Consider the two computers a and b with the clock cycle times 100 ps and 150 ps respectively for some program. the number of cycles per instruction (cpi) for a and b are 2.0 and 1.0 respectively for the same program. which computer is faster and how much? a) a is 1.33 times faster than b b) bis 1.22 times faster than a c) a is 1.23 times faster than b d) b is 1.33 times faster than a

Clock cycle times of Computer A  =100ps

Number of cycle per instruction (CPI) = 2

Time Required Per instruction = Clock times * CPI =100 * 2 = 200 ps

clock cycle times of Computer B =150ps

Number of cycle per instruction(CPI) = 1

Time Required Per instruction = Clock times * CPI 150 * 1 = 150 ps

Computer B requires less time than Computer A per instruction

So, Computer B is Faster

Computer B is Faster 200 - 150 = 50ps Faster Per instruction