techtud

Question

In a two level memory hierarchy, the access time of cache memory is 12 ns and the access time of the main memory is 1500 ns. The hit ratio is 0.98, the average access time of two-level memory system is

(A) 44.05 ns
(B) 41.76 ns
(C) 46.94 ns
(D) 40.25 ns

Comments

41.76 answer

Answer 1

In the case of simultaneous memory access organization, Average access time( T_avg) =H₁T₁+(1-H₁)H₂T_{2 ,}in the case of two level memory organization H₂ =1 (there will be 100% hit)

     so ,Tavg =.98 *12 +(1-.98)*1*1500  ns =41.76 ns

In the case of hierarchical memory access organizationn _,Average access time( T_avg) =H₁T₁+(1-H₁)H₂(T₁+T₂),in the case of two level memory organization H₂ =1 (there will be 100% hit)

     so,    T_avg =.98 *12 +(1-.98)*1*(12+1500 ) ns =42 ns

so, multi-level memory organization can be either simultaneous memory access organization or hierarchical memory access organizationn,it depends on information given in the question but if not clear from question ifself then solve for both cases and then check the right answer from options(sometimes this strategy needed).

Answer 2

T_{AVG= hit ratio * cache access time +(1-hit ratio)*main memory accesstime
          =0.98*12+0.02*1500
          =11.76+30}

         _{=41.76 nsec}

Comments

sir?

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it is already mentioned in question

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yes sir wt sudsho said that is true .as i also solved previous year question using simultaneously .access  method if nothing given in the question .??