Chapter 6, Problem 13E

a)

Explanation of Solution

In direct mapped cache the address of the main memory is being subdivided into 3 parts and they are:

• Tag
• Block
• Word.

Given:

Byte addressable memory= $\text{4K×8 bytes}$ is required.

${\text{log}}_{\text{2}}\left(\text{4K×8}\right){\text{=2}}^{\text{15}}$ ; 15 will be the total size of the address.

Number of blocks present in the cache is 8 and it is equivalent to ${\text{2}}^3\text{blocks}\text{.}$

It requires 3 bits for a block field

b)

Explanation of Solution

Calculating the hit ratio:

• The first address is 0=(000000000000000)₂and is considered to be a miss.
• Thus the blocks containing in this address is moved to the cache.
• The block that is moved to the cache is from 0₁₀ to 7₁₀.
• According to the bit size of the block field (000)₂ is equivalent to 0 and it is being moved to the block 0.
• All the bytes are now moved to the block 0.
• The above mentioned process of actions will be continued repeatedly until the address 63₁₀
• The process will be continued in such a way for every miss there would be 7 hits because the starting block is considered to be as miss.
• After continuing this process, the cache will be found to be full at this stage.
• Therefore, there will be 8 misses and 56 hits.
• At the address is 64₁₀=(000000001000000)₂and is considered to be a miss.
• Thus the blocks containing 8 bytes are moved to the main memory.
• The block 0₁₀ to 7₁₀ that resides in the main memory is now being replaced to be 64₁₀ to 67₁₀
• Therefore, there will be 1 miss and 3 hits.
• At the end of first loop there will be 9 miss and 59 hits.
• The above mentioned process will be repeated again and again for the remaining loops.
• The block that resides in the memory will be again gets replaced.
• Thus 7 hits will be followed for the bytes that remain in the block.
• For the remaining bytes till 63₁₀ it will be all hit.
• There will be 1miss followed by 63 hits which are followed by 1 miss and 3 hits.
• Therefore, in total there will be 2 misses and 66 hits in the second iteration.
• The process will be repeated for the next third and fourth loop...

c)

Explanation of Solution

Effective Access time (EAT):

• EAT is a measure of Hierarchal memory performance.
• A weighted average that utilizes the hit ratio and the relative access time of the levels present in successful order of the memory hierarchy is called as effective access time.

Formula for the effective access time:

The formula for effective access time where the memory contains two levels that includes a main memory and the cache memory is obtained as shown below:

$\text{EAT}\text{=}{\text{H×Access}}_{\text{c}}\text{+}\left(\text{1-H}\right){\text{×Access}}_{\text{MM}}$

In the above formula,

$\begin{array}{l}\text{H}\text{=}\text{cache}\text{hit}\\ {\text{Access}}_{\text{c}}\text{=}\text{cache}\text{access}\text{time}\end{array}$