Skip to content
GATETECH
Official PYQ · 2010Q48Memory Hierarchy: Cache, Main Memory and Secondary StorageMCQ1 mark

GATE CS 2010 Q48 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

Explain with AI

A computer system has an L1L1 cache, an L2L2 cache, and a main memory unit connected as shown below. The block size in L1L1 cache is 44 words. The block size in L2L2 cache is 1616 words. The memory access times are 22 nanoseconds, 2020 nanoseconds and 200200 nanoseconds for L1L1 cache, L2L2 cache and the main memory unit respectively.

 

When there is a miss in L1L1 cache and a hit in L2L2 cache, a block is transferred from L2L2 cache to L1L1 cache. What is the time taken for this transfer?

Choose one option
View worked solution

Answer: C

To find the total time taken for the transfer when there is a miss in the L1L1 cache and a hit in the L2L2 cache, we follow the hierarchical access path:

  1. L1 Cache Search: The CPU first searches the L1L1 cache. Since it is a miss, this step takes the L1L1 access time: TimeL1=2 ns\text{Time}_{L1} = 2\text{ ns}
  2. L2 Cache Access: Upon a miss in L1L1, the request is forwarded to the L2L2 cache. Since it is a hit in L2L2, the block is accessed and transferred to L1L1. This takes the L2L2 access time: TimeL2=20 ns\text{Time}_{L2} = 20\text{ ns}

Therefore, the total time taken for this transfer is:

Total Time=TimeL1+TimeL2=2 ns+20 ns=22 ns\text{Total Time} = \text{Time}_{L1} + \text{Time}_{L2} = 2\text{ ns} + 20\text{ ns} = 22\text{ ns}

Hence, the correct option is C.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Memory Hierarchy: Cache, Main Memory and Secondary Storage questions