COA: Cache Accesss Time

Question

Little confusion with these questions.

What will be o/p for these two questions one with Write back and the other is Write through.

1.)A 128 word cache and main memory are divided into 8 word blocks. The access time of a cache and main memory are 10ns/word and 25ns/word respectively.
The hit ratio for read operation is 90% and writes operation is 80%. Let there are 60% references for read operations, using write through. What is the average access time?(https://gateoverflow.in/256999/coa-cache-average-access-time)

2.)A 128 word cache and main memory are divided into 8 word blocks. The access time of a cache and main memory are 10ns/word and 25ns/word respectively. 
The hit ratio for read operation is 90% and writes operation is 80%. Let there are 60% references for read operations, using write back. What is the average access time?

What will be the default technique ( write-allocate and no write-allocate)followed for Hierarchal and Simultaneous access ?

Answer 1

System by default use write with no allocation, if specified then solve using write with allocation.

Answer 2

a)  FOR WRITE THROUGH POLICY

read access time= 0.6{0.9(10)+0.1(10+200)}=18ns

here 0.6 represent the read access time and 0.9X10 represent the read access time on hit and also 0.1(10+200) represent the time time to transfer whole block of 8 words from memory to cache on cache miss

write access time =0.4{0.8(25)+0.2(200+25)} =26ns

here 0.4 is write access and 0.8 X 25 is time to write a word in memory(this is case of simultaneous writes so max(10,25) is choosen that is 25) and 0.2(200+25) is time to transfer the whole block from memory to cache (200) and then writing word in both cache and memory simultaneously  (25) on cache write miss(0.2)

Average access time= read access time + write acess time=18+26 = 44 ns

 

b) solution is not possible without percentage of fraction of dirty bits in cache, i will provide you solution if you give me % of dirty bit