Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
||
| Line 4: | Line 4: | ||
if (i% 2==0) { | if (i% 2==0) { | ||
if (i<n) request R_i ; | if (i<n) request R_i ; | ||
| − | if (i+2<n)request | + | if (i+2<n)request R_{i+2} ; |
} | } | ||
else { | else { | ||
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: <math> if (i% 2==0) {
GATECSE if (i<n) request R_i ;
if (i+2<n)request R_{i+2} ;
} else {
if (i<n) request R_{n-i} ;
if (i+2<n)request 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
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.
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: <math> if (i% 2==0) {
if (i<n) request R_i ;
if (i+2<n)request <math>R_{i+2}</math> ;
} else {
if (i<n) request R_{n-i} ;
if (i+2<n)request 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
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.