Test ques.

Question

Answer 1

i am getting  228 
 

Comments

=> 60+(0.3)*[400+(0.1)(1600)]

=> 60+120+48

=> 228

Less time 

Answer 2

If cache hit, no need to go to higher memories;

If cache miss, then goes to main memory; if the main memory miss (word/block) is unavailable in the main memory then it goes to secondary memory (where the hit ratio or hit confidence is 1.00 or 100%).

Answer 3

Let's break this down.

 

In case of a hit in cache, time =

$$=0.7*60 = 42ns$$

 

In case of a miss in cache, time =

$$=0.3* (\text{time to get it from main memory})$$

 

The time to get it from main memory can broken down into 2 cases

1. Hit in main memory

2. Miss in main memory

 

Hit in main memory

$$=0.9 * ( 60 + (50*8) ) = 414$$

50*8 is to transfer from main memory to cache

60 is to fetch from cache

 

Miss in main memory

$$=0.1*(60+(50*8)+(16*100)) = 206$$

16*100 is to transfer from secondary memory to main memory

50*8 is to transfer from main memory to cache

60 is to fetch from cache

Now to add them all up

$$=\text{Hit in cache} + \text{Miss in cache}$$

$$=42 + 0.3*(\text{Hit in main memory} + \text{Miss in main memory})$$

$$= 42 + 0.3*(414+206)$$

$$=228$$