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3 years ago

Assume the system with 256B memory and 64B cache and the block size of 16 bytes. I.e., there are 4 blocks in the cache.

Computers and Technology

1 answer:

Arada [10]3 years ago

6 0

Answer:

a. 100%

b. 50%

c. 66.6%

Explanation:

The memory references D6, 58, D8, 9A, DE and 5C all results in misses and the miss ratio is 100 (that is , 6 misses/ 6 references*100 = 100%)

If fully associative cache is used, there will be a 3 cache misses with miss ratio of (3/6*100) = 50%.

With a 2-way set-associative cache, 4 memory reference misses will occur, with a miss ratio of (4/6*100)= 66.6% misses.

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3 years ago

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Ratling [72]

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3 years ago

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3 years ago

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3 years ago

// This pseudocode segment is intended to compute and display

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Inclusion of unusable segments
Incorrect variables
Incorrect loops

<h3>The correct pseudocode</h3>

The correct pseudocode where all errors are corrected and the unusable segments are removed is as follows:

start

Declarations

num test1

num test2

num test3

num average

output "Enter score for test 1 or a negative number to quit"

input test1

while test1 >= 0

output "Enter score for test 2"

input test2

output "Enter score for test 3"

input test3

average = (test1 + test2 + test3) / 3

output "Average is ", average

output "Enter score for test 1 or a negative number to quit"

input test1

endwhile

output "End of program"

stop