The same solution is available in the test series but my doubt is, in page fault why didn't you add the memory access time?
Isn't the formula like this:
P*( Page fault service time + Memory access time ) + (1-P)( Memory access time )
If there is no page fault then we can get it directly from memory but if there is page fault then we have to service the page fault looking if modified or not then we have to eventually access it from the memory.
When there is no page fault means that page is present in the main memory itself and we don't have to do anything, but if there is a page fault then it means first we are accessing the main memory to know that the page is not present there and then we have to bring that page from the disk which the page fault service time. Correct me if I am wrong here.
@jhaanuj2108 you are right,formula will be,EMAT =P*( Page fault service time + Memory access time ) + (1-P)( Memory access time ). page fault service time is much much greater than memory access time generally page fault service time is considered in milliseconds and memory access time considered in nanosecond ,in this case we can ignore Memory access time in comparison to Page fault service time. in Galvin ,memory access time is not considered because of above reason but in above question both are comparable so we can't ignore that.page fault service time does not include memory access time .There are three major components of the page fault service time-(1) service the page fault interrupt (2) read in the page (3) restart the process (according to Galvin).after it we have to access memory again for getting required page.
Are u sure?
@Arjun sir, please clear the confusion
which one is the correct formula
P*( Page fault service time ) + (1-P)( Memory access time )