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Previous Exams

Here you can find several previous exams with solutions. Also, here you will find the solutions of your exam and make-up exam.

If you have received Fx contact the teacher (vfodor@kth.se) as soon as possible. You have the opportunity to complement to E, but you have to do it within 6 weeks after the Fx has been reported (until ca. end of February for the main exam and end of September for the make-up exam).

The make-up exam is in June, see KTH Schema for date, time, location. Note, that you have to register for the make-up exam.

If you need to complain about the grading, you can submit complaints through STEX. Please ask them about the exact process.

Make-up exam 2017-06-10

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 21-24  
F: 0-20 

Exam 2017-03-24

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 21-24  
F: 0-20 

Make-up exam 2016-06-11

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 21-24  
F: 0-20 

Exam 2016-03-23

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 21-24  
F: 0-20 

Make-up exam 2015-06-13

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 21-24  
F: 0-20 

Exam 2015-03-16

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 22-24  
F: 0-21 

Make-up exam 2014-05-20

A: 45-50  
B: 40-44  
C: 35-39  
D: 30-34  
E: 25-29  
Fx: 22-24  
F: 0-21 

Exam 2014-03-18

Grading (with the number of students with a given grade). the exam was easier than the earlier ones, that is why limits are bit higher.

A: 45-50  (5) 
B: 40-44 (5) 
C: 35-39 (10) 
D: 30-34 (3) 
E: 25-29 (2) 
Fx: 22-24 (3) 
F: 0-21 (5)

Make-Up Exam 2013-06-13

Grading (with the number of students with a given grade)

A: 45-50  (1)
B: 39-44 (1)
C: 33-38 (1)
D: 27-32 (0)
E: 21-26 (5)
Fx: 18-20 (2)
F: 0-17 (5)

Exam 2012-12-14

Grading (with the number of students with a given grade)

A: 45-50  (1)
B: 39-44 (6)
C: 33-38 (4)
D: 27-32 (10)
E: 21-26 (13)
Fx: 18-20 (2)
F: 0-17 (9)

Make-up exam 2012-06-11

Grading (with the number of students with a given grade)

A: 45-50  (0)
B: 39-44.5 (1)
C: 32-38.5 (4)
D: 26-31.5 (1)
E: 21-25.5 (2)
Fx: 19-20.5 (0)
F: 0-18.5 (3)

Exam from 2011-12-15

A: 45-50  (0)
B: 39-44.5 (4)
C: 32-38.5 (4)
D: 26-31.5 (5)
E: 21-25.5 (5)
Fx: 19-20.5 (4)
F: 0-18.5 (8, out of these 4 not higher than 10)

Make-up exam from 2011-06-10

Exam from 2010-12-18

Exam from 2004-12-16

Exam from 2009-12-17

Viktoria Fodor skapade sidan 14 december 2016

Assistent kommenterade 11 mars 2017

Hi all!

Here are the answers to some of the questions from the consultations:

*Collection of the exam problems (problem 8):
- Question: Why system B is modeled as a system with a finite population and system A is not?
Answer: When should we consider a system as finite population system? Rule of tumb: C < 10m, where m is the number of servers (lecture 7, slide 6).

*Exam March, 2015 (problem 4 (d)):
-Arrivals still happen according to a Poisson distribution, but samples (the moments when the jobs arrive) are uniformly at random distributed over the fixed garbage collection time.
-Regarding the formula for the remaining garbage collection time, please check slide 6 , lecture 11, the first formula from which we started the derivation. In our problem vi is always the same and equal to v, and we consider these jobs that arrive during the garbage collection period, so T in our problem is n*v (total length of the garbage collection period).

*Exam March, 2016 (4 (a)):

-The average length of the idle periods (the solution is correct!).

-Explanation: The resulting process is a Poisson(lambda1 + lambda2). We are interested in the periods when a server is idle. Hence, let us look at the moment when a server becomes idle. The server remains in the idle state until the next arrival happens and the next arrival will happen according to a Poisson(lambda1 + lambda2) distribution. Therefore, the idle periods are exponentially distributed with a mean of 1/(lambda1 + lambda2) . Note that the type of the service time does not affect the distribution of the idle periods! This is because the interarrival time is exponential(lambda1 + lambda2) , and therefore the memoryless property holds (we do not care which type of the service was going on before the server became idle).

*Exam March, 2016 (problem 3(d)) and Exam June, 2016 (problem 2(e)):

Both solutions are correct and acceptable.