.686
2.498
4.617
1.501
4.617
1.501
4.617
1.501
8
8.745
2.946
5.001
1.685
5.001
1.685
5.001
1.685
9
9.865
3.418
5.369
1.861
5.369
1.861
5.369
1.861
10
11.046
3.916
5.721
2.028
5.721
2.028
5.721
2.028
11
12.289
4.438
6.056
2.187
6.056
2.187
6.056
2.187
12
13.596
4.986
6.374
2.338
6.374
2.338
6.374
2.338
CT(w)
8.500
10.647
12.804
14.971
17.149
19.337
21.537
23.749
25.973
28.209
30.458
32.719
TH(w)
0
0.118
0.188
0.234
0.267
0.292
0.310
0.325
0.337
0.347
0.354
0.361
0.367
(ii)
j
1
2
3
4
te(j)
2
2
2.5
2
ce(j)
1
1
1
1
w CT1(w) WIP1(w) CT2(w) WIP2(w) CT3(w) WIP3(w) CT4(w) WIP4(w)
0
0
0
0
0
1
2.000
0.235
2.000
0.235
2.500
0.294
2.000
0.235
2
2.471
0.464
2.471
0.464
3.235
0.608
2.471
0.464
3
2.928
0.686
2.928
0.686
4.019
0.942
2.928
0.686
4
3.372
0.901
3.372
0.901
4.854
1.297
3.372
0.901
5
3.802
1.109
3.802
1.109
5.743
1.674
3.802
1.109
6
4.217
1.308
4.217
1.308
6.686
2.075
4.217
1.308
7
4.617
1.501
4.617
1.501
7.686
2.498
4.617
1.501
8
5.001
1.685
5.001
1.685
8.745
2.946
5.001
1.685
9
5.369
1.861
5.369
1.861
9.865
3.418
5.369
1.861
10
5.721
2.028
5.721
2.028 11.046
3.916
5.721
2.028
11
6.056
2.187
6.056
2.187 12.289
4.438
6.056
2.187
12
6.374
2.338
6.374
2.338 13.596
4.986
6.374
2.338
CT(w)
8.500
10.647
12.804
14.971
17.149
19.337
21.537
23.749
25.973
28.209
30.458
32.719
TH(w)
0
0.118
0.188
0.234
0.267
0.292
0.310
0.325
0.337
0.347
0.354
0.361
0.367
Individual stations are affected by having bottleneck at station 1 or 3, but overall line
performance is not.
8
ISYE 6201 Spring 2006
Homework 6 Solution
b. (i)
j
1
2
3
4
te(j)
2
2
2.5
2
ce(j)
0.5
1
1
1
w CT1(w) WIP1(w) CT2(w) WIP2(w) CT3(w) WIP3(w) CT4(w) WIP4(w)
0
0
0
0
0
1
2.000
0.235
2.000
0.235
2.500
0.294
2.000
0.235
2
2.294
0.438
2.471
0.472
3.235
0.618
2.471
0.472
3
2.590
0.620
2.944
0.705
4.045
0.969
2.944
0.705
4
2.882
0.788
3.410
0.933
4.923
1.346
3.410
0.933
5
3.166
0.944
3.866
1.153
5.866
1.750
3.866
1.153
6
3.441
1.091
4.306
1.365
6.874
2.179
4.306
1.365
7
3.706
1.229
4.730
1.568
7.948
2.635
4.730
1.568
8
3.960
1.359
5.136
1.762
9.087
3.117
5.136
1.762
9
4.203
1.481
5.524
1.946 10.294
3.627
5.524
1.946
10
4.433
1.595
5.893
2.121 11.567
4.163
5.893
2.121
11
4.651
1.703
6.242
2.285 12.908
4.726
6.242
2.285
12
4.857
1.804
6.571
2.440 14.315
5.316
6.571
2.440
CT(w)
8.500
10.471
12.522
14.625
16.763
18.927
21.114
23.320
25.544
27.785
30.042
32.314
TH(w)
0
0.118
0.191
0.240
0.274
0.298
0.317
0.332
0.343
0.352
0.360
0.366
0.371
(ii)
j
1
2
3
4
te(j)
2
2
2.5
2
ce(j)
1
1
0.5
1
w CT1(w) WIP1(w) CT2(w) WIP2(w) CT3(w) WIP3(w) CT4(w) WIP4(w)
0
0
0
0
0
1
2.000
0.235
2.000
0.235
2.500
0.294
2.000
0.235
2
2.471
0.476
2.471
0.476
2.960
0.571
2.471
0.476
3
2.953
0.718
2.953
0.718
3.475
0.845
2.953
0.718
4
3.437
0.958
3.437
0.958
4.043
1.127
3.437
0.958
5
3.915
1.193
3.915
1.193
4.664
1.421
3.915
1.193
6
4.386
1.423
4.386
1.423
5.338
1.732
4.386
1.423
7
4.846
1.646
4.846
1.646
6.069
2.062
4.846
1.646
8
5.292
1.862
5.292
1.862
6.858
2.413
5.292
1.862
9
5.724
2.071
5.724
2.071
7.708
2.788
5.724
2.071
10
6.141
2.271
6.141
2.271
8.623
3.188
6.141
2.271
11
6.541
2.462
6.541
2.462
9.604
3.614
6.541
2.462
12
6.924
2.644
6.924
2.644 10.654
4.068
6.924
2.644
CT(w)
8.500
10.371
12.333
14.353
16.410
18.496
20.606
22.735
24.882
27.046
29.227
31.425
TH(w)
0
0.118
0.193
0.243
0.279
0.305
0.324
0.340
0.352
0.362
0.370
0.376
0.382
It is more effective (i.e., it improves TH more) to reduce variability at bottleneck.
9
ISYE 6201 Spring 2006
Homework 6 Solution
c. (i)
j
1
2
3
4
te(j)
2
0.25
2.5
2
ce(j)
1
1
1
1
w CT1(w) WIP1(w) CT2(w) WIP2(w) CT3(w) WIP3(w) CT4(w) WIP4(w)
0
0
0
0
0
1
2.000
0.296
0.250
0.037
2.500
0.370
2.000
0.296
2
2.593
0.585
0.259
0.058
3.426
0.772
2.593
0.585
3
3.169
0.862
0.265
0.072
4.431
1.205
3.169
0.862
CT(w)
TH(w)
6.750
8.870
11.034
0
0.148
0.225
0.272
4
3.723
1.126
0.268
0.081
5.512
1.667
3.723
1.126
13.226
0.302
5
6
7
8
9
10
11
12
4.252
4.754
5.227
5.672
6.088
6.476
6.836
7.168
1.377
1.614
1.836
2.044
2.238
2.418
2.584
2.736
0.270
0.272
0.273
0.274
0.275
0.275
0.276
0.276
0.088
0.092
0.096
0.099
0.101
0.103
0.104
0.105
6.667
7.897
9.202
10.580
12.032
13.558
15.154
16.820
2.159
2.681
3.232
3.813
4.423
5.062
5.728
6.422
4.252
4.754
5.227
5.672
6.088
6.476
6.836
7.168
1.377
1.614
1.836
2.044
2.238
2.418
2.584
2.736
15.442
17.676
19.929
22.198
24.484
26.785
29.101
31.432
0.324
0.339
0.351
0.360
0.368
0.373
0.378
0.382
(ii)
j
1
2
3
4
te(j)
2
2
2.5
2
ce(j)
0.5
0.5
1
0.5
w CT1(w) WIP1(w) CT2(w) WIP2(w) CT3(w) WIP3(w) CT4(w) WIP4(w)
0
0
0
0
0
1
2.000
0.235
2.000
0.235
2.500
0.294
2.000
0.235
2
2.294
0.453
2.294
0.453
3.235
0.640
2.294
0.453
3
2.610
0.656
2.610
0.656
4.099
1.031
2.610
0.656
CT(w)
TH(w)
8.500
10.118
11.930
0
0.118
0.198
0.251
4
2.936
0.846
2.936
0.846
5.077
1.463
2.936
0.846
13.884
0.288
5
6
7
8
9
10
11
12
3.259
3.575
3.878
4.165
4.436
4.690
4.926
5.145
1.023
1.188
1.342
1.486
1.619
1.742
1.855
1.958
3.259
3.575
3.878
4.165
4.436
4.690
4.926
5.145
1.023
1.188
1.342
1.486
1.619
1.742
1.855
1.958
6.157
7.330
8.590
9.933
11.358
12.860
14.438
16.090
1.932
2.436
2.973
3.543
4.144
4.775
5.436
6.125
3.259
3.575
3.878
4.165
4.436
4.690
4.926
5.145
1.023
1.188
1.342
1.486
1.619
1.742
1.855
1.958
15.935
18.054
20.222
22.429
24.666
26.930
29.217
31.524
0.314
0.332
0.346
0.357
0.365
0.371
0.376
0.381
Speeding up non-bottleneck here is better than reducing variability at nonbottlenecks.
Note that while this will often be the case, it will not always occur. Depending on the
specifics of the problem, including the costs, reducing variability can be more attractive
than increasing capacity.
C. Extra Credit – Chapter 9 Problem 6
The demand is 160 parts per day. With a 16-hour day, this becomes 160/16 = 10 parts
per hours.
(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.
10
ISYE 6201 Spring 2006
(b) The wait-to-batch time (WTBT) would be
Homework 6 Solution
k 1
125  1

 6.2 hours
2ra
2(10 pph)
10 pph
 0.48 .
20.833 pph
1
 0.008 and
The SCV of arrivals would be
125
the SCV of process time is (3h/6h)2 = 0.25.
 0.008  0.25  0.48 
Using the VUT equation we get CTq  

6  0.714h
2

 1  0.48 
Add these together with the process time and you get 6.2h + 0.714h + 6h = 12.91h.
The utilization would be u 
(c) The minimum batch size to meet demand is kmin/6h > 10 pph. Therefore kmin = 61
10
1
61  1
 0.984 , c a2 
 0.016 . Hence
 3h , u 
61 / 6
61
2(10 pph)
 0.016  0.25  0.984 
CTq  

6  47.95h
2

 1  0.984 
Average cycle time = 3 + 47.95 + 6 = 56.95 hours
(d) WTBT61 =
(e) Searching over the range reveals a minimum at batch size of 91 with a cycle time of
12.02 hours.
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