Difference between revisions of "GATE2010 q46"

Latest revision as of 12:01, 15 July 2014

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 by Arjun Suresh

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)

@@ Line 1: / Line 1: @@
 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:
-<syntaxhighlight lang="c">
+ <span class="nocode" style="color:#48484c">if (<math>i\% 2==0</math>) {
-if (i% 2==0) {
+  if (<math>i<n</math>) request <math>R_i</math> ;
-  if (i<n) request $R_i$ ;
+  if (<math>i+2<n</math>)request <math>R_{i+2}</math> ;
-  if (i+2<n)request R i+2 ;
+ }
-}
+ else {
-else {
+  if (<math>i<n</math>) request <math>R_{n-i}</math> ;
-  if (i<n) request R n-i ;
+  if (<math>i+2<n</math>) request <math>R_{n-i-2}</math> ;
-  if (i+2<n)request R n-i-2 ;
+ }
-}
+</span>
-</syntaxhighlight>
 In which one of the following situations is a deadlock possible?
@@ Line 21: / Line 21: @@
 (D) n = 41,k = 19
-===Solution===
+==={{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. 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.
-{{Template:FB}}
+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)
+{{Template:FBD}}
-<disqus/>
-[[Category:Operating Systems]]
 [[Category:GATE2010]]
-[[Category: Previous year GATE questions]]
+[[Category: OS questions from GATE]]

Difference between revisions of "GATE2010 q46"

Latest revision as of 12:01, 15 July 2014

Solution by Arjun Suresh

Solution[edit]