ISyE 3104: Introduction to Supply Chain Modeling: Manufacturing and Warehousing
Instructor: Spyros Reveliotis
Spring 2002
Homework #2
Due Date: Wednesday, 2/6/02
Problem Set:
1. Problems from textbook:
1.5 What is the purpose of a process plan? List the types of information normally included in the plan.
 The purpose of process plan is to transform part or product definition data into detail production
instructions; it can also be used to show step-by-step directions for process achievement.
 In general, the process plan includes the following types of information:
Sequence of operations to transform raw material into finished goods. Operations can be
fabrication, assembly, kitting, separation, testing, etc.
Production time information: setup time, unit-processing time and/or batch-processing
time.
Material handling information: amount of unit load, process batch, transfer batch, etc.
2.6 A product is produced in batches of 100 units. The machine requires 1 hour of setup time. Unit
processing time is 10 minutes but the machine can process up to 5 units at a time. After processing, each
unit must spend 2 hours on a cooling rack before it can be used. There is no limit to the number of units
that can be cooled at a time. Assuming there is no additional waiting time for the machine, find the
production time for batches of this product.
 Production time = Setup time + Processing time + Cooling time
= 1 hr. + (100/5)*10min + 2 hrs.
= 6 hrs and 20 min.
2.7 Describe the various types of physical layouts used by production systems and for each give an
example where it would be used.
 There are four types of physical layouts: product, process, cellular, and fixed position layouts,
each defined as follows:
Product layout – the product goes through a sequence of processes inside the plant
designed for and dedicated to that product. This type of layout is suitable for repetitive
manufacturing where demand is high and can support high utilization of the underlying
equipment. Product layout is also typical for the continuous-flow production of
commodity products processed in liquid or in bulk form, e.g., the production of paper
from pulp, the production of gas and other derivatives from crude oil, the production of
aluminum from bauxite.
Process layout – this layout groups the same or similar machines together according to
the operation they provide, establishing functional departments. Products go through the
different department in different sequences indicated by the corresponding process plans.
This type of layout provides more flexibility than product layout; however, it gives lower
throughput rate, needs skilled work at production processes, and, in general, it is more
difficult to schedule and control the underlying material flow. It might be suitable for low
volume, high variety, and/or high production cost. It is used extensively for the
fabrication of the various metallic or plastic parts of modern cars and appliances.
Cellular layout – separate cells are created within the production system, with each cell
producing a family of related parts or products. Each cell is essentially a mini-factory,
with short material moves and easier coordination. This layout is a compromise between
product and process layouts. An example can be garment manufacturing in which cell can
be classified by product types, i.e. shirt cell, pants cell, etc.
Fixed position layout – resources are allocated next to the ongoing product, instead of the
product going through a set of resources. An example of this type of layout might be ship
or space shuttle manufacturing.
2.23 A batch of material must visit five workstations in series to be converted into finished product. Each
batch makes 100 units. Setup time at each workstation is 1 hour. Variable processing time is 1.2 minutes
per unit. After completing an operation the batch has to wait an average of 1 hour for the material handler
to move it to the next machine. Assuming there is no time lost waiting for machines; find the time it takes
for a batch of material to go through the system.
We consider two different scenarios:
1. There will occur a setup time upon the arrival of a batch on each machine
 The total time for a batch = Setup time + Processing time + Material handling time
= 5 hr. + (1.2 min/unit/machine)(100 units)(5 machines) + 4 hrs.
= 300 + 600 + 240 = 1140 min. = 19 hrs.
2. All five workstations are setup together at the beginning of the process
 The total time for a batch = Setup time + Processing time + Material handling time
= 1 hr. + (1.2 min/unit/machine)(100 units)(5 machines) + 4 hrs.
= 60 + 600 + 240 = 900 min. = 15 hrs.
You can see that the second scenario results in a shorter total processing time for the considered batch, but
on the other hand, it enforces substantial idleness on the downstream machines (these machines could have
been used for some other jobs while the considered lot is processed in its first stages).
Repeat the previous problem assuming units are moved between operations in transfer batches of size 10
and that operations are close together so no time is lost waiting for material handling.
 The point here is that by reducing the transfer batch size from 100 to 10, multiple machines can
work on different sub-batches at the same time. This element of concurrency or parallelization
will eventually reduce the required total processing time for the entire batch. Again we consider
two different scenaria regarding the machine set-ups:
1. Each machine is set up only upon the arrival of the first sub-batch:
 Let’s consider at the end of the process line, the time until the first sub-batch is completed is
5(12+60) = 360 min. However, the time until the second batch is completed is the previous
amount of time plus the processing time of the last station = 360 + 12 = 372 min. Therefore, the
amount of processing time (the time until the last batch is completed) is 360 + 12*9 = 468 min.
Thus, the total processing time is 468 min = 7 hrs and 48 min.
Units
First 10 units
Second 10 units
Third 10 units
Fourth 10 units
Fifth 10 units
Sixth 10 units
Seventh 10 units
Eight 10 units
Ninth 10 units
Tenth 10 units
2.
Time until completed (min)
5(12+60) = 360
360 + 12 = 372
372 + 12 = 384
384 + 12 = 396
396 + 12 = 408
408 + 12 = 420
420 + 12 = 432
432 + 12 = 444
444 + 12 = 456
456 + 12 = 468
All five workstations are setup together upon the release of the entire batch of 100 parts into the
system:
 Now the required time for the completion of the first transfer batch is 60+5*12 = 120 min.
Similarly to the previous case, the completion time of the second batch is equal to the completion
time of the first transfer batch plus the processing time of the second batch on the last station =
120+12 = 132 min. The time for the completion of the last transfer batch is 120 + 12*9 = 228 min.
or 3 hrs and 48 min. The different between this and previous scenario is 4 hrs, which is the
additional setup time on the last four machines.
Units
First 10 units
Second 10 units
Third 10 units
Fourth 10 units
Fifth 10 units
Sixth 10 units
Seventh 10 units
Eight 10 units
Ninth 10 units
Tenth 10 units
Time until completed (min)
60+5*12 = 120
120 + 12 = 132
132 + 12 = 144
144 + 12 = 156
156 + 12 = 168
168 + 12 = 180
180 + 12 = 192
192 + 12 = 204
204 + 12 = 216
216 + 12 = 228
3.1 Why is demand forecasting important? How are forecasts used?
 Demand forecasting is important because it allows the company to see early on the various trends
developing into the market and plan accordingly. More specifically, demand forecasts taken over a
short to medium term, can predict how the product demand will behave in the future, and allow
the company to develop a set production plans describing the timing and quantity for
manufacturing products. Forecasts taken over a longer horizon can determine the new products to
be designed and technologies to be developed, the needs for part and material supply contracts,
financing plans, and capacity expansion.
 We notice that good forecasts must communicate also the level of uncertainty regarding the
forecasted quantities (typically in the form of confidence intervals around the forecasted values).
3.4 For checking model adequacy, why is the sum of absolute forecast errors a better measure than simply
the sum of forecast errors?
 The sum of absolute forecast errors will sum the total amount of errors; in contrast, in the sum of
forecast errors, the sums of positive and negative amounts of errors will tend to cancel each other,
often leading to a misleading low total/average error. Another way to avoid this canceling effect is
by using the sum of squared errors.
2. The Bill of Materials (BOM) for two end items, X and Y, and their sub-assemblies are as follows.
Classify the various components and sub-assemblies appearing in the above BOM’s into levels according
to the classification logic presented in class (hint: start by placing the end items X and Y at level 0)
Expanding the complete structure for each of the two end items we get:
X
Y
A
B
D
B
Therefore,
B
A
B
C
C
D
B
C
Level 0: X, Y
Level 1: components that feed only into level 0 items => A
Level 2: components that feed into level 1 and possibly level 0 items => D
Level 3: components that feed into level 2 and possibly level 1 and 0 items => B, C
Extra credit (30%)
Revise the theory on linear regression from your statistics classes, and apply it to problem 3.22 in your
textbook.
3.22 U.S. production of hydrochloric acid over a twelve-year period is shown in Table. Plot the data.
Hypothesize a model form. Estimate parameter values. Check adequacy of the model. Repeat until you are
satisfied with your model, then forecast 1993 demand.
Year (Xi)
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
Sum
Average
Xi
1
2
3
4
5
6
7
8
9
10
11
12
78
6.5
Sxx =
Sxy =


Production (Yi)
2574
2450
2468
2693
2803
2392
2996
2640
3268
3140
3381
3566
34371
2864.25
Xi^2
1
4
9
16
25
36
49
64
81
100
121
144
650
650-(78^2)/12 =
236904-78*34371/12 =
Xi*Yi
2574
4900
7404
10772
14015
14352
20972
21120
29412
31400
37191
42792
236904
143
13492.5
Sxy/Sxx =
13492.5/143 = 94.35315
E[Y]-1*E[X] = 2864.25 - 94.35315*(6.5) =
Simple linear regression model:
  o   1 xi   i
yi
Regression line:
2250.955
Y = 2251 + 94.35*X
Regression on Domestic Hygrochloric Acid Production
4000
3500
3000
2500
y = 94.353x + 2251
2000
R2 = 0.7337
1500
1000
500
Ye
92
90
91
19
19
88
89
19
19
87
19
85
86
19
19
83
84
19
19
81
82
19
19
19
ar
(
Xi
)
0
Forecasted demand for 1993:
ANOVA
Y = 2251 + 94.35*13 = 3477.55
df
Regression
Residual
Total
SS
MS
F
Significance F
1 1273060 1273060 27.55049
0.000374002
10 462082.4 46208.24
11 1735142
Coefficients Standard Error
t Stat
P-value
Intercept
2250.954545
132.2993644 17.0141 1.04E-08
X Variable 194.35314685
17.97594383 5.248856 0.000374
Lower 95%
Upper 95%
1956.173141 2545.73595
54.30024107 134.4060526
Hypothesis:
Ho: o = 0
Since the p-value = 1.04 E-08 is so small, the null hypothesis is rejected, which means that
intercept is not equal to zero.
Ho: 1 = 0
Since the p-value = 0.000374 is so small, the null hypothesis is rejected, which means that
the slop is not equal to zero
Adequacy:
X Variable 1 Residual Plot
4000
400
3000
200
Residuals
Y
Normal Probability Plot
2000
1000
0
-200
0
5
10
-400
0
-600
0
50
100
Sample Percentile
150
X Variable 1
Regression Statistics
Multiple R
0.856558177
R Square
0.733691911
Adjusted R Square
0.707061103
Standard Error
214.9610235
From the Normal plot, residual plot, and the R-square = 0.73, we can say that this model is “fairly”
accurate.
15