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GATE | GATE-CS-2003 | Question 60

  • Difficulty Level : Hard
  • Last Updated : 14 Feb, 2018

A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by :

         
A) f_{1}(t)+f_{2}(t)
B) \int_{0}^{t}f_{1}(x)f_{2}(x)dx
C) \int_{0}^{t}f_{1}(x)f_{2}(t-x)dx
D) max\left \{ f_{1}(t),f_{2}(t) \right \}

(A) A
(B) B
(C) C
(D) D


Answer: (C)

Explanation:
We assume the total time to be ‘t’ units and f1 executes for ‘x’ units.

Since, f1(t) and f2(t) are executed sequentially.
So, f2 is executed for ‘t – x’ units.

We apply convolution on the sum of two independent random variables to get probability density function of the overall time taken to execute the program.

f1(t) * f2(t – x) =
  \int_{0}^{t}f_{1}(x)f_{2}(t-x)dx

 
Thus, option (C) is correct.

 
Please comment below if you find anything wrong in the above post.


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