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Effective Memory Access Time (EMAT) with Page Fault rate

Before seeing this chapter please follow the previous chapter: Demand paging and page fault

So, from what we know from the previous chapter due to page fault, the overhead is very high, it takes a lot of time for the whole procedure to update the page table or page replacement operation.

Now we calculate Effective Memory Access Time (EMAT) using page fault ratio:

Two situations can occur to get page number from the page table:

(a) Page Fault occurred: It will take extra time for a page fault.

(b) No Page Fault occurred: It will not take extra time.

Now, a page fault will occur for every certain amount of main memory access, which is called page fault rate.

For example, if one page fault has occurred for every memory access ${\mathbf{10}}^{\mathbf{5}}$ instruction i.e. one-page fault occur after every ${\mathbf{10}}^{\mathbf{5}}$  instruction in memory then 

Or we can calculate it another way like if the hit ratio to access memory is 80% means the page available at memory is 80%. So, the miss ratio means the page fault ratio is 20%.

Let,

Page fault rate = p,  memory access time = m, and extra time is taken for page fault service = PF (because we know if page fault occurred, we need some extra time for it).

So,

Where (1 - p) is no page fault rate. 

Note: Page fault service time is comparatively higher than memory access time. So, it makes a big difference in EMAT if a page fault occurred. 

Now, we can minimize the formula of EMAT to make calculation easy:

So, it is easy to use this formula to calculate

Note: In this minimize formula basically we neglect main memory access time when page fault occurred because main memory access time is very low compare to page fault service time.

Example Type 1: 

Consider a paging system where memory access time is 60ns and additional page fault service time is 20ms. If a one-page fault occurred after every instruction access in memory, then find effective memory access time (EMAT).

Solution:

As we know, 

Now, we can use optimize formula EMAT =  p(pf)+  m which same as above

Here memory access time (m) = 60ns, page fault service time = 20ms and page fault rate = $\mathbf{1}\mathbf{/}{\mathbf{10}}^{\mathbf{6}}$

So, Effective Memory Access Time (EMAT) for this above example is 80ns.

Example Type 2: 

In a paging system, page fault service time required 10ms and main memory access time is 4 microseconds. If the hit ratio is 90%, find the Effective Memory Access Time (EMAT).

Solution:

As we know, 

Now, we can use optimize formula EMAT =  p(pf)+  m which same as above

Here memory access time (m) = 4μsec, page fault service time = 20ms and hit ratio = 99.9% so, miss ratio or page fault ratio = (1 - 99.99%) = 0.01%, so page fault rate 0.01/100 = 0.0001 i.e. ${\mathbf{10}}^{\mathbf{4}}$.

So, Effective Memory Access Time (EMAT) for this above example is 6μsec.