In a pipelined processor, each TLB entry consists of 32 bits of data and 36 bits

Question

In a pipelined processor, each TLB entry consists of 32 bits of data and 36 bits of tag. The page size used by the processor is 16 kB and the block size is 64 B. Each of the L1 caches consists of 32 kB of data and are 2-way set-associative; the L2 cache is 512 kB large and is 4-way set-associative; and, the L3 cache has 8 MB and is 8-way set-associative. All read/write caches employ write back and write allocate strategies. The clock frequency is 3 GHz. Accessing a block from L2 consumes 10 ns, from L3 takes 20 ns, and from main memory 30 ns.

a.Calculate the numbers of tag, index and offset bits in each cache.

b.Calculate the execution time of a program that involves completion of 3 millioninstructions,40% of which are load or store. The instruction cache has a hit rate of 99%,data cache 85%, L2 cache 70%,and L3 cache 50%.

Explanation / Answer

I have created and designed the calculation of tag, index, and execution time of a pipelined processor.I have included the comments for each part of the section and atatched the final output.

Let me explain you in simpler and step-by-step procedure:-

A)

Step-1:

The initial part is to make use of TLB which is basically helpful in caching the translation from a virtual page to a Physical page number. So, A page size must be bigger and larger than any one of the normal caching size.

The main points is to remember is the L1 data cache is consist of 32 Kbytes, 2-way set associative,  64 byte blocks,indexed and the tagged with the physical address of the given memory.

B)

Step-2:

The next step is to find the execution time of a program which consist of 3 million instructions, 40% of comprises of Load or Store, Having a hit rate of 99%, data cache as 85%,.

The L1 cache value is 70% and the L2 cache value is 50%.

Let us consider the Number of sets present in the cache, 8192 / 16 = 1024,

Next step is to calculate the number of misses = 512 ,

Following this, The calculated number of Hits = 512,

The Final step is to find the total execution time which can be obtained as follows :-

Execution Time of a program = 100 * 1024 + 10 * 1024 = 112640

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