effective memory access time

Question

Page tables are stored in memory , which has access time of 100 ns. The TLB holding 8 page table entries, has an access time of 10 ns. Using execution of process , it is found that 85 % of the time, a required page table entry exist in TLB and only 2 % of the total references causes page fault. Page replacement time is 2 ms . Calculate the effective memory access time , assuming page memory access requires 2 memory accesses and TLB requires one memory access.

A) 38120 ns B) 40000 ns C) 40120 ns D) None

Answer 1

0.85*[100+10]+0.15[0.02{10+100+2000000+100}+0.98{10+100+100}]

= 6125.00 ns

If there's TLB hit just refer main memory directly.

If there's TLB miss ,refer memory if page fault occurs replace page and again refer memory. 

Answer 2

What i get from the question is
Page fault rate is 2% means 98% reference of memory found either in TLB or in page to main memory reference.
after page fault occurred we need the page service time.

so total
0.98(0.85(10+100)+0.15(2*100)) + .02*2000000 = 40121.03ns

it is closer to C and hence C.