Please answer ALL the parts 8 . Suppose you are the head of a computer architect
ID: 3816686 • Letter: P
Question
Please answer ALL the parts
8. Suppose you are the head of a computer architecture group that has produced a computer system whose main memory has an access time of 250ns. Careful statistical sampling has indicated that each instruction requires an average of 2 memory accesses and has an average execution time (exclusive of memory accesses) of 150ns. Three members of your group each present proposals to speed up this system and their proposals are given below. Rank these proposals assuming each costs the same. Justify your ranking.
(a) Use a main memory that is twice as fast.
(b) Add enough CPU registers to reduce memory accesses to an average of 1.5 memory accesses per instruction. Assume tm = 250ns.
(c) Add a cache to the system with tc = 50ns and a hit ratio h = 0.9. Assume that upon a cache miss, a word is first transferred from main memory to cache and then from cache to the CPU. Assume tm = 250 ns and an average of 2 memory accesses per instruction.
Explanation / Answer
Given Main Memory Access Time is = 250ns
Total Execution Time of an instruction is = 2 memory accesses + 150ns
= 500 + 150
Total Execution Time of an instruction is = 650ns
(a) Use a main memory that is twice as fast.
Now Main Memory Access Time is = 125ns
Total Execution Time of an instruction is = 2 memory accesses + 150ns
= 2*125 + 150
= 250 + 150
Total Execution Time of an instruction is = 400ns
(b) Add enough CPU registers to reduce memory accesses to an average of
1.5 memory accesses per instruction. Assume tm = 250ns.
Now Main Memory Access Time is = 250ns
Total Execution Time of an instruction is = 1.5 memory accesses + 150ns
= 1.5*250 + 150
= 375 + 150
Total Execution Time of an instruction is = 525.
c) Add a cache to the system with tc = 50ns and a hit ratio h = 0.9. Assume that upon
a cache miss, a word is first transferred from main memory to cache and then from
cache to the CPU. Assume tm = 250 ns and an average of 2 memory accesses per instruction.
Now Main Memory Access Time is = 250ns
Average Memory Access time is = h * tc + (1-h) * (tc + tm)
= 0.9 * 50 + 0.1 * (50 + 250)
= 45 + 30
Hence Average Memory Access time is = 75
Total Execution Time of an instruction is = 2 memory accesses + 150ns
= 150 + 150
Total Execution Time of an instruction is = 300ns
Hence According to the Total Execution Time,
c will stood 1st with Total Execution Time of 300ns
a will stood 2nd with Total Execution Time of 400ns
b will stood 3rd with Total Execution Time of 525ns
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