For each of the following memory hierarchies, calculate the average memory acces

Question

For each of the following memory hierarchies, calculate the average memory access time. If you end up with a fractional number of cycles, round up there isn't much you can do (besides read/write the register file) in half a cycle! The cache takes 1 cycle to access and has a 15% miss rate, main memory takes 1 cycles to access and has an 8% miss rate, and the disk takes 30,000 cycles to access. The cache takes 3 cycles to access and has a 92% hit rate, main memory takes 450 cycles to access and has a 96% hit rate, and the disk takes 15,000 cycles to access. This problem deals with a multi-level cache like those found in modem machines. The cache levels are listed in terms of their order in the memory hierarchy-an access initially goes to the level 1 (LI) cache. If there is a miss in the LI cache, you then check the level 2 (L2) cache, then the level 3 (L3) cache, and then main memory. The LI cache takes 1 cycle to access, with a 94% hit rate. The L2 cache takes 20 cycles on each access and has a 97% hit rate. The L3 cache takes 60 cycles to access and has a 99% hit rate. Main memory takes 500 cycles to access, with an 85% hit rate, while the disk takes 45,000 cycles to access.

Explanation / Answer

a) Average Mean Access Time (Amat) = (hit time) + (miss rate) × (miss penalty)
      where the miss penalty is simply the AMAT for the next level of the memory hierarchy.
      Therefore:

        AMAT = 1 + (0.15)(AMAT of Main Memory)
       = 1 + (0.15)(150 + (0.8)(AMAT of Disk)
       = 1 + (0.15)(150 + (0.8)(30000)
                   = 1 + (0.15)(150 + 2400)
                   = 1 + (0.15)(2550) = 1+382.5 = 383.5 Cycles round up to 384 cycles

b)   AMAT = 3 + (0.08)(AMAT of Main Memory)
       = 3 + (0.08)(450 + (0.04)(AMAT of Disk)
     = 3 + (0.08)(450 + (0.04)(15000)
     = 3 + (0.08)(450 + 600)
       = 3 + (0.08)(1050) = 3 + 84 = 87 Cycles

c) 1 + (0.06)(AMAt of L2 )
1 + (0.06)(20 + 0.03(Amat of L3)
1 + (0.06)(20 + 0.03( 60 + 0.01(amat of Main Memory)
1 + (0.06)(20 + 0.03( 60 + 0.01(500 + 0.15(AMAT of DISK)
1 + (0.06)(20 + 0.03( 60 + 0.01(500 + 0.15(45000)
1 + (0.06)(20 + 0.03( 60 + 0.01(500 + 6750)
1 + (0.06)(20 + 0.03( 60 + 0.01(7250)
1 + (0.06)(20 + 0.03( 60 + 72.5)
1 + (0.06)(20 + 0.03( 60 + 73)    72.5 rounded off to 73
1 + (0.06)(20 + 0.03( 133)
1 + (0.06)(20 + 3.99)
1 + (0.06)(24)     3.99 rounded off to 4 and added to 20
1 + 1.44 = 2.44cycles   round up to 3 cycles

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