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Cache characteristics:

● Cache size (capacity): 4KB
● Cache-line size: 4B
● Address width: 16 bits.
● Sequence of load addresses: 0x0, 0x4, 0x1ff8, 0x0, 0x4000, 0x0
1) What is the structure of the cache?
2) How many bits required to allocate offset, index and tag?
3) What is the hit rate of the cache?

User Hustnzj
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2 Answers

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Answer:

To determine the structure of the cache, we need to consider the cache size, cache-line size, and address width.

1. Structure of the cache:

Given the cache size of 4KB (4 kilobytes) and a cache-line size of 4B (4 bytes), we can calculate the number of cache lines in the cache:

Number of cache lines = Cache size / Cache-line size

Number of cache lines = 4KB / 4B

Number of cache lines = 1024

So, the cache has 1024 cache lines.

2. Number of bits required for offset, index, and tag:

To determine the number of bits required for offset, index, and tag, we need to consider the cache-line size and the number of cache lines.

Offset:

The cache-line size is 4B, which requires 2 bits to represent all possible offsets (2^2 = 4).

Index:

The number of cache lines is 1024, which requires 10 bits to represent all possible indices (2^10 = 1024).

Tag:

The remaining bits after considering offset and index are used for the tag. The address width is 16 bits, and we have already used 2 bits for the offset and 10 bits for the index. So, the remaining bits for the tag are 16 - 2 - 10 = 4 bits.

Therefore, the number of bits required for offset, index, and tag are as follows:

Offset: 2 bits

Index: 10 bits

Tag: 4 bits

3. Hit rate of the cache:

To calculate the hit rate of the cache, we need to analyze the sequence of load addresses and determine if each address is a hit or a miss.

Given the sequence of load addresses: 0x0, 0x4, 0x1ff8, 0x0, 0x4000, 0x0

First, let's convert the load addresses to binary form:

0x0 = 0000000000000000

0x4 = 0000000000000100

0x1ff8 = 0001111111111000

0x0 = 0000000000000000

0x4000 = 0100000000000000

0x0 = 0000000000000000

Now, let's analyze each load address:

- 0000000000000000 (0x0): This is a hit because it falls within the cache line with an index of 0.

- 0000000000000100 (0x4): This is a hit because it falls within the cache line with an index of 1.

- 0001111111111000 (0x1ff8): This is a hit because it falls within the cache line with an index of 1023.

- 0000000000000000 (0x0): This is a hit because it falls within the cache line with an index of 0.

- 0100000000000000 (0x4000): This is a miss because it falls outside the cache lines.

- 0000000000000000 (0x0): This is a hit because it falls within the cache line with an index of 0.

Out of the 6 load addresses, 5 are hits and 1 is a miss.

Therefore, the hit rate of the cache is 5/6, which is approximately 0.8333 or 83.33%.

Please note that this analysis assumes a simplified direct-mapped cache structure and does not consider other cache replacement policies or complexities.

User Meta Fan
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5 votes

Final answer:

The cache structure consists of 1024 cache lines and 1024 sets. The offset requires 2 bits, the index requires 10 bits, and the tag requires 4 bits. The hit rate of the cache is 33.33%.

Step-by-step explanation:

Structure of the cache:

The given cache has a capacity of 4KB, with each cache line having a size of 4B. The address width is 16 bits. To determine the structure of the cache, we need to calculate the number of cache lines and the number of sets.

Number of cache lines = Cache size / Cache-line size = 4KB / 4B = 1024 cache lines

Number of sets = Number of cache lines / Associativity = 1024 / 1 = 1024 sets

Bits required to allocate offset, index, and tag:

The offset is the number of bits required to represent the byte within a cache line. It can be calculated as log2(Cache-line size) = log2(4B) = 2 bits

The index is the number of bits required to represent the set. It can be calculated as log2(Number of sets) = log2(1024) = 10 bits

The tag is the remaining bits after accounting for offset and index. It can be calculated as Address width - (Offset bits + Index bits) = 16 - (2 + 10) = 4 bits

Hit rate of the cache:

Given the sequence of load addresses, we can determine the hit rate by counting the number of hits (addresses that are already present in the cache) and dividing it by the total number of load addresses.

The hit rate = Number of hits / Total number of load addresses

In the given sequence, the hits are: 0x0, 0x0

The total number of load addresses is 6.

Therefore, the hit rate = 2 / 6 = 1/3 = 33.33%

User SnellyBigoda
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