Management Notes

# Management Notes

Reference Notes for Management

# Suppose we have a logical address space of 64 pages of 1024 words each, mapped onto a physical memory of 256 frames. How many bits are there in the logical address?

## Suppose we have a logical address space of 64 pages of 1024 words each, mapped onto a physical memory of 256 frames. How many bits are there in the logical address?

Options:

(a) 15
(b) 16
(c) 17
(d) 18
(e) 19.

(b) 16

The correct answer is (b) 16 bits for the logical address. Let’s break down the reasoning behind this answer and then explain why the other options are not correct.

Explanation for Correct Answer (b) 16 bits:

In a paging system, the logical address is divided into two parts: the page number and the offset within the page.

To determine the number of bits required for each part, we can use the given information:

1. Logical address space: 64 pages.
2. Page size: 1024 words.
3. Physical memory: 256 frames.

Number of bits for the offset: To represent 1024 words, you need log2(1024) bits, which is 10 bits. This is because 2^10 = 1024.

Number of bits for the page number: The total number of pages in the logical address space is 64. To represent 64 pages, you need log2(64) bits, which is 6 bits. This is because 2^6 = 64.

So, the logical address is divided into a 6-bit page number and a 10-bit offset within the page, which sums up to a total of 16 bits (6 bits for the page number and 10 bits for the offset). Therefore, the correct answer is (b) 16 bits.

Explanation for why the other options are not correct:

(a) 15 bits:

This option suggests that the logical address space can be represented using 15 bits. However, this is insufficient for the given scenario. We have 64 pages in the logical address space. To represent 64 unique pages, we need at least log2(64) = 6 bits, because 2^6 = 64.

If we only use 15 bits, we would have only 2^15 = 32,768 possible unique addresses. This is not enough to represent the 64 pages.

(c) 17 bits:

This option proposes that the logical address can be represented using 17 bits. While this would be enough to represent the 64 pages, it would be an overestimation. We’ve already determined that 16 bits (6 bits for the page number and 10 bits for the offset) are sufficient for this scenario.

Using 17 bits would provide more address space than necessary, which could lead to wasted resources and inefficient memory management.

(d) 18 bits:

This option suggests using 18 bits for the logical address. However, this is excessive for the given scenario. As we established, 16 bits (6 bits for the page number and 10 bits for the offset) are sufficient to represent the 64 pages and 1024 words per page.

Using 18 bits would provide even more address space than required, which would be inefficient and wasteful in terms of memory allocation.

(e) 19 bits:

This option proposes using 19 bits for the logical address. Again, this is excessive and unnecessary. With 16 bits, we can efficiently represent the 64 pages and 1024 words per page.

Adding an extra 3 bits would provide far more address space than needed, which would be inefficient and could lead to memory wastage.

In this scenario, the logical address space encompasses 64 pages, with each page containing 1024 words. To represent the offset within a page, we require 10 bits since 2^10 equals 1024. Moving on to the page number, we must be able to uniquely identify each of the 64 pages.

To do this, we need log2(64) bits, which amounts to 6 bits. This is because 2^6 gives us 64. Therefore, the total number of bits required for the logical address is the sum of the bits needed for the page number and the offset, which results in 16 bits in total.

The other options provided (a, c, d, and e) are not correct for various reasons. Option (a) suggests 15 bits, which is insufficient to represent the 64 pages. Option (c) proposes 17 bits, which, while adequate, is an overestimation and could lead to inefficient memory usage.

Options (d) and (e) suggest 18 and 19 bits, respectively, both of which are excessive for the given scenario. These would offer more address space than needed, leading to potential memory wastage and inefficiency.

Therefore, after careful consideration, it is evident that the correct answer is indeed (b) 16 bits, as it efficiently represents the 64 pages and 1024 words per page in the logical address space.

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