CHAPTER 4: FLOW TIME ANALYSIS
4.3 Solutions to the Problem Set
Problem 4.1
[a] Draw a process flow diagram.
[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.
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
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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 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
14) Refer to the table. What is the average time a customer spends waiting in line and being served?
Thus we get 6+2*(2+1+2)+1=17 min of valuable time for two orders. For three orders we get 6+3*(2+1+2)+1=22 min of valuable time. A further assumption is that the costs of my and my roommate’s time are L per hour, then the average costs for the first dozen are: (12/60)*L= 0,2L, for the second dozen: ((17/60)*L)/2=0,142L, for the third dozen: ((22/60)*L)/3=0,122. We see that the costs for each the second and the third dozens are less than for the first. Thus, we can give discount on orders for more than one dozen. These can be for the first dozen up to (0,2-0.141)*L=0.058L for the second dozen, and (0,2-0,122)*L=0,078L for the third order.
Therefore, we would only be willing to pay at the most, (27-22) orders worth of rent, which = 5 * (2.00 + 0.70) = $13.50 per day.
B cannot be correct, because the freight rates were reduced from one hundred dollars to five dollars. It was a dramatic reduction in expense and a speedy success. The travel time from New York City to Buffalo decreased from twenty days to six days along with the cost. However, the correct numbers for the reduction of freight rates is one hundred dollars to five dollars, not two hundred dollars to two dollars.
As we saw in the earlier section the critical path activities are not delayed. But some of the tasks that are delayed are mentioned in the table below. The table below shows the free slack and the total slacks of the tasks that are delayed.
Activity B: This activity is not started yet; it will start in 2 weeks. It is planned to cost $400,000. Its costs to date are $0.
b. The branch manager wants to improve the service and suggests dispatching buses every 0.5 min. She argues that this will reduce the average traveling time from the airport to Avis offices to 2.5 min. Is she correct? If your answer is negative then what will the average traveling time be?
Using the intercompany-billing rate per hour of $400 and the intercompany demand of 205 hours we calculate revenue and expenses.
Diagram 1 displays the Activity-on-Node diagram. The critical path is identified by the blue dash line. The total shortest time to finish the project is 786 hours, equivalent to 98.25 days. EOC requested to finish the project in 6 month, equivalent to 120 working days.
Dr. Shreffler is staying at the non-conference hotel which is lower in cost even with her request for a rental car. The car is needed because transportation from her hotel to the event conference is $95.40 one-way for $95.60 for 6 days. Thus, the round-trip for 6 days is $41,144.80 so a rental car is the most advantageous method of
d) Q = 2,529.8, R = 60 The number of operating days to receive the entire order is Q = 2,529.8 = 42.16 days R 60
5. If the company continues to use one technician when the customer base expands to 20 customers, the average time in the waiting line will increase to 6.9454 hours. With an average travel time of 1 hour, the average total waiting time will be 6.9454 + 1 = 7.9454 hours. The total cost will be $397.78 per hour. This average total waiting time is too long and a second technician is definitely necessary. Using output from The Management Scientist, two service technicians provide the following:
The second element is work center loading database. This database presented data via series of loading graphs. It also allowed the bottleneck to be identified and remedial action to be taken and the calculation of the overtime to be easier. Moreover, it allowed Task to be selected in the correct order by operators and Capacity utilization to be improved. ( Bettley,2005)
Every activity has three attributes; its duration, the work effort and the resources. The secret is to estimate the work effort first, then the resources, and finally able to estimate activity durations.
With the help of this diagram critical path of the project is determined. This path tells us about the spare times in the project available for these activities thus they should be carried out very carefully to avoid delays in the project.