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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: | ||
| − | <span class="nocode" style="color:#48484c"> | + | <span class="nocode" style="color:#48484c">if (<math>i\% 2==0</math>) { |
| − | |||
if (<math>i<n</math>) request <math>R_i</math> ; | if (<math>i<n</math>) request <math>R_i</math> ; | ||
if (<math>i+2<n</math>)request <math>R_{i+2}</math> ; | if (<math>i+2<n</math>)request <math>R_{i+2}</math> ; | ||
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(D) n = 41,k = 19 | (D) n = 41,k = 19 | ||
| − | === | + | ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}=== |
| − | From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered processes share no more than 1 resource | + | From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered processes share no more than 1 resource. Now, if we make sure that all odd numbered processes take odd numbered resources without a cycle, then deadlock cannot occur. The "else" case of the resource allocation logic, is trying to do that. But, if n is odd, <math>R_{n-i}</math> and <math>R_{n-i-2}</math> will be even and there is possibility of deadlock, when two processes requests the same <math>R_i</math> and <math>R_j</math>. So, only <math>B</math> and <math>D</math> are the possible answers. |
| − | Now, in <math>D</math>, we can see that <math>P_0</math> requests <math>R_0</math> and <math>R_2</math>, <math>P_2</math> requests <math>R_2</math> and <math>R_4</math> | + | Now, in <math>D</math>, we can see that <math>P_0</math> requests <math>R_0</math> and <math>R_2</math>, <math>P_2</math> requests <math>R_2</math> and <math>R_4</math>, so on until, <math>P_{18}</math> requests <math>R_{18}</math> and <math>R_{20}</math>. At the same time <math>P_1</math> requests <math>R_{40}</math> and <math>R_{38}</math>, <math>P_3</math> requests <math>R_{38}</math> and <math>R_{36}</math>, so on until, <math>P_{19}</math> requests <math>R_{22}</math> and <math>R_{20}</math>. i.e.; there are no two processes requesting the same two resources and hence there can't be a cycle of dependencies which means, no deadlock is possible. |
But for <math>B</math>, <math>P_8</math> requests <math>R_8</math> and <math>R_{10}</math> and <math>P_{11}</math> also requests <math>R_{10}</math> and <math>R_8</math>. Hence, a deadlock is possible. (Suppose <math>P_8</math> comes first and occupies <math>R_8</math>. Then <math>P_{11}</math> comes and occupies <math>R_{10}</math>. Now, if <math>P_8</math> requests <math>R_{10}</math> and <math>P_{11}</math> requests <math>R_8</math>, there will be deadlock) | But for <math>B</math>, <math>P_8</math> requests <math>R_8</math> and <math>R_{10}</math> and <math>P_{11}</math> also requests <math>R_{10}</math> and <math>R_8</math>. Hence, a deadlock is possible. (Suppose <math>P_8</math> comes first and occupies <math>R_8</math>. Then <math>P_{11}</math> comes and occupies <math>R_{10}</math>. Now, if <math>P_8</math> requests <math>R_{10}</math> and <math>P_{11}</math> requests <math>R_8</math>, there will be deadlock) | ||
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{{Template:FBD}} | {{Template:FBD}} | ||
| − | + | ||
[[Category:GATE2010]] | [[Category:GATE2010]] | ||
| − | [[Category: | + | [[Category: OS questions from GATE]] |
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:
GATECSEif (<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
From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered processes share no more than 1 resource. Now, if we make sure that all odd numbered processes take odd numbered resources without a cycle, then deadlock cannot occur. The "else" case of the resource allocation logic, is trying to do that. But, if n is odd, <math>R_{n-i}</math> and <math>R_{n-i-2}</math> will be even and there is possibility of deadlock, when two processes requests the same <math>R_i</math> and <math>R_j</math>. So, only <math>B</math> and <math>D</math> are the possible answers.
Now, in <math>D</math>, we can see that <math>P_0</math> requests <math>R_0</math> and <math>R_2</math>, <math>P_2</math> requests <math>R_2</math> and <math>R_4</math>, so on until, <math>P_{18}</math> requests <math>R_{18}</math> and <math>R_{20}</math>. At the same time <math>P_1</math> requests <math>R_{40}</math> and <math>R_{38}</math>, <math>P_3</math> requests <math>R_{38}</math> and <math>R_{36}</math>, so on until, <math>P_{19}</math> requests <math>R_{22}</math> and <math>R_{20}</math>. i.e.; there are no two processes requesting the same two resources and hence there can't be a cycle of dependencies which means, no deadlock is possible.
But for <math>B</math>, <math>P_8</math> requests <math>R_8</math> and <math>R_{10}</math> and <math>P_{11}</math> also requests <math>R_{10}</math> and <math>R_8</math>. Hence, a deadlock is possible. (Suppose <math>P_8</math> comes first and occupies <math>R_8</math>. Then <math>P_{11}</math> comes and occupies <math>R_{10}</math>. Now, if <math>P_8</math> requests <math>R_{10}</math> and <math>P_{11}</math> requests <math>R_8</math>, there will be 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:
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
From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered processes share no more than 1 resource without a cycle. Now, if we make sure that all odd numbered processes take odd numbered resources and share no more than 1 resource without a cycle, then deadlock cannot occur. The "else" case of the resource allocation logic, is trying to do that. But, if n is odd, <math>R_{n-i}</math> and <math>R_{n-i-2}</math> will be even and there is possibility of deadlock, when two processes requests the same <math>R_i</math> and <math>R_j</math>. So, only <math>B</math> and <math>D</math> are the possible answers.
Now, in <math>D</math>, we can see that <math>P_0</math> requests <math>R_0</math> and <math>R_2</math>, <math>P_2</math> requests <math>R_2</math> and <math>R_4</math> and so on until, <math>P_{18}</math> requests <math>R_{18}</math> and <math>R_{20}</math>. At the same time <math>P_1</math> requests <math>R_{40}</math> and <math>R_{38}</math>, so on until <math>P_{19}</math> requests <math>R_{22}</math> and <math>R_{20}</math>. i.e.; there are no two processes requesting the same two resources and hence there can't be a cycle of dependencies and hence no deadlock is possible.
But for <math>B</math>, <math>P_8</math> requests <math>R_8</math> and <math>R_{10}</math> and <math>P_{11}</math> also requests <math>R_{10}</math> and <math>R_8</math>. Hence, a deadlock is possible. (Suppose <math>P_8</math> comes first and occupies <math>R_8</math>. Then <math>P_{11}</math> comes and occupies <math>R_{10}</math>. Now, if <math>P_8</math> requests <math>R_{10}</math> and <math>P_{11}</math> requests <math>R_8</math>, there will be deadlock)