Exam 2II75 answer
1.Comment on the problem sketched below and suggest improvements, extensions and additions to thetwoquestions, with a brief indication of the reason why. (2 pts)
Process P experiences queueing problems for a certain resource X. On several occasions, this has led to unacceptable lead times. A brainstorm session has led to two tentative improvements, called Q and R. Due to expected costs, Q is the preferred solution. Management has decided that Q can be implemented if (w.r.t. the current P) it is expected to yield at least a 15% reduction for the average waiting time for resource X. If the expected waiting time reduction due toimprovement R is at least 5% higher than the reduction due to Q, it is worthwile to study the cost of R and then decide which of the two to implement. A simulation study should investigate the following two questions.
i) Is the average queueing timefor resource X in the Q model at least 15% less than in the current (P) model?
ii) Is the average queueing time for resource X in the R model at least 5% less than in the Q model?
ANSWER:
The definition above focuses too much on the queues for X. This may be indeed a cause of the unacceptable lead times, but there may well be more causes. Solutions Q and R may reduce the queues for X while increasing other queues. So the effect on the lead time should be considered as well. Another comment is that reducing an average may not necessarily reduce unacceptably long waiting times. It is wise to consider the standard deviation too. Question i) could be replaced by
Is the average lead time in model Q at least x% less than in model P?
where x is the reduction in lead time that would result from the desired waiting time reduction for X.
Question ii) is similar.
Finally, the situation could arise that question i) is answered negatively (so Q will not be implemented) and ii) positively (so the cost of R should be studied). It seems that it is only worthwile to study the cost of R if the expected lead time reduction is 19% of the X-queueing time.
EVALUATION:
Many exam answers addressed confidence. It should be pointed out that questions to be answered by simulation may not refer to confidence. It is the answers to these questions that should contain a confidence. Also, many answers seemed besides the point by referring to the tasks and obligations of the simulation people. The exam question was to assess whether the questions i,ii) were adequate, not how they need to be answered. The average score was very low: BIS: 26%, non-BIS: 29%.
2.A certain production step in a job shop requires two consecutive processing steps. The first step
S1 needs resource R1 and has a duration distribution D1. The second step S2 needs another resource R2 and has a duration distribution D2. Jobs often have to queue for R2 after finishing S1 before S2 can start. Due to space problems, the maximum queue size for S2 equals five. If a sixth job enters the queue, it is transported to a storage area instead. When the queue becomes empty (by starting S2 with the last queued job) and the storage area contains one or more jobs transported earlier, one of them is transported back to the queue. Create conceptual and Arena models for the sketched situation. Give conditions for any modeled “choice” or “assign” blocks.
ANSWER:
We give a Petri net model and its Arena implementation. In the Petri net, transition t fires iff place p contains six tokens and transition u fires iff p is empty (and q not), which can be modeled in various ways, e.g. arcs with multiplicities.
In the Arena model, the condition for the choice block is that the queue size is less than five. For bringing back, there are several possibilities. The depicted Arena model uses a “hold” block, which can be replaced by “busy wait” or by introducing a dummy resource that becomes available when the queue is empty. The condition for the decide block is [NQ(step2.Queue)<5] and for the hold block [NQ(step2.Queue)==0]
EVALUATION:
Various conceptual constructions were used, many of them containing an “available queue” place containing intially five (or six) tokens. This construction allows a “classical” model. Others used priorities, colors or inhibitor arcs. All these models were accepted if correct.
Many Arena models did have a busy wait construction, repeatedly checking whether the queue has become empty. If there is not a delay in the cycle, this yields a livelock. About half a point was subtracted in this case. Average score: BIS: 63%, non-BIS: 57%.
3.In the depicted Arena model, the blocks arrive1 and arrive2 have negative exponential
distributions with average interarrival times 20, resp. 30 minutes. Step 1 seizes resource A, step2 is delay-only, step 3 releases A. The durations step 1 and 2 are normally distributed, with averages of respectively 4 and 6 minutes and standard deviations of 1 minute each. Step 3 has a constant duration of 1 minute.
a) What occupation rate for resource A do you derive from the model?
b) Derive an equivalent simpler (less blocks) model that cannot be simplified further. Give the parameters for the remaining block(s).
ANSWER:
a)
On average, five arrivals occur per hour: three due to arrive1 and two due to arrive2, so the total average interarrival time is twelve minutes. The average occupation time of resource A is eleven minutes, so the occupation rate is 11/12, almost 92%.
b)
arrive: exponential, interarrival time 12 minutes. step: S/D/R for resource A, duration normal with average 11 minutes and stdev . Durations in original “step” blocks independent!
EVALUATION
I witnessed some strange mistakes, such as 1+1=1. Average score: BIS: 64%, non-BIS: 46%.
4.Study the following excerptfrom a simulation report, that refersback to exam question (1). The simulation questions formulated there are assumed correct, irrespective of necessary additions and modifications. Comment on the conclusions and give improvements and additions when necessary.The report contains no miscalculationsand all rounding-off is permitted.The z-table for the normal distribution is included.
ANSWER:
Sentences that are to be criticized are underlined and discussed.
In order to answer the questions, simulation models have been constructed for P, Q and R. All three models have been validated and found correct by experts; the simulation results from model P have also been found to match recorded measurements from the current process. In order to assess sensitivity, the create block parameters for models P, Q and R have been adapted to reflect a 5% demand increase, yielding respectively models P+,Q+ and R+ .
Simulation experiments have been conducted to address questions i) and ii). For each of the models, the average numberT of hours spent by jobs in queues for resource X has been measured, giving the following averages and 95% half widths for the performance indicator T.
modelThalf width
P1.790.132
P+1.850.173
Q1.340.085
Q+1.370.088
R1.230.091
R+1.270.096
The following hypotheses are derived from the questions. T(M)denotes the “true” value of T for model M.
H1: T(Q) is less than 0.85 * T(P),H2: T(R) is less than 0.95 * T(Q).
We derive confidence intervals for H1 as follows. The measured difference equals 0.85*1.79 – 1.34 = 0.1815.Assuming independence, the standard deviation of this difference equals .
The standard deviation of 0.85*P is 0.85*0.132, which equals 0.1122. So the sdev should be 0.072 instead, which would lead to 99.5% confidence.
The quotient 0.1815/0.080 equals 2.26.Therefore, assuming no change in the demand, H1 is accepted with 97% confidence.
Two errors in the underlined sentence: H1 is not influenced by a demand change, since the demand is part of the model. The confidence should have been 0.9881, almost 99% (table lookup error).
A similar derivation yields acceptance of H1 with 96% confidence in case of a demand increase.
Intended is a new hypothesis H1+: T(Q+) is less than 0.85 * T(P+). Calculation would reveal a higher confidence (98%)as well.
The answer to question i) is thus affirmative.
A confidence interval is needed! In this case it would be appropriate to give a proviso, e.g.: assuming no change in the demand, the answer to i) is affirmative with almost 99% confidence (which is a paraphrase of H1). In this case the hypotheses H1 and H1+ are not likely to be independent; a solution that gives an improvement for the current demand will most likely also give an improvement for a slightly increased demand, so there is a positive correlation.This observation increases the confidence in H1 and H1+ somewhat, but the amount cannot be assessed.
In the same fashion as above,confidence intervals for H2 are derived as follows.
0.95*1.34 – 1.23 = 0.043,, 0.023/0.0635 = 0.362.
0.95*0.091 must be used.Wrong intermediate result copied; this should have been 0.043/0.0635 = 0.667.
An affirmative answer to question ii) has only 75% confidence if the demand does not change. The same confidence is 69% in case of a 5% demand increase.
Same remarks as before: define H2+. This time, the table lookup is right (taking 0.667).
Confidence in an affirmative answer of question ii) is thus at most 75%.
An answer to question ii) without provisos is not possible. You cannot combine confidences like this. Due to the positive correlation, the 75% confidence may be higher.
We therefore recommend to skip an investigation of the cost of improvement R and to implement solution Q immediately.
This recommendation is not allowed.Confidence may be too low to accept H2, but is is plausible nonetheless, so studying the cost of R is recommended still. It may be better to recommend toperform additional prolonged runs to increase confidence.
EVALUATION
The sentence “no miscalculations” did give a wrong impression. It was meant to refer to the divisions and multiplications only, but not table lookup and so forth. For many students, the term “calculation” apparently has a wider scope. In the light of this, I did not subtract anything when such errors were missed, but I nevertheless rewarded those that saw them. The conceptual errors that remain are:
- use of half widths 0.132 and 0.088 for the stochastic expressions 0.85*T(P) and 0.95*T(Q).
- Mixing H1 and a demand increase of 5%.
- Combining 69% and 75% confidences to “at most 75%”.
- Recommending to skip an investigation of the cost of R.
Error 4 was noticed by quite a few, whereas only two students saw 1. Errors 2 and 3 were not noticed by anyone. I rewarded 4) with one point, 1) with 1.5 point and the lookup/substitution errors with one point. Points were subtracted when e.g. hypothesis H1 was questioned for the wrong reason. When management wants to test question i) by simulation, it is not relevant to test whether Q (or R) is better than P. This hypothesis may be confirmed with 99.999% confidence, but this is not relevant at all. In fact, it is not set out as a hypothesis to be tested by the simulation experiment, so computing its confidence after the experiment is exactly the deadly methodogical sin that has been exposed in the lectures. Quite a few students falsely accused the report of this methodological error. A correct methodology approach first sets out hypotheses (H1, H2), then conducts a simulation experiment, from which their likelyhood can be assessed. This approach has apparently been used. The error is that a “hidden” hypothesis H1+ has been used, but that this has been formulated as “H1 with a demand increase”. I am prepared to believe that H1+ was an intended hypothesis when the models P+ and Q+ were constructed, so deadly sins seem to be absent in the report.Some students did not like the assessment of confidence intervals without subruns. Simulation tools such as Arena allow this, if you are prepared to trust their results. It is assumed that a correct statistical analysis has been performed to obtain the listed half widths.
The average score was extremey low: BIS: 20%, non-BIS: 29%.