Assume all instructions of the computer A design will execut

Assume all instructions of the computer A design will execute in 8 clock cycles and all instructions of the Computer B design will execute in 7 clock cycles. Also assume that clock period for design A is 66.25 ns while clock period for design B is 75 ns. Compute the speedup to be expected for the computer of design B.

Solution

The time to execute an instruction on each of these computers is #cycles per instruction * time per cycle. Once you have this for each computer you can compare their speeds.

For A:8*66.25=530

For B:7*75=525

Hence B is faster than A