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kvasek [131]
4 years ago
7

You are asked to optimize a cache design for the given references. Th ere are three direct-mapped cache designs possible, all wi

th a total of 8 words of data: C1 has 1-word blocks, C2 has 2-word blocks, and C3 has 4-word blocks. In terms of miss rate, which cache design is the best? If the miss stall time is 25 cycles, and C1 has an access time of 2 cycles, C2 takes 3 cycles, and C3 takes 5 cycles, which is the best cache design?
Computers and Technology
1 answer:
Alja [10]4 years ago
8 0

Answer:

Explanation:

From the given data, Direct-Mapped cache with 8words of data means 2^3 of data.

There are three caches C1, C2 and C3.

C1 has 1 word blocks

C2 has 2 word blocks

C3 has 4 word blocks

also, Miss stall = 25 cycles.

We have,

Miss rate = 3% for current block size

Thus, for cache C1, miss rate = 3%

cache C2, miss rate = 2%

cache C3, miss rate = 1.2%

So, the cache C3 design is the best. In terms of miss rate. The cache performance increases with decrease in miss penalty.

That means if miss penalty is less, then the performance of cache increases. And for cache C3, the miss penalty is less.

From given data

C1 acess time is 2 Cycles.

C2 acess time is 3 Cycles.

C3 acess time is 5 Cycles.

Then

C1 stall time is 25*11+ 2*12 = 299.

C2 stall time is 25*9+ 3*12 = 261.

C3 stall time is 25*10+ 5*12 = 310.

In this case, C2 is the best cache design.

And

C1 stall time is 25*11+ 2*12 = 299.

C2 stall time is 25*8+ 3*12 = 236.

C3 stall time is 25*8+ 5*12 = 260.

In this case also, C2 is the best cache design.

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Suppose that a 64MB system memory is built from 64 1MB RAM chips. How many address lines are needed to select one of the memory
alexdok [17]

Answer:

6 address lines

Explanation:

The computation of the number of address lines needed is shown below:

Given that

Total memory = 64MB

= 2^6MB

=2^{26} bytes

Also we know that in 1MB RAM the number of chips is 6

So, the number of address lines is 2^{26} address i..e 26 address lines

And, the size of one chip is equivalent to 1 MB i.e. 2^{20} bytes

For a single 1MB chips of RAM, the number of address lines is

\frac{2^{26}}{2^{20}} \\\\= 2^6 address

Therefore 6 address lines needed

5 0
3 years ago
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter
trapecia [35]

Answer:

5 Letters

Explanation:

So we need 12 distinct codes made of a single letter or a pair of letters.

What would be the least number of letters?

Lets try with 3 letters A, B and C

The possible combinations are: A, B, C, AB, AC, BC

These are 6 codes and we need 12 so lets try more A, B, C and D

The possible combinations are: A, B, C, D, AB, AC, AD, BC, BD, CD

These are 10 codes and we need 12 so lets try more A, B, C, D and E

The possible combinations are:

A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

Finally we got 15 distinct codes which are more than 12 so the least number of letters needed are 5.

Using formula:

Four letters = 4C1 + 4C2 = 4 + 6 = 10

Five letters = 5C1 + 5C2 = 5 + 10 = 15

5 0
3 years ago
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WARRIOR [948]

Answer:

tamera like from sister sister hehe

Explanation:

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3 0
3 years ago
Which of the following does not reflect the second step of effective communication?
Ilya [14]

Answer:

c

Explanation:

3 0
3 years ago
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quester [9]

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But since it is now Post Mortem all say here is...

Unas Annus. Memento Mori.

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3 0
3 years ago
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