A CPU is equipped with a cache. If it takes 5 ns to access the data from the cache and 60 ns to access data from the main memory. What should be the hit ratio if the effective memory access time is 8 ns? Answer=
A CPU is equipped with a cache. If it takes 5 ns to access the data from the cache and 60 ns to access data from the main memory. What should be the hit ratio if the effective memory access time is 8 ns? Answer=
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![**Question:**
A CPU is equipped with a cache. If it takes 5 ns to access the data from the cache and 60 ns to access data from the main memory, what should be the hit ratio if the effective memory access time is 8 ns?
**Answer = ________________%**
**Explanation:**
To find the hit ratio, you can use the formula for effective memory access time:
\[ \text{Effective Access Time (EAT)} = (\text{Hit Ratio} \times \text{Cache Access Time}) + ((1 - \text{Hit Ratio}) \times \text{Main Memory Access Time}) \]
Given:
- Cache Access Time = 5 ns
- Main Memory Access Time = 60 ns
- Effective Access Time = 8 ns
Substitute the values:
\[ 8 = (\text{Hit Ratio} \times 5) + ((1 - \text{Hit Ratio}) \times 60) \]
Solve for Hit Ratio:
1. Expand the equation:
\[ 8 = 5 \times \text{Hit Ratio} + 60 - 60 \times \text{Hit Ratio} \]
2. Combine like terms:
\[ 8 = 60 - 55 \times \text{Hit Ratio} \]
3. Rearrange:
\[ 55 \times \text{Hit Ratio} = 60 - 8 \]
4. Simplify:
\[ 55 \times \text{Hit Ratio} = 52 \]
5. Divide both sides by 55:
\[ \text{Hit Ratio} = \frac{52}{55} \approx 0.9455 \]
Therefore, the hit ratio should be approximately 94.55%.
Fill in the answer:
**Answer = 94.55%**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44c3b237-e4ca-4e38-a278-251c2495f1b6%2Fcac094de-c32e-4f6f-ab71-a115f606d50b%2Fsg35nx_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A CPU is equipped with a cache. If it takes 5 ns to access the data from the cache and 60 ns to access data from the main memory, what should be the hit ratio if the effective memory access time is 8 ns?
**Answer = ________________%**
**Explanation:**
To find the hit ratio, you can use the formula for effective memory access time:
\[ \text{Effective Access Time (EAT)} = (\text{Hit Ratio} \times \text{Cache Access Time}) + ((1 - \text{Hit Ratio}) \times \text{Main Memory Access Time}) \]
Given:
- Cache Access Time = 5 ns
- Main Memory Access Time = 60 ns
- Effective Access Time = 8 ns
Substitute the values:
\[ 8 = (\text{Hit Ratio} \times 5) + ((1 - \text{Hit Ratio}) \times 60) \]
Solve for Hit Ratio:
1. Expand the equation:
\[ 8 = 5 \times \text{Hit Ratio} + 60 - 60 \times \text{Hit Ratio} \]
2. Combine like terms:
\[ 8 = 60 - 55 \times \text{Hit Ratio} \]
3. Rearrange:
\[ 55 \times \text{Hit Ratio} = 60 - 8 \]
4. Simplify:
\[ 55 \times \text{Hit Ratio} = 52 \]
5. Divide both sides by 55:
\[ \text{Hit Ratio} = \frac{52}{55} \approx 0.9455 \]
Therefore, the hit ratio should be approximately 94.55%.
Fill in the answer:
**Answer = 94.55%**
![A CPU is equipped with a cache. If it takes 5 ns to access the data from the cache and 60 ns to access data from the main memory, what is the effective memory access time if the hit ratio is 90%?
Answer = _______________ ns
[Input box]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44c3b237-e4ca-4e38-a278-251c2495f1b6%2Fcac094de-c32e-4f6f-ab71-a115f606d50b%2Fu5uyc2r_processed.png&w=3840&q=75)
Transcribed Image Text:A CPU is equipped with a cache. If it takes 5 ns to access the data from the cache and 60 ns to access data from the main memory, what is the effective memory access time if the hit ratio is 90%?
Answer = _______________ ns
[Input box]
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