Given-
- Number of levels of page table = 2
- Main memory access time = 150 ns
- Page fault service time = 8 msec
- Average instruction takes 100 ns of CPU time and 2 memory accesses
- TLB Hit ratio = 90% = 0.9
- Page fault rate = 1 / 104 = 10-4
Assume TLB access time = 0 since it is not given in the question.
Also, TLB access time is much less as compared to the memory access time.
Effective Access Time Without Page Fault-
Effective memory access time without page fault
= 0.9 x { 0 + 150 ns } + 0.1 x { 0 + (2+1) x 150 ns }
= { 0.9 x 150 ns } + { 0.1 x 450 ns }
= 135 ns + 45 ns
= 180 ns
Effective Access Time With Page Fault-
Effective access time with page fault
= 10^-4 x { 180 ns + 8 msec } + (1 – 10^-4) x 180 ns
= 8 x 10^-4 msec + 180 ns
= 8 x 10^-7 sec + 180 ns
= 800 ns + 180 ns
= 980 ns
Effective Average Instruction Execution Time-
Effective Average Instruction Execution Time
= 100 ns + 2 x Effective memory access time with page fault
= 100 ns + 2 x 980 ns
= 100 ns + 1960 ns
= 2060 ns
resource- gatevidyalya