Question

2. A computer uses virtual memory implemented by paging. The TLB lookup takes 150 ns and the update takes 300 ns. The PT lookup takes 2 us and the update takes 4 us. Loading a word from main memory onto the CPU takes 25 us and loading a page from the disk into main memory takes 20 ms. The TLB hit ratio is 0.3 and the main memory hit ratio is 0.4. Compute the average access time for a referenced word: i.e., the time it takes • to find out the physical address • to load the referenced word onto the CPU and • to perform the necessary updates to the page tables. Hint: Compute the access time for the possible scenarios (TLB-hit, etc.) and take the weighted average according to the hit ratios.

Answer 1

Solution:

Given that :-

TLB lookup = 150 ns

TLB update = 300 ns

PT lookup = 2 $\mu s$

PT update = 4 $\mu s$

Reading/ loading a word from main memory = 25 $\mu s$

Loading a page from disk to memory = 20 ms

TLB hit ratio = 0.3

Main memory hit ratio = 0.4

• To find the address : -

Average access Time = TLB hit ratio* TLB Look up +TLB miss ratio[ TLB update + PT Look up ]

= 0.3 + 150 +0.71300 + 2000

= 45+ 1610 = 1655 ns

• To load referenced word on CPU :-

Average access time = Time to find physical address + Main memory hit ratio * read time from memory + Memory miss ratio * disk read time.

= 1655 ns +0.4(25 pus) + 0.6(20 ms)

= 1655 ns +0.4(25 pus) + 0.6(20000 pus)

= 1655 ns + 10 us + 12000 us

= 1655 ns + 12010 us

= 12.011 us

• Perform necessary update on page table = This happens when we have referred something from page table.

Average access time = Time to get physical address + PT update time.

= 1655 ns + 4000 ns = 5655 ns

Answer 2

Why are you using Main memory hit ratio is 0.4 ? 
I am wondering if it should be 0.7 *0.4