CHAPTER 4: FLOW TIME ANALYSIS
4.3 Solutions to the Problem Set
Problem 4.1
[a]
Draw a process flow diagram.
Prepare Food
Kitchen
18 Minutes
Take Order
Deliver Order
Prepare Wine
80%
Bill Guest
Manager
Sommelier
Waiter
Waiter
4 Minutes
Average Time= 0.8 ×
12 Minutes
2 Minutes
6 = 4.8Minutes
Prepare Cart
Waiter
10 Minutes
[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
[a] The flow chart of the process is as shown in Figure TM-4.1.
25
26
Chapter 4
Wash
& Mix
Take
Order
You
6min / 1-3doz
Spoon
You
2min /
Load &
Set Timer
Bake
Roommate, Oven
Oven
1 min / doz.
9 min/doz.
doz.
Order
Ready
Accept
Payment
Pack
Roommate
Roommate
1 min / order
2 min /doz
Cool
-
Unload
Roommate
5 min
Figure TM-4.1. 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.
[b] 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 TM-4.2. 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 8th 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 9th minute and completes baking at the
18th minute. Meanwhile, You spoon the second dozen into another tray. At the 18th 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 19th 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 28th 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 35th minute. Finally, payment for the order and delivery to
customer takes place in the 36th 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.
Chapter 4 27
You
RM
Oven
5
8 10
15
20
25
30
35
Minutes
Figure TM-4.2: Gantt Chart for Kristen’s Cookie; order size = 2 dozen.
1. With two ovens, the baking of the two dozen can be overlapped. So You can spoon the second
dozen into another tray by the 10th minute. At this time, the RM is ready to load the second oven.
The second oven will finish baking the second dozen by the 20th minute. The first dozen is out of
the oven at the 18th minute after which it cools for 5 minutes and then the RM packs them in 2
minutes finishing at 25th 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.
2. With one big oven, both trays get loaded into the oven at the end of the10th minute. The timer is
set only once; the baking starts at the 11th 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.
3. 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 9th minute and finishes
baking at the 15th 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 16th minute and
ends at the 22nd minute. Meanwhile, there is just enough time for the first dozen to cool and be
packed. At the end of 22nd 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 29th minute at which time it is
delivered and payment is accepted, giving a total theoretical flow time of 30 minutes.
Problem 4.3
[a] From Table 4.9 of the Problem set, we compute the work content of various activities as shown in
Table TM-4.2. 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
28
Chapter 4
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.
[b] 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 TM-4.2. 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] A process flow chart identifying resources, activity times and any potential storage buffers is given in
Figure TM-4.3.
[b] We have 4 paths, the critical path is the longest path = 37 minutes (going through seat assembly).
Therefore, the theoretical flow time = 37 minutes.
[c] 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.
Chapter 4 29
Activity
1 Separate
2 Punch the base
3 Punch the roof
4 Form the base
5 Form the roof
6 Sub-assemble
7 Assemble
8 Inspect
Deluxe Garage
Work Content (minutes)
Activity
Number
75% Standard
Time
of
and
(Minutes)
Visits
Deluxe Standard 25% Deluxe
10
1
10
10
10
30
1.5
45
30
33.75
24
1.25
30
22
24
5
1.2
6
6
6
10
1.2
12
12
12
15
1.2
18
13
14.25
12
1
12
10
10.5
30
1.2
36
36
36
Table TM-4.2: Work Content computation for Problem 4.2
2 min
1 min
3 min
molding
aut. molding machine
start
eng. ass.
1 molding operator
welding
1 min
3 min
stamping
1 engine assembler
30 min
continuous welding machine
FA
1 welder
2 min
10,000lbs press
1 press operator
10 final assemblers
7 min
seat ass.
2 seat assemblers
Figure TM-4.3: Flow Chart for Problem Set 4.4 (a)
end
30
Chapter 4
4.5
[a]
A process map
Activity
A
F1
15
min
Process
And
Separate
10
min
Activity
B
F2
10
min
Activity
C
Loan
Officer
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
Chapter 4 31
[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%