Question

Consider a demand-paging system with a paging disk that has an average access and transfer time of 20 milliseconds. Addresses are translated through a page table in main memory, with an access time of 1 microsecond per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added an associative memory that reduces access time to one memory reference, if the page-table entry is in the associative memory.

Assume that 80 percent of the accesses are in the associative memory and that, of the remaining, 10 percent (or 2 percent of the total) cause page faults. What is the effective memory access time?

in this question i know we must use

effective access time =(1-p)*ma+p*page fault time.

effective access time = (0.8) (1 microsec)+ (0.1) (2 microsec) +(0.1) (5002 microsec)

                               = 501.2 microsec

                               = 0.5 millisec

however, could you explain where did each number come from?

for example,

0.8 come from 80 percent of the accesses

1 microsec come from.....

Answer 1

It seems that you have written wrong formula for effective memory access time. It should be:-

Effective Memory Access Time = 80% of Reference from associative memory + 18% form page table + 2% from page fault

a) Reference from associative memory required only one memory access = 1 microsecond

b) Reference from page table required two memory access, one for page table which is itself in memory and other for actual data which is also stored in memory. i.e. total time = 2 microsecond

c) If there is a page fault then page will be brought back from paging disk and will be loaded in memory and page table is updated. Hence it required one paging disk access and two memory access, one for page table and other for actual data.

i.e. total time = 20 millisecond + 2 microsecond = 20000 microsecond + 2 microsecond = 20002 microsecond

Hence,

Effective Memory Access Time = 80% of Reference from associative memory + 18% form page table + 2% from page fault

= 80% *(1 microsecond) + 18%*(2 microsecond) + 2%*(20002 microsecond)

=0.8 microsecond + 0.36 microsecond + 400.04 microsecond

=401.2 microsecond (Ans)