#effective memory access time

Question

Comments

2620.9395 nano sec. ??

Answer 1

EAT= TLB hit*(TLB time+MM time)+TLB miss*(Page fault rate*Service time+No page fault *MM time)

EAT=0.99*(20+100)+0.01*(0.05*5*$10^{6}$+0.95*100)

       =118.8+2500.95

      =2619.75 nsec. 

If page fault occurs OS generates a system call to swap in the required process until that current execution stops.After that execution of a process continues.

Comments

i was also doing the same way ...but solution given is beyond my understanding

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Remember one thing when page is not found in TLB then we are not consider TLB lookup time.

when page is found in TLB then only we access the page it as TLB lookup time.

The given solution was wrong.

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so the one whose answer is 2619.75ns is right approach???

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Yes,