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GATE 1993 EAT = ?

GATE 1993 EAT = ?

5Comments
TarGate @tar_gate
4 Jan 2015 03:06 pm

I am getting around 1.1 X 10^-5 sec.

Need someone to concur !

Thanks :)

Arul @innovwelt
4 Jan 2015 08:59 pm

Is this the formula you have used? *all calculations are in microseconds

EAT = hit_ratio * access_time + (1 - hit_ratio) (penalty)

M3 EAT = (1* 100) = 100 us

M2 EAT = 0.998 * (10) + 0.002 * (10 + 100 + 100000) = 210.18 us

M1 EAT = 0.99*(1)+0.01*(1+210.18+1000) = 13.102 us

TarGate @tar_gate
4 Jan 2015 10:02 pm

Apparently you took a wrong value for M2 hit :) It is 0.00998

All calculations are in seconds

M3 EAT = 0.00002 [ 10^-4 (To access) ] + 0.1 (To swap with upper level)  = 0.1 X 10^-9

M2 EAT = [ 0.00998 ( 10^-5 ) + 0.9902 (10^-5 + M3 EAT) ]+ 0.001(To swap with upper level) = 1.01 X 10-^3

M1 EAT = 0.99 X 10^-6 + 0.01 (10^-6 + M2 EAT) = 1.11 X 10^-5 seconds

Arul @innovwelt
4 Jan 2015 10:09 pm

well, I took the correct value , I guess :)

during the miss of M1, M2 hit is 99.8% & M2 miss is 0.2%.

during the miss of M2, M3 hit is 100%

 

given access percentage is as a whole (M1 + M2 + M3).

but I think, We have to take the hit percentage, only with respect to the access coming to M2, not based on the whole range of access (M1, M2).

what do you think?

TarGate @tar_gate
5 Jan 2015 12:01 am

Well, I am convinced :) Thanks

But calculations will be as below I suppose

I am co-relating this question with concept of collectively exhaustive events. The union of all the probabilities is 1. And it may be the case that the intersection of two events be non-empty.

Now I think it should be simply = 0.99 X 10^-6 + 0.00998 ( 10^-6 + 10^-5 + 0.001) + 0.00002 (10^-6 + 10^-5 + 10^-4 + 0.1) = 13.08 us

Either of the event can happen.

More about exhaustive events - http://statistics.about.com/od/ProbHelpandTutorials/a/What-Are-Exhaustiv...

Let me know what you think.