Name:
Program:

PROD 2100 /PROD 2110

Production and Operations Management

Solutions: June 1999

Grading :

Question / 1 / 2 / 3 / 4 / 5 / 6 / Total
Grade
Over / 4 / 3 / 4 / 4 / 3 / 2 / 20

The solutions presented here are syntheticand just aim at giving the kind of answers that was expected.

Question 1 (4 points)

In the POM course, we reviewed three main types of organizations: the job-shop, the line and the individual organization. We also tried to compare these organizations on the basis of different performance indicators.

In his article “Mechanization Takes Command”, G. Morgan describes the organization theory named “Scientific Management” that has been developed by F. Taylor. If we decide to apply the “Scientific Management” theory to an organization, which performance indicator(s) can we clearly expect to improve and for which one(s) can we clearly expect to have problems ?

SOL:

1. high specialization => high production rate, high potential for learning, good quality (if the human problems are solved and if there is a feedback mechanism)

2. tight control of the worker; task specialization => the process can be rather well controlled

3. Human aspects are the main problem (demotivation, dehumanization; people can be expected to look for another job, ...)

4. Lack of process flexibility and difficulty to customize the products because of the high specialization (the process can produce what it has been designed for and nothing else).

Question 2 (3 points)

The “Boucherie du centre” manufactures sandwiches. The service of a customer order consists in three operations: take the order; prepare the sandwich according to the order specifications and cash the money. The preparation of the sandwich is the longest operation.

In addition to these sandwiches assembled on order, the “Boucherie du centre” decided to prepare a given set of standard sandwiches in advance.

Two policies are now considered to operate the shop and serve the customers.

In the first policy, a single waiting queue is built and all the customers are served in the order they arrive in the shop.

The second policy consists in building two queues: one for the already prepared sandwiches and one for the sandwiches that have to be assembled on order. The customers in the queue for already prepared sandwiches can then be served in priority, that is, before the customers asking for sandwiches to be assembled.

Comment on the waiting time experienced by the customers with these two policies.

SOL:

- 1 line. FIFO. This policy minimizes the maximum wait time. In other words, nobody will experience a too long waiting time.

- 2 lines. Here, the shortest operations are processed first. This means that we try to minimize the average wait time (this is the objective of the scheduling policy “shortest processing times first”). In other words, the customers wait, on the average, less than in the first solution. However, some customers could wait much more than in the first solution.

Question 3 (4 points)

For my shopping, I use a unique service company. I call this company once a week, on Thursday evening, and order all the goods (bread, meat, vegetables, drinks, ...) I need for the next week. The company prepares my order and delivers all the goods on the next day (Friday evening).

How many liters of milk shall I order if I do not want to be short of milk more than once a year ? (explain how you proceed)

Data:

  • fixed fee charged by the company: 200 Fb
    (What I pay is thus the cost of the goods (bread, meat, vegetables, drinks, ...) plus a fixed fee of 200 Fb that is independent of what I ordered);
  • cost of one liter of milk: 30 Fb;
  • holding cost rate : 20% per year;
  • daily milk consumption: 1 liter with a standard variation of 1/2 liter.

SOL:

  1. periodic system : Vp = 7+1=8
  2. standard deviation of the demand during the vulnerability period:

(8 days) = sqrt (8) * 1/2 = sqrt(2)  1.4.

  1. Assumption: the consumption of milk over 8 days has a distribution whose shape is that of a normal distribution.
  2. The objective is a stockout frequency of once a year. So, stockout probability per cycle = 1/52  0.02  2 standard deviations (by looking up in the normal table)
  3. Qmaxis equal to the demand during the vulnerability period plus the safety stock. So, Qmax= 8 + 2 * 1.4  11
  4. In conclusion, I order 11 liters minus what is left in my fridge.

Question 4 (4 points)

A company manufactures 3 products called P1, P2 and P3 by transforming 3 different raw materials: RM1, RM2 and RM3. The following table shows how each of the finished good is manufactured.

Machine
Raw Material / A / B / C / Finished good
RM1 / 2 hrs / 2 hrs / 2 hrs / P1
RM2 / 4 hrs / 2 hrs / 0 hrs / P2
RM3 / 0 hrs / 1 hrs / 2 hrs / P3

For example, one P1 is obtained by transforming one unit of RM1 as follows: 2 hours of processing on machine A, then 2 hours on machine B and finally, 2 hours on machine C. Product P2 requires processing on machine A and B only. Product P3 requires processing on machine B and C only. All these machines operate 24 hours per day, 25 days per month.

Here below is a table that gives the demand for the next 6 months.

Month / P1 / P2 / P3
1 / 0 / 50 / 0
2 / 100 / 50 / 250
3 / 100 / 50 / 250
4 / 100 / 50 / 250
5 / 100 / 50 / 250
6 / 100 / 50 / 250

A. Which is the most critical resource (machine) for the production over the next 6 months ?

B. Give an aggregate plan for the use of this machine over the next 6 months.

C. Propose a feasible production plan for the product P1, P2 and P3 for the next 6 months.

SOL:

A. The machine C is the most critical because the needed production time over the next six months is 3500hrs (for machine A : 2200hrs ; for machine B : 2850hrs). Note that the total available time is 3600 hours for each machine = 24*25*6).

B. An aggregate plan for machine C (maximum capacity for machine C = 24*25 = 600hrs / month):

MONTH / PRODUCTION TIME
FOR MACHINE C
1
2
3
4
5
6
7 / 500
600
600
600
600
600
600

C. You can find different plans that meet the capacity constraint.

Here is a possible solution for P1, P2, P3 that meetsthe capacity constraints for all the machines.

Month / P1 / P2 / P3
1 / 0 / 50 / 250
2 / 100 / 50 / 200
3 / 100 / 50 / 200
4 / 100 / 50 / 200
5 / 100 / 50 / 200
6 / 100 / 50 / 200

Question 5 (3 points)

Assume that for an oral examination, a professor proceeds as follows. A first set of two questions are asked to the student. If the student answers both questions correctly, then he passes. If none of the answers is correct then he fails. If one answer is correct and the other is wrong, then the professor asks a third question. If the student answers correctly this third question, he succeeds otherwise he fails.

A. To what kind of quality technique does this exam procedure refers to ? (Explain)

Assume now that the students complain. They think that too many good students fail because they are unlucky in the questions they get.

B. To which quality concept does this refer to ? and how can you improve the situation ?

SOL:

A. Double sampling plan (n1, n2, c1, c2, c3). Lot = knowledge of the student.

Sample size (n) = number of questions asked to the student.

Defective items (X) = wrong answers.

Decision threshold (c) = maximum number of wrong answers accepted.

Here n1 = size of the first sample = 2. If number of wrong answers among n1 is less than or equal to c1 (if X(n1) <= c1), we accept the student (the lot). If X(n1) > c2, we reject the lot. If c1 < X(n1) <= c2, we take another sample n2 (another question). If X(n1 + n2) <= c3, we accept the lot, otherwise we reject it.

Here, n1 = 2; n2 = 1; c1 = 0; c2 = 1; c3 = 1.

B. Producer and consumer risks. Here, the students = the producer (their risk is to have good students rejected), the teacher = the consumer (his risk is to accept bad students).

We can improve the situation of the students (the producer) by:

-increasing c, the number of wrong answers tolerated before rejecting a student (this solution will decrease the producer’s risk but the consumer’s risk (that of the teacher) will increase),

-increasing n, the number of questions (the sample size) (this will reduce both risks but could be costly).

Question 6 (2 points)

Assume that a manager is planning a small project in which many tasks could be performed in parallel. However, because of a lack of resource, only one employee can be allocated to this project. This employee will thus be in charge of all the tasks.

Assume that the manager uses the Pert method to estimate the duration of the project and its variability.

A. The manager has to choose the order in which the tasks will be performed. How does this order influence the results of the method ?

B. The Pert method is based on several approximations / assumptions. Which one is, in this particular case, relaxed ?

SOL:

A. Since there is only one employee, all the tasks will be performed sequentially. The critical path will thus reduce to the chosen sequence of tasks. The PERT method evaluates the total duration of the critical path by summing up the duration of its different tasks. Their order is therefore irrelevant, because whatever order is used, the sum will remain the same (all the tasks are critical).

B. The Pert method approximates the total duration of the project by focusing on the critical path only. In this case, there is no approximation since no parallel path exists.