Data-Interpretation
October 4, 2023Data-Communication
October 4, 2023Operating-Systems
| Question 38 |
A 1000 Kbyte memory is managed using variable partitions but to 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
| 151 | |
| 181 | |
| 231 | |
| 541 |
Question 38 Explanation:
200 and 260 are already hold by some other processes. Now we have to model the partition in such a way so that smallest allocation request could be denied. So, we can do the division as,
So, smallest allocation request which can be denied is 181 KB.
So, smallest allocation request which can be denied is 181 KB.
Correct Answer: B
Question 38 Explanation:
200 and 260 are already hold by some other processes. Now we have to model the partition in such a way so that smallest allocation request could be denied. So, we can do the division as,
So, smallest allocation request which can be denied is 181 KB.
So, smallest allocation request which can be denied is 181 KB.
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