GATE2010 q46 Revision as of 12:57, 11 December 2013 by Arjun Suresh (talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

A system has <math>n</math> resources <math>R_0 ,..., R_n-1</math> , and <math>k</math> processes <math>P_0 ,.....P_{k-1}</math> . The implementation of the resource request logic of each process <math>P_i</math> . is as follows:

if (<math>i% 2==0</math>) {
  if (<math>i<n</math>) request <math>R_i</math> ;
  if (<math>i+2<n</math>)request <math>R_{i+2}</math> ;
}
else {
  if (<math>i<n</math>) request <math>R_{n-i}</math> ;
  if (<math>i+2<n</math>) request <math>R_{n-i-2}</math> ;
}

In which one of the following situations is a deadlock possible?

(A) n = 40,k = 26

(B) n = 21,k = 12

(C) n = 20,k = 10

(D) n = 41,k = 19

Solution

From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered resources are taken by 2 processes. Now, if we make sure that all odd numbered processes take odd numbered resources, then dead lock cannot occur. The "else" case of the resource allocation logic, is trying to do that. But, if n is odd, Rn-i and Rn-i-2 will be even and there is possibility of deadlock.

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