1. The document discusses two machines that are identical except for their CPUs. One machine has a CISC CPU while the other has a RISC CPU. 2. The average program for the CISC machine has 1/5 the number of instructions as the same program on the RISC machine. 3. For the execution time to be the same on both machines, the relationship between their average cycles per instruction and time per cycle must be that the CISC CPU's cycles per instruction is 5 times greater than the RISC CPU's, while the RISC CPU's time per cycle is 5 times greater than the CISC CPU's.

1. 1. Consider the fundamental practitioner equation that allows for an approximate estimate of
computer performance:
TimeCP U = seconds /complete program execution = instructions /program × average cycles/
instruction × average seconds /cycle
1.1 Assume you need to decide between two machines that are otherwise identical except for the
actual CPU – that is, the machines have the same RAM, internal bus, I/O systems, and the like
– only the CPUs change but that the throughput of the various other components balance with
either CPU. One CPU is a CISC machine such an X86-64, and the other is a RISC machine such
as a MIPS . The average executable program of the CISC machine has 1/5 the instructions used
for the same program on the RISC machine (that is, the same program source code when
compiled by equivalent compilers on each of the two platforms is used to 1 produce the
executable program for each platform). What must be the relationship between the average
number of cycles per instruction and the average time per cycle for the two machines to take the
same T imeCP U ?
Please show all work and reasoning.
Solution
CPU execution time :The CPU Equation
CPU time = seconds/program=(instructions/program)*(cycles/instruction)*(seconds/cycle)
T=I*CPI*C
Where ,
T= executed time per program in seconds,
I=number of instructions executed
CPI=average CPI for program
C=CPU clock cycle

2. Now, assuming 2 machines have the same RAM,internal bus,I/O systems and so but except the
CPU.
Machine A average executable program has 1/5 the instructions used for the same program on
machine B.
That is ,let I(number of instructions executed ) for machine a is x,and for machine B this will be
5x.
When T ,executed time per program is same for both the machines T1=T2=T.
T1=I1*CPI1*C1=T and T2=I2*CPI2*C2=T
Here ,
I1*CPI1*C1=I2*CPI2*C2
x*CPI1*C1=5x*CPI2*C2
CPI1/CPI2=5C2/C1