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ISRO CSE 2011 | Question: 10

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13 13 votes

Below is the precedence graph for a set of tasks to be executed on a parallel processing system $S$.

What is the efficiency of this precedence graph on $S$ if each of the tasks $T_1, \dots, T_8$ takes the same time and the system $S$ has five processors?

  1. $25\%$
  2. $40\%$
  3. $50\%$
  4. $90\%$

6 Answers

Best answer
19 19 votes

Maximum number of tasks that could have been executed in four unit time using five processors = 4*5 =20

Number of tasks executed = 1+1+3+3=8

Efficiency = 8/20 =40%

Answer is B: 40%

• selected by
14 14 votes
System has 5 processors. Time to complete each process i.e. Ti is same.

Reasoning :
T1 executes on Processor 1. other 4 processors are idle.
T2 executes on Processor 1, other 4 Processors are idle.
T3, T4, T5 can run parallely on different different processors. Now 3 processors are working and 2 are still unused.
T6, T7, T8 can run parallely on different different processors. now 3 processors are working and 2 are still idle.

Total 5*4 =40 instance of processors. 8 out of 20 are working.
Efficiency = (8/20)*100 = 40%
2 2 votes

Answer is C  = 50%

As per the below question paper and key:

http://www.eduers.com/gre/CompSci.pdf

1 flag:
✌ Low quality (Jagadeesh_Reddyy)
1 1 vote
Here Number of tasks =8

Number of processors =5

Efficiency is ratio of speed up and number of processors

Speed up is number of task executed in single processor system

Efficiency =((8/4)/5)⨉100=40%
1 1 vote

Total processes that ran = 8

Total processes that could've had ran = 20.

Efficiency = 8/20 = 0.4 = 40%

 

Option B

0 0 votes
T1 exec. first and take 1 out of 5 processor to execute.. Efficiency   = 1/5 * 100 = 20%

T2 takes 1 out of 5 processor to execute.. Eff.  = 1/5*100 = 20%

T3,T4 and T5 will execute in parallel and all will take same amt. Of time... So they will use 3 out of 5 processors to execute.. Eff.  = 3/5*100 = 60%

T6,T7 and T8 will also do the same... And will take 3 out of 5 processors to execute... Eff. = 3/5* 100 = 60%

Avg eff.  = (20+20+60+60)/4 = 160/4 = 40% ans
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