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Operations Management Re-exam August 2022

Solutions

MA(5) scores lowest on both the MFE and the MAD, so it has the lowest bias and the lowest deviation among the alternatives. This method is preferred.

b)   It is better to use the joint forecasts. The motivation in the textbook is in one of the forecast principles: ‘aggregate forecasts are always better’ (BH p. 278). The reason is that variations in         larger populations tend to be smaller than the sum of the variations within their subpopulations, as is stated in the square root law.

Problem 2

Question 2.1:

First-come-first-served rule: As jobs have arrived to the processing centre in the same order that they appear in Table 2.1, they should be processed in the same order according to the FCFS rule. Calculations are done in Excel and screenshots of formulas used for calculations and numerical results are presented below.

Let STi , ETi , EPTi, and DUDi  denote start time, end time, estimated processing time and days until due of job i, respectively.

To make the calculations, we start from the first job – job U. The STU  is set to 0. Given the start time, the end time can simply be calculated as the sum of the start time and the estimated processing time. For job U, ETU   = STU  + EPTU  = 0  +  10  =  10. The start time of the next job, i.e. job V, is the end time of the previous job, i.e. job U. For job V, STV   =  ETU  = 10. As we did for job U, the end time of job V can be calculated as ETV   = STV  + EPTV  = 10  +  8  =  18. The same logic holds for the rest of the jobs. See columns E and F for more details.

If the end time of a job is less than or equal to days until due, then the job is not late, i.e. the lateness is       equal to zero. However, if the end time exceeds days until due, then the lateness can be calculated as the    difference between these two values. To capture both cases, we use Excel MAX() function and compare the difference between end time and day until due with zero. See column G for more details. To indicate if a      job is late for more than 10 days, we use Excel IF() function and pass 1 if the lateness is greater than 10 and 0 otherwise. See column H for more details.

Calculating the average lateness and the number of jobs with more than 10 days of lateness is straightforward as shown in G13 and H13, respectively. For the FCFS rule, the average lateness is 5.8 days and 2 jobs experience a lateness longer than 10 days.

Earliest Due Date rule: According to the earliest due date rule, the jobs should be processed in the order of the days they have until their due dates. Rearranging the jobs based on the days until due date, we get the following sequence: U, W, V, Z, X, Y. The calculations are the same as we did for the FCFS rule. Numerical results and formulas are presented below.

For the earliest due date rule, the average lateness is 5.3 days and only 1 job will be delayed for more than 10 days.

Question 2.4:

Based on the results in Question 2.3, the requirement is not met as the estimated probability of the average lateness not exceeding 10 days is about 58% (= 100% - 42%) which is much less than 95%.

Problem 3

Question 3.1:

The EOQs of items A and B are respectively

and

The order size of item A has a duration of 160/40 = 4 weeks, while the order size of item B has a duration of 1000/50 = 20 weeks.

Question 3.2:

a)

If we consider item A in isolation, the optimal review period is 4. Thus, total costs (order and holding costs) of item A are higher for any review period below 4 and increasing again for any review period above 4.

Likewise, if we consider item B in isolation, the optimal review period is 20, and total costs of item B are higher for any review period below 20 and increasing for any review period above 20.

Therefore, the total cost function is decreasing for any review period below 4 and increasing for any review period above 20. This implies that the optimal review period is between 4 and 20.

b)

Using Ronny’s proposal, the common review period is (4+20)/2 = 12 weeks. This means the order size of item A will be 40*12 = 480. This gives total annual order and holding costs:

The order size of item B will be 50*12 = 600 giving total annual order and holding costs:

The total annual costs will then be 4172.67 + 1133.33 = 5306.

Question 3.3:

Taking the derivative of the cost function gives

Setting this derivative to zero and isolating RP, gives the square-root formula:

When putting numbers in, we get RP = 0.1332 years. Measured in weeks, this gives RP = 50 * 0.1332 = 6.66. It probably makes most practical sense to let RP be an integer (though in principle there is nothing wrong in using a decimal-valued RP), so let us round it to RP = 7 weeks. This means that the order size of item A should be 40*7 = 280, giving annual costs of

The order size of item B is 50*7 = 350 giving total annual order and holding costs:

The total annual costs become 2905.29 + 1603.57 = 4508.86.

The CEO’s proposal is better. The reason that Ronny calls it “overkill” is probably because he has learned that the cost functions (when doing this sort of EOQ analysis) are in general “flat” . That is, you can deviate quite much from the optimal solution without being penalised very much cost-wise. However, here Ronny deviates quite much from the optimal solution (from 7 to 12). Furthermore, we add two “flat” functions together meaning the total cost function is less “flat”, thereby making it cost more to deviate from the optimal solution.