Bangladesh Bank
Post: Assistant maintenance engineer (AME),
Exam Date: 04.02.2023, Exam Taker: BIBM
In direct mapping physical address is divided into three parts i.e., Tag bits, Cache Line Number and Byte offset.
| Tag | Number of Cache Lines/Index bit | Offset Bit |
Given:-
Cache size: 16 KB
Block size: 4 words
Word size: 4 bytes (assuming a typical 32-bit architecture where each word is 4 bytes)
Address size: 32 bits
Calculate the Block Size in Bytes:
Since each block contains 4 words and each word is 4 bytes:
Block size = \( 4 \text{ words} \times 4 \text{ bytes/word} = 16 = 2^4 \text{ bytes} \)
Number of offset bits = \( \log_2 2^4 \) = 4 bits
Calculate the Number of Cache Blocks:
Block size = Line size = \( 2^4 \text{ bytes} \)
The total cache size is 16 KB = \( 2^{14} \text{ bytes} \) [ 1 KB = 210B]
Number of Lines = \( \frac{\text{Total cache size}}{\text{Line size}} = \frac{2^{14}}{2^4} = 2^{10} \text{ lines} \)
Number of Lines in the cache:
Number of index bits = \( \log_2 \text{(Number of lines)} = \log_2 2^{10} = \) 10 bits
Number of tag bits = Total Address bit − (Index bits + Offset bits) = \( 32 – (10 + 4) = \) 18 bits.
Solution: Here, The source uses 480 x 500 pixels/picture And, 30 pictures are sent per second.
So, we can find the source rate in pixels/s as:
\[ (30 \, \text{pictures/s}) \times (480 \times 500 \, \text{pixels/picture})\] \[= 7.2 \times 10^6 \, \text{pixels/s} \]
Now, each pixel can take on one of 32 intensity values and these 32 intensity values can be represented by:
\[ \log_2 32 = \log_2 2^5 = 5 \, \text{bits} \] We can find the source rate in bits/s (bps) as:
R = \( 7.2 \times 10^6 \, \text{pixels/s} \times 5 \, \text{bits/pixel} \)
\[ = 36 \times 10^6 \, \text{bps} = 36 \, \text{Mbps} \]
source:[brainly.in]
