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21 for read as 32 for write.total  53
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For read, memory references:   168 (0.9(0) + 0.1(1) )  = 16.8

For write, memory refernces :   32 (0.9(1) + 0.1(1) ) = 32

Total memory references, interms of bytes = (16.8 + 32)*128 = 6246.4 ~ 6246

As one line has 128 Bytes

Since write through is used , there is no need to consider dirty bits.

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There is no use of dirty bit.

168 read out of which 10% needs memory access. 17 memory accesses for read.

32 write instructions. in case of cache miss we need 4 memory access. Because cache is write through we need 32 more memory acess for writing.

Total = 17 + 4 + 32 = 53

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