960 views
0 votes
0 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. Which of the following is the smallest allocation request that could be denied?

Why isnt the answer to this question 238?In options 475,285,355 and none are given and none is chosen,saying allocation can be at any place.

Here I can show atleast one arrangement where 238 is denied.,

Any other way of interpreting this

2 Answers

0 votes
0 votes

Answer of this question is 475 kb

Total space = 900kb

Occupied space=212+114+100=426 kb

Actually we are using variable size partitioning 

Total remaining space=900-426=474kb

Hence smallest request that is denied=475bKb

0 votes
0 votes

Answer is 119KB

A 212kb B 114kb C 100 D

Occupied space=212+114+100=426KB

remaining space=900-426=474KB

In the question they Given that there is no compaction.So,if we try to distribute remaining space to A,B,C,D equally(474/4) each will get 118.5KB   Hence smallest request that is denied is 119KB  

Related questions

357
views
1 answers
0 votes
Veer123 asked Sep 2, 2022
357 views
If the main memory is divided into 4 fixed sized partitions of 5KB each and each of them is occupied by processes of 4KB each can I say that the total 4KB of space unused as external fragmentation?
771
views
1 answers
2 votes
A_i_$_h asked Sep 20, 2017
771 views
Consider a scenario having fixed partition allocation scheme where each partition is of size 100KB , 500KB , 200Kb , 300KB and 600KB.In these ... How may partitions will remain unallocated after placing the proceses by firstbit approach
931
views
1 answers
1 votes
vamp_vaibhav asked Sep 1, 2017
931 views
Consider a dynamic partitioning scheme.Show that on an avg. memory contains half as many holes as segments??i am not able to understand the concept of relation bw holes and segments in context to this question..