A CPU has a 2 level cache with hit ratio of L1 and L2 cache as 90% and 95% respecrtively. Also the access time of L1 and L2 is 1ns and 5ns respectivelly.If access time of main memory is 50ns .then EAT(Effective Access Time ) is ....
My Solution : Since nothing is mentioned in the questin about the condition of the cache I took the deafult case
0.90*1 + 0.10*0.95(5) + 0.10*0.05*50 = 1.625
But the Solution given has assumed it to be Hierarchail Access where we add Cache time in Miss part also Solution given is : 0.90*1 + 0.10*0.95*(5+1) + 0.10*0.05*(50+5+1) =1.75
So My Doubt is If nothing is mentioned in the question do we always have to take Hierarchial case or not ????