Suppose we wanted Computer B to have the same CPU time as Computer A. 1. What wo

Question

Suppose we wanted Computer B to have the same CPU time as Computer A.
1. What would the cycle time for Computer B need to be (to nearest integer) for the cycle time to be the same as Computer A?
2. What would the CPI for Computer B need to be for the cycle time to be the same as Computer A? CPI Example " Computer A: Cycle Time-250ps, CPI = 2.0 , Computer B: Cycle Time = 500ps, CPI = 1.2 Same ISA .Which is faster, and by how much? CPU Time A-Instruction Count× CPIA × Cycle TimeA =lx2.0x250ps-I×500ps CPU Time B = Instruction Count × CPIB x Cycle TimeB = I × 1·2x 500ps-I × 600ps / Ais faster.. CPU Time Ix600ps CPU Time A 1x500ps ...by this much 39

Explanation / Answer

1.

we need to find the cycle time of the computer B

Given that cpu times of computr A and B are same
we know that
  
   (CPUTime)ofmachine = (Instruction count)of machine * (CPI)of machine * (Clock cycle Time)of machine
Here

(CPUTime)A = (Instruction count)A * (CPI)A * (Clock cycle Time)A
(CPUTime)B = (Instruction count)B * (CPI)B * (Clock cycle Time)B

we know that cpu times of two computers are same so

(Instruction count)A * (CPI)A * (Clock cycle Time)A=(Instruction count)B * (CPI)B * (Clock cycle Time)B
we have same ISA so
   (CPI)A* (Clock cycle Time)A=(CPI)B* (Clock cycle Time)B
   2* (Clock cycle Time)A=1.2* (Clock cycle Time)B
Let us assume (Clock cycle Time)A is 300ps then
   2*300=1.2* (Clock cycle Time)B
   (Clock cycle Time)B=500ps

2.

we need to find the CPI of the computer B

Given that cpu times of computr A and B are same
we know that
  
   (CPUTime)ofmachine = (Instruction count)of machine * (CPI)of machine * (Clock cycle Time)of machine
Here

(CPUTime)A = (Instruction count)A * (CPI)A * (Clock cycle Time)A
  
(CPUTime)B = (Instruction count)B * (CPI)B * (Clock cycle Time)B
   
we know that cpu times of two computers are same so

(Instruction count)A * (CPI)A * (Clock cycle Time)A=(Instruction count)B * (CPI)B * (Clock cycle Time)B
we have same ISA and same clock cycles for both computers A and B so
       (CPI)A=(CPI)B*
  
Let us assume (CPI)A is 2.5 then (CPI)B also 2.5

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