Question

Computer A executes the MIPS ISA and computer B executes the x86 ISA. On average, programs execute 1.5 times as many MIPS instructions as x86 instructions. Computer A has an average CPI of 1.5 and computer B an average CPI of 3. If computer B runs at a 3GHz clock frequency, what speed does computer A have to run at to be at least as fast as computer B?

Answer 1

B runs at 3 GHz clock frequency means, it runs 3* 10⁹ clock cycles per second

B has CPI(clock per instruction) 3.

hence number of instruction exectued per second by B is 3*10⁹/3 = 10⁹

Given, A executes 1.5 times as many instructions as B

hence A should execute 1.5*10⁹ instructions per second.

Hence, number of clock cycles required in one second is 1.5*1.5*10⁹ = 2.25 *10⁹

Hence, Clock frequency of A should be 2.25 GHz