Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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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 | 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: | request logic of each process <math>P_i</math> . is as follows: | ||
| − | <math> | + | |
| − | + | if (<math>i% 2==0</math>) { | |
| − | if (i<n) request | + | if (<math>i<n</math>) request <math>R_i</math> ; |
| − | if (i+2<n)request | + | if (<math>i+2<n</math>)request <math>R_{i+2}</math> ; |
} | } | ||
else { | else { | ||
| − | if (i<n) request | + | if (<math>i<n</math>) request <math>R_{n-i}</math> ; |
| − | if (i+2<n) request | + | if (<math>i+2<n</math>) request <math>R_{n-i-2}</math> ; |
} | } | ||
| − | + | ||
In which one of the following situations is a deadlock possible? | In which one of the following situations is a deadlock possible? | ||
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>) {
GATECSE 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
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 Ri ; if (i+2<n)request Ri+2 ;
} else {
if (i<n) request Rn-i ; if (i+2<n) request Rn-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.