GATE Overflow Test Series | Operating Systems | Test 2 | Question: 13

Question

A $2048\;\text{Kbyte}$ memory is managed using variable partitions but no compaction. It currently has two partitions of sizes $840\;\text{Kbyte}$ and $518\;\text{Kbyte}$ respectively. The smallest allocation request in $\text{Kbyte}$ that could be denied is for ________

Answer 1

With the two given partitions we can view memory like below

$a - 840 - b - 518 -c,$

where, $a + b + c = 2048 - 518 -840 = 690.$

Now, to minimize the size of request to be denied, the holes should be of minimum size. This happens when all of $a,b,c$ are the same. So, $a = b = c = 690/3 = 230$ and the smallest allocation request that could be denied $ = 231 KB.$

Comments

@gatecse  Please explain, I think I am misunderstanding something? We already have allocated that 840 and 518KB correct?? Now we have empty space 690KB. Correct??

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@gatecse @Arjun Sir, why (231K + 1) Byte is wrong ? as this will be the smallest allocation request that could be denied. Plz correct me if I’m wrong