Redirected
2,607 views
3 votes
3 votes
Consider 900 KB memory is managed using variable partitions but no compaction. It currently has three process occupied partition of sizes 212 KB, 114 KB and 100 KB respectively. What is the smallest allocation request that could be denied?

1 Answer

Best answer
9 votes
9 votes

Complexity lies in this question is.... how much size of small holes can you created?

 

with n processes ==> you can create minimum 1 hole ===> you can identify The maximum Process can be denied

P1 P2 P3 hole
212 114 100 474

Therefore if a process comes with size 475 KB you can't allocate.

 

with n processes ==> you can create maximum n+1 holes ===> you can identify The minimum Process can be denied

hole P1 hole P2 hole P3 hole
  212   114   100  

 

     hole size = $\frac{Total \; size\; available}{no.of holes}$ = $\frac{474}{4}$ = 118.5

 Therefore two holes get 118 KB and two holes can get 119 KB

Therefore if a process comes with size 120 KB you can't allocate.

 

( if you have doubt that why holes must be equal size, then if you decrease size of one hole then one of the remaining holes size must be increased then you can allocate 120 KB process )

edited by

Related questions

0 votes
0 votes
0 answers
2
Nidhi Budhraja asked Oct 1, 2018
616 views
Is it true that we can allocate only one process to one block?In First fit memory allocation scheme, can we allocate two processes to the same block of memory if block ha...
0 votes
0 votes
1 answer
4
agoh asked Dec 11, 2016
510 views
Given following data, find average access time. Please explain the solution.