A 2000 KB physical memory is managed using variable partitioning scheme with no compaction. currently , it has three partitions of sizes 200KB , 400KB ,and 600KB respectively. Then , calculate the SMALLEST allocation request that could be denied by the system . Explain your answer?
Answers
Answer:
ueutuisjrguitirhyirhrueufuufurx
Explanation:
jejwiajajdgjzsjhduxisiiduic
euufijdjdjrifiidiwit
jrrjidiruuririei
ruuroussudkkdjg
uwuiqkwkskek
iwis
Answer:
To determine the smallest allocation request that could be denied by the system, we need to consider the remaining free space after the current partitions are created.
Explanation:
The total size of the physical memory is 2000 KB, and the current partitions are 200 KB, 400 KB, and 600 KB. Therefore, the remaining free space is:
2000 KB - 200 KB - 400 KB - 600 KB = 800 KB
Now, let's assume that the system receives an allocation request of 500 KB. The system will search for a partition that can accommodate the request. However, the largest partition available is 600 KB, which is not enough to satisfy the request. Therefore, the system will deny the allocation request.
In this scenario, the smallest allocation request that could be denied by the system is 500 KB. Any allocation request larger than 500 KB will definitely be denied since there is no single partition that can accommodate it.
In summary, the smallest allocation request that could be denied by the system is determined by the size of the largest partition available after the current partitions are created. Any allocation request larger than the remaining free space will be denied by the system.
Learn more about Physical Memory :
https://brainly.in/question/23165902
#SPJ2