4.3 Solutions to the Problem Set
Problem 4.1
80%
[b]. The theoretical flow time is 36 minutes:
There are three paths through the system:
A: Take Order – Food – Deliver – Bill → 4 + 18 + 12 + 2 = 36 mins
B: Take Order – Wine – Deliver – Bill → 4 + 4.8 + 12 + 2 = 22.8 mins
C: Take Order – Cart – Deliver – Bill → 4 + 10 + 12 + 2 = 28 mins
Path A is critical so the TFT is 36 minutes
[c]. The flow time efficiency is 36/60 = 60% Problem 4.2

The flow chart of the process is as shown in Figure TM4.1.
Figure TM4.1. Flow chart of the Kristen Cookies Process flow
Flow unit = 1 order of 1 dozen the theoretical flow time is 26 minutes. This is determined by adding the activity times from start to finish for 1 dozen cookies.

Flow unit = 1 order of 2 dozen. To determine the theoretical flow time for this flow unit, we first observe that certain activities can be performed in parallel. For example, while the oven is baking the first dozen, You can spoon the dough for the second dozen into another tray. Therefore, the flow time of such an order is not simply the sum of the activity times. A useful tool is a Gantt chart that shows the times during which different resources of interest are occupied for various activities. A Gantt chart for the three resources executing an order of 2 dozen cookies is shown in Figure TM4.2. The dough for the 2 dozen cookies is mixed by You in 6 minutes and subsequently you spoon dough for 1 dozen in 2 minutes. Therefore in the 8^{th} minute, the RM is ready to load the oven and set timer, which takes 1 minute. The oven starts baking the first dozen at the 9^{th} minute and completes baking at the 18^{th} minute. Meanwhile, You spoon the second dozen into another tray. At the 18^{th} minute, the RM unloads the first tray from the oven and loads the second tray into the oven and sets the timer. So the second dozen starts baking at the 19^{th} minute. While the second dozen bakes, the first dozen cookies cool and RM packs them into a bag, which takes a total of 7 minutes. At the 28^{th} minute the second dozen finishes baking at which time the RM unloads the tray. After cooling for 5 minutes, the RM packs the second dozen in 2 minutes by the 35^{th} minute. Finally, payment for the order and delivery to customer takes place in the 36^{th} minute. Therefore the theoretical flow time for an order of 2 dozen cookies is 36 minutes.
An alternate way to compute the theoretical flow time is using the concept of bottleneck resources (discussed in Chapter 5). Oven is the bottleneck resource with 10 minutes of activity per dozen cookies. Observe that the first dozen goes into the oven at the end of 8 minutes. The second dozen will go into the oven 10 minutes after the first dozen. Therefore, the theoretical flow time of the second dozen = theoretical flow time of the first dozen + activity time at the bottleneck = 26 + 10 = 36 minutes. Similarly, theoretical flow time of 3 dozen = theoretical flow time of the first dozen + 2 * activity time at the bottleneck = 26 + 2*10 = 46 minutes.
Figure TM4.2: Gantt Chart for Kristen’s Cookie; order size = 2 dozen.

With two ovens, the baking of the two dozen can be overlapped. So You can spoon the second dozen into another tray by the 10^{th} minute. At this time, the RM is ready to load the second oven. The second oven will finish baking the second dozen by the 20^{th} minute. The first dozen is out of the oven at the 18^{th} minute after which it cools for 5 minutes and then the RM packs them in 2 minutes finishing at 25^{th} minute. By this time, the second dozen has cooled and ready to be packed which takes another 2 minutes. Payment and delivery takes another 1 minute finishing the order in 28 minutes.

With one big oven, both trays get loaded into the oven at the end of the10^{th} minute. The timer is set only once; the baking starts at the 11^{th} minute and finishes at minute 20. The two trays cool simultaneously by minute 25. Packing takes a total of 4 minutes and payment and delivery of order another 1 minute completing the process in 30 minutes.

With one faster convection oven, we can repeat the procedure as for a single oven (using the Gantt chart). The first dozen starts baking in the convection oven at the 9^{th} minute and finishes baking at the 15^{th} minute. Meanwhile the second dozen is ready to go into the oven. The RM sets the timer in 1 minute and the baking process for the second dozen starts at the 16^{th} minute and ends at the 22^{nd} minute. Meanwhile, there is just enough time for the first dozen to cool and be packed. At the end of 22^{nd} minute, the RM unloads the second dozen, lets it cool for 5 minutes and packs them taking 2 more minutes. The order is ready by 29^{th} minute at which time it is delivered and payment is accepted, giving a total theoretical flow time of 30 minutes.
Problem 4.3

From Table 4.9 of the Problem set, we compute the work content of various activities as shown in Table TM4.2. As in Example 4.3, the two paths are:
Path 1 (roof): Start 1 3 5 7 8 End
Path 2 (base): Start 1 2 4 6 7 8 End
The theoretical flow time of each path is determined by adding the work contents of the activities along that path. Thus the theoretical flow time of path 1 is 100 minutes and that of path 2 is 127 minutes. Path 2 is then the critical path and the theoretical flow time for the Deluxe model is 127 minutes.

Table 4.2 of the chapter gives the work content of the standard garage. Taking a weighted combination of the work contents of the standard garage and the deluxe garage, we compute the work content of a 75% Standard and 25% Deluxe product mix as shown in Table TM4.2. The length of paths 1 and 2 are computed to be equal to 92.5 and 110.5 minutes respectively. Therefore the theoretical flow time for a mixture of 75% standard and 25% Deluxe is 110.5 minutes.
[c] The flow time efficiency is 110.5/4800 = 2.3%
Problem 4.4

A process flow chart identifying resources, activity times and any potential storage buffers is given in Figure TM4.3.

We have 4 paths, the critical path is the longest path = 37 minutes (going through seat assembly). Therefore, the theoretical flow time = 37 minutes.

We decrease the theoretical flow time only by decreasing the length of the critical path. Thus we need to decrease the theoretical flow times of seat assembly and/or final assembly: try to change the design and process so that we can perform subtasks in parallel.


Deluxe

Garage

Work

Content

(minutes)


Activity

Activity Time

Number
of



75% Standard
and



(Minutes)

Visits

Deluxe

Standard

25% Deluxe

1

Separate

10

1

10

10

10

2

Punch the base

30

1.5

45

30

33.75

3

Punch the roof

24

1.25

30

22

24

4

Form the base

5

1.2

6

6

6

5

Form the roof

10

1.2

12

12

12

6

Subassemble

15

1.2

18

13

14.25

7

Assemble

12

1

12

10

10.5

8

Inspect

30

1.2

36

36

36

Table TM4.2: Work Content computation for Problem 4.2
Figure TM4.3: Flow Chart for Problem Set 4.4 (a)
4.5
Activity
B
Activity
A
F1
Loan Officer
Process
And
Separate
15 min
10 min
F2
Activity
C
10 min
15 min
20 min
[b] The theoretical flow time is 50 minutes. The critical path is the one of F1 (process and separate, Activity A, Activity B, Loan Officer). [c] The flow time efficiency is 50/7*480 = 50/ 3360= 1.49% 4.6 [a]. The theoretical flow time of the process is 17 seconds: For passenger without additional check (80% of passangers), the theoretical flow time is (1.5/18)*60 = 5 seconds
For passengers with additional check the theoretical flow time is 5 +60 = 65 seconds
The theoretical flow time for an average passenger is 80% *5 + 20% * 65 = 17 seconds
[b] The flow time efficiency is 17/ 135 = 12.5% 4.7 [a] The theoretical flow time can be obtained by averaging the activitity time of the two activities, “time with judge” and “pay fine”. These amount to 1.55minutes and 2.4 minutes respectively. Thus the theoretical flow time is 1.55 + 2.4 = 3.95 minutes. [b] The average flow time of the process can be estimated by taking average of the total time spent is system. This come out to 120 minutes. [c] The flow time efficiency is 3.95/120 = 3.3%
