Operations Management Assignment - RMIT, Semester 1, 2020

This Operations Management assignment presents solutions to two case studies: ZenoFex and FORTech Inc. The ZenoFex case explores forecasting methods, specifically simple moving average (SMA) and exponential smoothing, to predict semiconductor demand. The solution includes calculations, graphs, and a comparison of the two methods, recommending SMA based on accuracy. The FORTech Inc. case delves into process capability analysis, using control charts and statistical calculations (Cpk) to assess a production process. The solution analyzes data from two sets of samples, determining process capability and explaining the impact of sample size and data variations. The assignment also addresses the implications of identifying and correcting process problems, computing potential process capability improvements, and discussing the trade-offs of sample size in statistical analysis. The student uses the Stevenson (2015) Operations Management textbook to support the analysis.

Operations Management 1
Operations Management
By (Name)
Course
Professor’s Name
Institution
Date

Operations Management 2
Operations Management
ZenoFex Case Study
The simple moving average forecasts demand by getting the moving average over a given
interval in this case. The forecast average is taken to be the expected demand for the next
demand period. In this case the SMA (Simple moving average is calculated in 3 period intervals.
The average is calculated as = sum of 3 consecutive periods /3. These calculations can be done
using excel SMA tool, as shown below.
To obtain this computational table
using excel, the data is copied into an
excel workbook. Once the data is ready,
launch the data analysis tool and select the
moving average from the options. Then
give the range of input, the interval, and
select chart output. Click okay to get the
results.
Plotted on a graph, the forecast will be as
Period Semiconductor sum 3/3 Difference
1 40 #N/A #N/A
2 39 #N/A #N/A
3 55 44.67 -10.33
4 88 60.67 -27.33
5 51 64.67 13.67
6 54 64.33 10.33
7 66 57.00 -9.00
8 63 61.00 -2.00
9 57 62.00 5.00
10 68 62.67 -5.33
11 77 67.33 -9.67
12 72.50 72.50
Average -3.85

Operations Management 3
shown in the graph below.
1 2 3 4 5 6 7 8 9 10 11
0
20
40
60
80
100
Moving Average
Actual
Forecast
Period
Semiconductors
Exponential smoothing is considered a rule of thumb tool. It is used to smooth out variate
data. The difference from SMA is that exponential smoothing assigns a function to allocate
different weights over time.
Below is the forecast data representation for ZenoFex using exponential smoothing.
Using excel, the data is analyzed as follows. Given the damping factor is 0.4. To obtain this
computational table using excel, the data is copied into an excel workbook. Once the data is
ready, launch the data analysis tool and select exponential smoothing from the options. Then
give the range of input, range of output, the damping factor, and select chart output. Click okay
to get the results.
Period Semiconductor Difference
1 40 #N/A #N/A
2 39 40.00 1.00
3 55 39.40 -15.60
4 88 48.76 -39.24
5 51 72.30 21.30

Operations Management 4
6 54 59.52 5.52
7 66 56.21 -9.79
8 63 62.08 -0.92
9 57 62.63 5.63
10 68 59.25 -8.75
11 77 64.50 -12.50
12 72.00 72.00
Average -5.33
Represented in a chart the results are as follows
1 2 3 4 5 6 7 8 9 10 11
0
20
40
60
80
100
Exponential Smoothing
Actual
Forecast
Data Point
Value
The company should adopt the Simple Moving average forecast method. The reason for
choosing the simple Moving Average method is because it is more accurate, as can be seen from
the calculation, the average deviation of the forecast values using the SMA (-3.85) is less than
the deviation when using exponential smoothing (-5.33).

Operations Management 5
FORTech Inc. Case Study
1. How did Scott conclude that the process was not
capable based on his first set of samples?
Scott used 18 samples. For a sample size n = 18, the
value of A2 (constant is 0.194)
For the process to be viable, the process capacity
coefficient must be greater than 1.33, CPK > 1.33.
Process capability = specification width / 6 * Std dev
From the excel table, the mean range is 0.8741 and the
mean of means is 44.99889
R- bar = 0.8741
A2 = 0.194
Scotts Analysis
Sa
mple
M
ean
R
ange
1
4
5.01
0
.8500
2
4
4.99
0
.8900
3
4
5.02
0
.8600
4
4
5
0
.9100
5
4
5.04
0
.8700
6
4
4.98
0
.9000
7
4
4.91
0
.8600
8
4
5.04
0
.8900
9
4
5
0
.8500
10
4
4.97
0
.9100
11
4
5.11
0
.8400

Operations Management 6
A2 R = 3 σ
❑
√ n
0.194 × 0.8741 = 3 σ
❑
√ 18
σ =0.239
Process capability = 1.44 /6 * 0.239
Cpk = 1.0008.
Scott used this calculation to conclude that the process was not capable; the process
capacity value was 1.0008, which is lower than the threshold for capacity 1.33; hence the process
is not capable.
2. Does the second set of samples show anything that the first set didn’t? Explain what
and why
The second set of samples shows significant variations in the range. For the first set of
samples, there was a very small range for the individual sample observations because the
margins were covered up in the mean of large sample sizes. A small sample size is more
sensitive to variations and easily highlights small errors within the data.
3. Assuming the problem can be found and corrected, what impact do you think this
would have on the capability of the process? Compute the potential process
capability using the second data set.
If the problem is identified and corrected, the process capacity will increase/improve, and
the process would become capable of more performance.
27 samples were used. For a sample size n = 27, the value of A2 (constant is 0.146)

Operations Management 7
For the process to be viable, the process capacity coefficient must be greater than 1.33,
CPK > 1.33.
Sample Mean Range
1 44.96 0.42
2 44.98 0.39
3 44.96 0.41
4 44.97 0.37
5 45.02 0.39
6 45.03 0.4
7 45.04 0.39
8 45.02 0.42
9 45.08 0.38
10 45.12 0.4
11 45.07 0.41
12 45.02 0.38
13 45.01 0.41
14 44.98 0.4
15 45 0.39
16 44.95 0.41
17 44.94 0.43
18 44.94 0.4
19 44.87 0.38
20 44.95 0.41
21 44.93 0.39
22 44.96 0.41
23 44.99 0.4
24 45 0.44
25 45.03 0.42
26 45.04 0.38
27 45.03 0.4
Average
44.9959
3
0.40111
1
Process capability = specification width / 6 * Std dev
From the excel table, the mean range is 0.4011 and the mean of means is 44.99593

Operations Management 8
R- bar = 0.4011
A2 = 0.146
A2 R = 3 σ
❑
√ n
0.1 46 × 0. 4011 = 3 σ
❑
√ 27
σ =0. 1014
Process capability = 1.44 /6 * 0.1014
Cpk = 2.3669.
Cpk is 2.3669, which is higher than 1.33; hence the process is capable.
4. If small samples can reveal something that large samples might not, why not just
take small samples in every situation?
Small sample size decreases statistical power and increases the risk of type 2 errors.
Reducing the sample size also increases the margin of error.

Operations Management 9
References
Stevenson, W.J. (2015) Operations Management, 12th edition, McGraw Hill.