Request edit access
System Modelling 2019/2  Theoretical Questions 2
Example theoretical questions for practising for the 2nd midterm exam in System Modelling. (In the midterm 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.)
* Required
2. Through refinement …
2/a … we can detail the behaviour of a model.
*
1 point
Yes
No
Maybe
2/b … we can detail the structure of a model.
*
1 point
Yes
No
Maybe
2/c … we can detail the data handled by the model.
*
1 point
Yes
No
Maybe
3. To define a test case …
3/a … a test input is required.
*
1 point
Yes
No
Maybe
3/b … a test coverage is required.
*
1 point
Yes
No
Maybe
3/c … an expected property of the output or a test oracle is required.
*
1 point
Yes
No
Maybe
3/d … a runtime monitor is required.
*
1 point
Yes
No
Maybe
4. Deadlock …
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
Yes
No
Maybe
4/b … can happen, when processes of the sytem are waiting for each other.
*
1 point
Yes
No
Maybe
4/c … differs from livelock, while in a livelock state changes may happen, while in a deadlock that is not the case.
*
1 point
Yes
No
Maybe
4/d … cannot happen in a deterministic process.
*
1 point
Yes
No
Maybe
5. The utilization of a resource …
5/a … is nonnegative.
*
1 point
Yes
No
Maybe
5/b … is always greater than or equal to the visitation number.
*
1 point
Yes
No
Maybe
5/c … is less than or equal to the maximum throughput.
*
1 point
Yes
No
Maybe
5/d … is the same as the arrival rate, when the system is in stable operation.
*
1 point
Yes
No
Maybe
6. The throughput of a process …
6/a … is nonnegative.
*
1 point
Yes
No
Maybe
6/b … is always greater than or equal to the visitation number.
*
1 point
Yes
No
Maybe
6/c … is less than or equal to the maximum throughput.
*
1 point
Yes
No
Maybe
6/d … is the same as the arrival rate, when the system is in stable operation.
*
1 point
Yes
No
Maybe
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
Yes
No
Maybe
7/b … in a given time unit, the same number of process instances is started and finished.
*
1 point
Yes
No
Maybe
7/c … the throughput and the maximum throughput are always the same.
*
1 point
Yes
No
Maybe
7/d … the Law of Little is always valid.
*
1 point
Yes
No
Maybe
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
Yes
No
Maybe
8/b … a similarity is, that both techniques take the number of suitable resources for the given activities into account.
*
1 point
Yes
No
Maybe
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
Yes
No
Maybe
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
Yes
No
Maybe
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
Yes
No
Maybe
9/b From the boxplot, it is always easy to find out the 40. percentile.
*
1 point
Yes
No
Maybe
9/c From the boxplot, it is always easy to find out the mode.
*
1 point
Yes
No
Maybe
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
Yes
No
Maybe
Submit
This content is neither created nor endorsed by Google.
Report Abuse

Terms of Service
Forms