Memory Management Quiz

If main memory access time is 400 μs, TLB access time is 50 μs, considering TLB hit as 90%, what will be the overall access time?

• 800 μs

• 490 μs

• 485 μs

• 450 μs

Answer explanation

EAT = TLB hit rate × (TLB access time + Main Memory access time) + (1−TLB hit rate) × (TLB access time + 2 × Main memory access time)

EAT = 0.9 X ( 50μs + 400μs ) + 0.1 X ( 50μs + 2 X 400μs )

EAT = (405 + 85) μs

EAT = 490 μs

If main memory access time is 100 ns, TLB access time is 20 ns, considering TLB hit as 80%, what will be the effective access time?

• 100 ns

• 120 ns

• 140 ns

• 160 ns

Answer explanation

EAT(effective access time)= P x hit memory time + (1-P) x miss memory time.

effective access time = 0.80 x (20 + 100) + 0.20 x (20 + 100 + 100) = 140 nanoseconds.

What will be the EAT if hit ratio is 70%, time for TLB is 30ns and access to main memory is 90ns?

• 45 ns

• 150 ns

• 175 ns

• 147 ns

Answer explanation

Hit ratio (P) = 70% = 70/100 = 0.7

Hit memory time = 30ns + 90ns = 120ns

Miss memory time = 30ns + 90ns + 90ns = 210ns

Therefore, EAT = P x Hit + (1-P) x Miss = 0.7 x 120 + 0.3 x 210 =840 + 63.0 =147 ns

If main memory access time is 100 ns, TLB access time is 20 ns, considering TLB hit as 98%, what will be the effective access time?

• 122 ns

• 125 ns

• 118 ns

• 21.6 ns

Answer explanation

EAT(effective access time)= P x hit memory time + (1-P) x miss memory time.

Where: P is Hit ratio.

effective access time = 0.98 x 120 + 0.02 x 220 = 122 nanoseconds.

Consider a paging hardware with a TLB. Assume that the entire page table and all the pages are in the physical memory. It takes 10 milliseconds to search the TLB and 80 milliseconds to access the physical memory. If the TLB hit ratio is 0.6, the effective memory access time (in milliseconds) is _________.

[GATE-CS-2014]

• 38
• 120

• 122

• 118

Answer explanation

EAT := TLB_miss_time (1- hit_ratio) + TLB_hit_time hit_ratio.

EAT := (TLB_search_time + 2*memory_access_time) (1- hit_ratio) + (TLB_search_time + memory_access_time) hit_ratio.

As both page table and page are in physical memory
EAT = hit ratio (TLB access time + Main memory access time) +
(1 – hit ratio) (TLB access time + 2 main memory access time)
= 0.6(10+80) + (1-0.6)*(10+2*80)
= 0.6 (90) + 0.4 (170)
= 122