Applied Course

Question

Consider a computer system in which cache memory write hit takes 5ns and and miss takes 80ns. Cache memory read hit takes 2ns and miss takes 40ns. The cache is having 95% hit. The system received 1000 instructions out of which, 700 operand fetch operations and 300 operand write operations. The average time taken to execute above 1000 instructions is___________?

 

 

what should be the Average access time for read operation(Tread) = 2+ (1-h1)*40 = 2 + (0.05)*4   OR   h1*2+ (1-h1)*40 =0.95*2+(0.05)*40 ?

Comments

@Shaik Masthan

Please check this ..correct if wrong

Assuming cache is WRITE BACK

Tread = Hread(Trh) + (1-Hread)(Trm) = (0.95)(2) + (0.05)(40) = 1.9 + 2 = 3.9ns

Twrite = Hwrite(Twh) + (1-Hwrite)(Twm) = (0.95)(5) + (0.05)(80) = 4.75 + 4 = 8.75ns

1000 instructions ==> 700 operand fetch and 300 operand write

what is operand write ? will it take Twrite or Tread + Twrite ?? I assume  1Operand write =Twrite

Hence total Read = 700

Total write = 300

Total time for 1000 instructions = 700(3.9) + 300(8.75) =  5355ns

THIS IS NOT ASKED IN QUES

AVG TIME ASKED NOT TOTAL

hence Avg time for 1000 instructions = Total time for 1000 ins / Total number of instructions = 5355 /1000 = 5.355 ns

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operand write means, just we have to write in that memory

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Okkk..rest part is OK ??