Andrew S. Tanenbaum (OS) Edition 4 Exercise 1 Question 15 (Page No. 82)

Question

Consider a computer system that has cache memory, main memory (RAM) and disk, and an operating system that uses virtual memory. It takes $1$ nsec to access a word from the cache, $10$ nsec to access a word from the RAM, and $10$ ms to access a word from the disk. If the cache hit rate is $95\%$ and main memory hit rate (after a cache miss) is $99\%$, what is the average time to access a word?

Answer 1

it will be =.95 *1 ns +0.05 [{0.9*10}+0.1*10*10^6}}=5001.445 ns

Answer 2

By default it should be hierarchical right ?

1ns + 0.05*(10ns +   0.1*(10 ms))

verify @Arjun Sir