CSci 4203 Fall 2018 Assignment 1 Solutions :
Problem 1:
280 x10^6 instructions in total.
Total cycles =( 60*2 + 120*1 + 80*5 +20*1)*10^6 = 660*10^6
For twice as fast, total cycles = 330*10^6
A. Impossible to to reach this speed by speeding up floating point instructions.
B. Load/Store instruction should have CPI of ⅞ to reach this speed, which is a speedup of 6x.
C. This results in a reduction of execution time by 40%.
Problem 2 :
A. Processor 2
B. 30, 25, 40 Gigacycles, 20, 25, 20 Gigainstructions
C. Execution time = instructions * CPI/ ClockRate
E2 = 0.8 * E1
CPI1*1.2 / ClockRate2 = 0.8 * CPI1 / ClockRate1
ClockRate2 = ClockRate1 * 1.2/0.8 = Clockrate1 * 1.5
So clockrate increases by 50 percent.
Problem 3 :
Part A :
Processor 1 :
C = 2 x DP/(V^2 x F)
= 2 x 90/(1.25^2 x 3.6x 10^9)
= 3.2 x 10^-8 F
Processor 2 :
C = 2 x DP/(V^2 x F)
= 2 x 40 / ( 0.9^2 * 3.4 )
= 2.9048 * 10 ^ -8 F
Part B :
(Snew + Dnew) /( Sold + Dold) = 0.90
Dnew =0.5* C x Vnew ^2 x F
Sold = Vold x I
Snew = Vnew x I
Therefore:
Vnew = [Dnew / 0.5*(CxF)]^0.5
Dnew = 0.90 x (Sold + Dold) - Snew
Snew = Vnew x (Sold/Vold)
