System Modelling 2019/2 -- Theoretical Questions 2
Example theoretical questions for practising for the 2nd mid-term exam in System Modelling. (In the mid-term exam, you get +1/4 point for each correct answer, -1/4 point for each incorrect ones, and 0 point for each question left out. You have to reach 50% of the possible points in this system. The Google Form is not able to implement this points system.)
2. Through refinement …
2/a … we can detail the behaviour of a model. *
1 point
2/b … we can detail the structure of a model. *
1 point
2/c … we can detail the data handled by the model. *
1 point
3. To define a test case …
3/a … a test input is required. *
1 point
3/b … a test coverage is required. *
1 point
3/c … an expected property of the output or a test oracle is required. *
1 point
3/d … a runtime monitor is required. *
1 point
4/a … is a state, which cannot be left by the system through the modelled inputs and events, only through (to the model) external help. *
1 point
4/b … can happen, when processes of the sytem are waiting for each other. *
1 point
4/c … differs from livelock, while in a livelock state changes may happen, while in a deadlock that is not the case. *
1 point
4/d … cannot happen in a deterministic process. *
1 point
5. The utilization of a resource …
5/a … is non-negative. *
1 point
5/b … is always greater than or equal to the visitation number. *
1 point
5/c … is less than or equal to the maximum throughput. *
1 point
5/d … is the same as the arrival rate, when the system is in stable operation. *
1 point
6. The throughput of a process …
6/a … is non-negative. *
1 point
6/b … is always greater than or equal to the visitation number. *
1 point
6/c … is less than or equal to the maximum throughput. *
1 point
6/d … is the same as the arrival rate, when the system is in stable operation. *
1 point
7. In case of a performance model of a system in stable operation …
7/a … the arrival rate can be higher than the maximum throughput. *
1 point
7/b … in a given time unit, the same number of process instances is started and finished. *
1 point
7/c … the throughput and the maximum throughput are always the same. *
1 point
7/d … the Law of Little is always valid. *
1 point
8. In case of a process simulation and a performance modelling …
8/a … a difference is, that the simulation estimates the run times for different cases based on the resource reservations, while the calculations on the performance model give the average run time of many run cases. *
1 point
8/b … a similarity is, that both techniques take the number of suitable resources for the given activities into account. *
1 point
8/c … a similarity is, that both support hierarchical models (so a composite activity may cover multiple elementary activities with different resource requirements). *
1 point
8/d … both mean the same system modelling step (one is the goal to achieve, the other is the used technique), and both can be applied for the same cases. *
1 point
9. Characteristic values of a continuous variable are represented with a boxplot and a histogram.
9/a From the boxplot, it is always easy to find out the first quartile. *
1 point
9/b From the boxplot, it is always easy to find out the 40. percentile. *
1 point
9/c From the boxplot, it is always easy to find out the mode. *
1 point
9/d Every information, which is easy to find out from the boxplot, is also easy to find out from the histogram. For this reason, we can view the boxplot as an abstraction of the histogram. *
1 point