# Comp.Architecture-3

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My work: $1+0.1*5+0.05*50=4ns$
Now please give me reasoning about :  missing in $L1$ i will access $L2$ and i did that now When i am missing in $L2$ isn`t this obvious that i have actually missed in $L1$ or should i mention it by $0.05*0.1*50$
Thanks!

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is answer is 1.750 nano sec??
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you can see this...

they said that its 2 level cache...

means one after another....

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Work done by me :for write it will always be $100ns$ Now for read :$20+0.2*100=40$ Total =$0.7*40+0.3*100=58$