Calculation of the average memory access time based on the following data?

Consider the following information

Assume the cache is physically addressed
TLB hit rate is 95%, with access time = 1 cycle
Cache hit rate is 90%, with access time of again = 1 cycle
Page fault is 1% and occurs when miss occurs in both TLB and Cache
The TLB access and cache access are sequential
Main memory access time is 5 cycles
Disk access time is 100 cycles
Page tables are always kept in main memory

What will be the Average memory access time based on the following information ?

My Approach => I am giving my approach of how i understood this question. Please check it.

Average memory access time

==>

Probability of NO page fault (Memory access time) 
+ 
Probability of page fault (Page fault service time)

==>

0.99 ( TLB hit (TLB access time + cache hit + cache miss) + TLB miss (TLB access time + Page table access time + cache hit + cache miss) )

0.01 (TLB access time + page table access time + Disk access time)

==>

0.99 ( 0.95 (1 + 0.90(1) + 0.10(1 + 5)) + 0.05(1 + 5 + 0.90(1) + 0.10(1 + 5)))

0.01 (1 + 5 + 100)

Is the given expression correct ?

Please let me know, that, is my approach right or have i committed some mistakes?

Can Anyone help me ?

PS : I am having my mid term next week and need to practice such questions

Solution

In your case the tricky line is cache is physically addressed meaning before hitting it we must perform the translation (as programs use virtual addresses)

I build the following probability tree to compute the average. We will be reducing it from leaves to calculate the whole average. Rules are super easy: we calculate branch costs and multiply them by the probability pretty much like you did in your calc. The value I get is 2.7225

Tree

1st leave reduction, page fault

after the reduction

cost for cache hit scenario

Note: that we pay 1 cycle for cache access anyway

before last reduction

Note: we pay 1 cycle for TLB anyway

1 + 0.95*1.5 + 0.05*5.95 = 1 + 1.425 + 0.2975 = 2.7225