GATE CS 2026 Set 2 Q.55: Memory Management and Virtual Memory

Consider contiguous allocation of physical memory to processes using variable partitioning scheme. Suppose there are 8 holes in the memory of sizes 20 KB, 4 KB, 25 KB, 18 KB, 7 KB, 9 KB, 15 KB, and 12 KB. Assume that no two holes are adjacent. Two processes P1 of size 16 KB and P2 of size 9 KB arrive in that order, and they are allocated memory using the best-fit technique. After allocating space to P1 and P2, the number of holes of size less than 8 KB is ____________. (answer in integer)

Note: $1\text{K}=2^{10}$

Your answer

Official answer and solution

Official answer: 3
From: Official GATE 2026 answer key

Solution

Allocate P1 (16 KB) and P2 (9 KB) using Best-Fit and count the remaining non-zero holes strictly smaller than 8 KB.

Initial holes: $20, 4, 25, 18, 7, 9, 15, 12$ KB.

P1 (16 KB) arrives: Under Best-Fit, the smallest hole $\ge 16\text{ KB}$ is the $18\text{ KB}$ hole, leaving a leftover hole of size $18 - 16 = 2\text{ KB}$ .
P2 (9 KB) arrives: The smallest hole $\ge 9\text{ KB}$ is the $9\text{ KB}$ hole, which is used completely ( $9 - 9 = 0\text{ KB}$ , no hole remains).

Remaining holes: $20, 4, 25, 2, 7, 15, 12$ KB.
The holes with size $< 8\text{ KB}$ are $4\text{ KB}, 2\text{ KB},$ and $7\text{ KB}$ , which gives a total of $3$ holes.

Final answer

Common trap

Counting the completely filled 9 KB partition (0 KB remainder) as a hole.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question: 55 of 65
Section: Computer Science and Information Technology

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