Test by Bikram | Operating Systems | Test 2 | Question: 10

Question

Consider a logical address space of eight pages of $1024$ words each mapped onto a physical memory of $32$ frames. The number of bits are there in the logical address is _______

Answer 1

The logical address looks as follows:
Page Number     Offset
 m-n bits            n bits

$$\begin{array}{|c|c|c|} \hline \text{Page Number} & \text{Offset} \\ \hline m-n\:\text{bits} & n\:\text{bits}\\ \hline  \end{array}$$
The size of the page, which is also equal to frame size is defined by hardware. If the logical address is m bit long, we have n bits for offset and higher-order m- n bits for pages.

1. We have in all 8 pages. To store info. about 8 pages we require 3 bits. This means page number needs 3 bits.
2. To address 1024 (= 2^10) words in each page, we need 10 bits, since 2^10 = 1024. Thus, the offset needs 10 bits.
The total address size = number of bits for page index + number of bits for offset = 10 + 3 = 13 bits