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A 1000 Kbyte memory is managed using variable partitions but no compaction. It currently has two partitions of sizes 200 Kbytes and 260 Kbytes respectively. The smallest allocation request in Kbytes that could be denied is for

  1. 151 
  2. 181
  3. 231
  4. 541

Won't the ans be 541Kb??

In this link the answer is given 181Kb.But 181 Kb can fit into both 200 Kbytesand 260 Kbytes partitions.

Help me ..I'm confused


closed as a duplicate of: GATE1996-2.18
in Operating System by (307 points)
closed by | 193 views
200 and 260 are already hold by some other processes...and even if they are not hold by any process then don't you think that we have 1000B free space and only 1001B request would be denied..

so we have 540B free memory and that may be continuous or may be scattered ...your goal is to find smallest request that can be denied , so if you suppose 540 continuous then 541 would be answer but now think more if any other allocation possible , as our goal is to fins MIN...

so we can see if partitions are like 180(free)-200-180(free)-260-180(free)then in that way 181 would be minimum that can't be satisfied...
"200 and 260 are already hold by some other processes" Its not mentioned in ques... Now my ques is, If this two partitions are empty then it can hold processes of size 151 Kb,181Kb, 231Kb. But it can not hold any process more than 260 Kb. So, ans is 541 Kb
Rahul brother you are seeming quite keen to set the answer 541 , as you are not looking other possibilities...

okay they didn't said 200 and 260 are allocated then if they are too empty then we have 1000B space vacant then every request upto 1000 can be your mentioned link guys mentioed everything in quite a clear manner
Yep I understood..


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