OPSMGT 258: Business Process Design Assignment 1 - Semester Two 2019

This document presents a comprehensive solution to Assignment 1 for the OPSMGT 258 Business Process Design course. The assignment delves into the analysis of various business processes, calculating key performance indicators such as cycle time, throughput time, and work in process (WIP). It explores the identification of bottlenecks and the impact of different process configurations on overall efficiency. The solution includes detailed calculations and explanations for scenarios involving both serial and parallel process flows, as well as real-world applications. Furthermore, the assignment analyzes a case study of Miami Children's Hospital, identifying strategic, tactical, and operational level objectives. The solution also demonstrates the application of lean principles such as value stream mapping and cost avoidance strategies. The document also includes tables and charts to illustrate the WIP storage and its changes during the day.

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OPSMGT 258
Business Process Design
Assignment 1 – Semester Two 2019

Table of Contents
Task 1.................................................................................................................... 3
Task 1 (a)........................................................................................................... 3
Task 1 (b)........................................................................................................... 3
Task 1 (c)............................................................................................................ 3
Task 1 (d)........................................................................................................... 3
Task 1 (e)........................................................................................................... 4
Task 1 (f)............................................................................................................ 4
Task 1 (g)........................................................................................................... 4
Task 1 (h)........................................................................................................... 4
Task 2.................................................................................................................... 5
Task 2 (a)........................................................................................................... 5
Task 2 (b)........................................................................................................... 5
Task 2 (c)............................................................................................................ 5
Task 2 (d)........................................................................................................... 5
Task 2 (e)........................................................................................................... 6
Task 2 (f)............................................................................................................ 6
Task 2 (g)........................................................................................................... 6
Task 5.................................................................................................................... 7
Task 6.................................................................................................................... 8
Task 6 (a)........................................................................................................... 8
Task 6 (b)........................................................................................................... 8

List of Tables
Table 1: WIP Storage during the day.....................................................................7
List of Figures
Figure 1: WIP Storage Chart.................................................................................. 6

Task 1
Task 1 (a)
The process capacity is equivalent to the maximum output rate or the minimum
throughput rate of all the tasks (Snyder & Shen, 2011). The lowest capacity task
in the process determines the capacity of the overall process. In this case,
“Cutting” task has the minimum flow rate or throughput rate. Therefore, the
capacity of this process = 75 shirts/hr
Task 1 (b)
The cycle time at “Cutting” task = 1/75 hr/shirt or 48 seconds/shirt
The cycle time at “Sewing” task = 1/140 hr/shirt or 26 seconds/shirt
Bottlenecks are obviously those tasks where materials ply up and process in the
next step stands idle (Kazemian & Aref, 2016). Hence, the slowest task is
determined as bottleneck. The identification of bottleneck is essential aspect of
process design analysis as it would provide opportunity to minimise the delay
and increase capacity of the overall process. The cycle time of “Cutting” is more
than “Sewing”. Cutting would be the bottleneck task. Hence, the non-bottleneck
task would be “Sewing” which has the cycle time of 26 seconds/shirt
Task 1 (c)
Starvation occurs when the step after the bottleneck task becomes idle because
its cycle time is less than the previous step. It happens because the upstream
delays and downstream activity becomes idle without any input to process
(Stadtler, 2015). It can be observed that “Sewing” task becomes idle after 26
seconds as it waits for WIP because the bottleneck process can supply materials
only after 48 seconds.
The starvation time at “Sewing” would be 48 – 26 = 22 seconds for each shirt.
Task 1 (d)
Capacity utilization determines that percentage of the process capacity which is
actually used (Hugos, 2018). Hence, utilization of a process activity gives the

proportion of the time that the operator is working on a particular task. The
utilization is computed using the formula:
Utilization of the process= Cycle time of task
Cycle time of bottleneck
Hence,
Utilization of the Cutting task= Cycle time of Cutting
Cycle time of bottleneck
Utilization of the Cutting task= 48 second
48 second =100 %
And,
Utilization of the Sewing task= Cycle time of Sewing
Cycle time of bottleneck
Utilization of the Sewing task= 26 second
48 second =54 %
Task 1 (e)
The total time in which a blank is pulled from “RMI” and is placed in “FGI” will be
equivalent to the total cycle time or throughput time = 48 + 26 = 74 seconds
Task 1 (f)
Throughput time is defined as the length of time from the release of a material
to the process and its arrival as finished goods inventory to the final customer
(Berrah & Foulloy, 2013).
Therefore, the throughput time in the process = average time a unit takes to go
through the entire process
Therefore, throughput time = 48 + 26 = 74 seconds
Task 1 (g)
Work in Process is the amount of partially finished inventory in the process which
awaits the completion of the process to be converted into finished goods. The

WIP is important characteristic of the process as if it increases without changing
the throughput, the cycle time increases. The performance is measured with the
average WIP which is proportional to the throughput time and throughput rate
and the relation is determined using Little’s Law (Axsäter, 2015). It is given by
the formula:
The work in process (WIP) = Throughput time X Throughput Rate
Where, Throughput time = 74 seconds
and Throughput Rate = 0.020833 shirt/second
Hence, WIP of this process = 74 X 0.020833 = 1.54 shirts
Task 1 (h)
i. True, because the cycle time of “Sewing” task is less than the “Cutting” task.
Inventory will be stored in the buffer after each task with the buffer after the final
task being the finished goods. The initial tasks will be assumed to have been
standardized by way of storing the inventory in the buffer (Graves &
Schoenmeyr, 2016).
ii. True, as it will prevent “Sewing” task to remain idle.
iii. False, as it will eliminate the waiting time.
iv. True, as the throughput time will increase

Task 2
Task 2 (a)
The minimum throughput rate among all the tasks is for “Sewing” in this case.
Therefore, the capacity of this process = 50 shirts/hr
Task 2 (b)
The cycle time at “Cutting” task = 1/75 hr/shirt or 48 seconds/shirt
The cycle time at “Sewing” task = 1/50 hr/shirt or 72 seconds/shirt
The cycle time of “Sewing” is more than “Cutting”. Hence, the non-bottleneck
task would be “Cutting” which has the cycle time of 48 seconds/shirt
Task 2 (c)
Blocking occurs when there is not any provision of buffer before the bottleneck
task. In this way the activity becomes idle as the next downstream task cannot
be ready to take the material (Fredendall & Hill, 2011). It can be observed that
“Cutting” task is blocked after 48 seconds until the task “Sewing” completes.
The blocking time at “Cutting” would be 72 – 48 = 24 seconds for each shirt.
Task 2 (d)
Utilization of a process activity gives the proportion of the time that the operator
is actually working on a particular task. The utilization is computed using the
formula:
Utilization of the process= Cycle time of task
Cycle time of bottleneck
Hence,
Utilization of the Cutting task= Cycle time of Cutting
Cycle time of bottleneck
Utilization of the Cutting task= 48 second
72 second =67 %

And, Utilization of the Sewing task= Cycle time of Sewing
Cycle time of bottleneck
Utilization of the Sewing task= 72 second
72 second =100 %
Task 2 (e)
The total time in which a blank is pulled from “RMI” and is placed in “FGI” will be
equivalent to the total cycle time = 48 + 72 + 24 = 142 seconds
Task 2 (f)
The work in process (WIP) = Throughput time X Throughput Rate
Where, Throughput time = 142 seconds
and Throughput Rate = 0.0139 shirt/second
Hence, WIP of this process = 142 X 0.0139 = 1.97 shirts
Task 2 (g)
i. The maximum level of WIP = (75 X 16) – (50 X 15) = 1250 – 750 = 450 shirts
(at the end of 2nd shift)
ii. The following chart present the change in WIP during the day
0 8 16 24
0
100
200
300
400
500
WIP Storage Graph
Time of Day (in Hours)
WIP Storage (shirts)
Figure 1: WIP Storage Chart
WIP storage has been calculated as per the given information and presented in
the following table:

Time of day (in
Hours)
WIP Storage
(shirts)
1 75
2 100
3 125
4 150
5 175
6 200
7 225
8 250
9 275
10 300
11 325
12 350
13 375
14 400
15 425
16 450
17 400
18 350
19 300
20 250
21 200
22 150
23 100
24 50
Sum = 6000
Table 1: WIP Storage during the day
iii. The average WIP = 250 shirts during the day
Task 5
The seven operational level processes at the Miami Children’s Hospital can be
identified as follows:
1. Paperwork (00:48 “…how we process paperwork…”)
2. Patient Scheduling (00:46 “…scheduling the patients actually coming in…”)
3. Patient Arrival (00:42 “…we have several different areas of responsibility from
prior to the patient arriving…”)
4. Preparation (1:10 “…sterile processing department where we worked on our
trays…”)

5. Operation Treatment (1:28 “…we could get our patient in the room on
time…”)
6. Post operation recovery (2:30 “…patient was in these bed boards are located
in inpatient nursing units…”)
7. Discharge (2:00 “The discharge process was probably our biggest one….”)
Task 6
Task 6 (a)
 Strategic level objectives:
o To reduce cost (00:55 “…what we have seen as bottom line impact
is cost avoidance…”)
o To deliver best care services (3:44 “…value back to the patient…the
satisfaction with them…”)
 Tactical level objectives:
o To reduce timeline (2:35 “…what the timeline of the patient…”)
o To streamline and integrate several processes (00:35 “lean brought
the ability to unite several different areas that once were
segregated…”)
 Operational level objectives:
o Eliminate batch scheduling (3:00 “…by eliminating batching we
probably save 20% of the whole process…”)
o To use value stream map (3:33 “…we need a value stream map…
for lean process improvement…”)
Task 6 (b)
The performance target at Miami Children’s Hospital case can be identified as
cost avoidance by eliminating wastage of time (00:57 “…cost avoidance of
$415,000…”). The patient timeline at the care centre is used as KPI to measure
the wastage of time (2:35 “…what the timeline of the patient while they are in
the hospital…”)

Refences
Axsäter, S. (2015). Inventory control (Vol. 225). Springer.
Berrah, L., & Foulloy, L. (2013). Towards a unified descriptive framework for
industrial objective declaration and performance measurement. Computers in
Industry, 64(6), 650-662.
Fredendall, L. D., & Hill, E. (2016). Basics of supply chain management. CRC
Press.
Graves, S. C., & Schoenmeyr, T. (2016). Strategic safety-stock placement in
supply chains with capacity constraints. Manufacturing & Service Operations
Management, 18(3), 445-460.
Hugos, M. H. (2018). Essentials of supply chain management. John Wiley & Sons.
Kazemian, I., & Aref, S. (2016). Multi-echelon supply chain flexibility
enhancement through detecting bottlenecks. Global Journal of Flexible Systems
Management, 17(4), 357-372.
Snyder, L. V., & Shen, Z. J. M. (2011). Fundamentals of supply chain theory (p.
367). Hoboken, NJ: Wiley.
Stadtler, H. (2015). Supply chain management: An overview. In Supply chain
management and advanced planning (pp. 3-28). Springer, Berlin, Heidelberg.