3. a. A one way set associative cache has a total size of 64 bytes with 2 bytes/

Question

3. a. A one way set associative cache has a total size of 64 bytes with 2 bytes/block. Find i. the mapping of main memory address 2EF4 H into cache. (3 pts.) the tag (2 pts) b. A certain 12 stage MIPS processor with a 2 nsec clock has a split cache for instruction and data. Assuming a base (ideal) time per instruction of2 nsec, hit time of 2 cycles, miss penalty of 100 cycles, a cache size of 1024 blocks with 16 words/block and miss rates for instruction and data 0.5% and 10% respectively a. What is the total size of the cache in MB? (3 pts.) b. Find the actual CPI for the following set of instructions add St1, Sa0, St1 lw St0, 4(St1) sw $t2, 0(St1) or $t3, Stl, $t2 beq $to, St1, Exit add Stl, Sa0, Stl lw $t0, 4(St1) sw $12, 0(St1) or $t3, St1, $12 beq St0, St1, Exit (5 pts.) a. In Virtual memory LRU approximation with hardware support is used to replace a miss. Explain. (2 pts.) Calculate the transfer rate of a certain storage disk with the following specifications b. 1024 byte sector 250 revolutions per second 1.25 msec actual seek time 0.25 msec controller overhead and 2.25 msec average read time. (3 pts.)

Explanation / Answer

a)

In a computer operating system that uses paging for virtual memory management, page replacement algorithms decide which memory pages to page out (swap out, write to disk) when a page of memory needs to be allocated. Paging happens when a page fault occurs and a free page cannot be used to satisfy the allocation, either because there are none, or because the number of free pages is lower than some threshold.

The prediction behind LRU, the Least Recently Used, and algorithm is that the page that has not been used in the longest time is the one that will not be used again in the near future. LRU is considered a good replacement policy, and is often used. Unfortunately full implementation of LRU requires hardware support, and few systems provide the full hardware support necessary.

           

b)

Rotational Delay = [(60 seconds/7500 rpm) * (1000ms/second)]/2 = 4ms + 8 ms seek time

                            = 12ms.

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