$1.$ Consider the following data about two computers X & Y running different compilers.

Instruction Type CPI $($Cycles Per Instruction$)$

Instruction Count $($in billions$)$

A $29,15\%$

B $515,25\%$

C $112,20\%$

D $424,40\%$

Instruction Type CPI

Instruction Count $($in billions$)$

A $516,20\%$

B $124,30\%$

C $312,15\%$

D $128,35\%$

The clock rate of Computer X & Y is $2\mathrm{.}5$ GHz & $3$ GHz respectively. Calculate

a$)$ average CPI of both machines.

b$)$ Which machine is faster based upon

i$)$ Execution Time

ii$)$ MIP rate.

$2.$ Suppose a program takes $25$ seconds to run on a particular hard ware. Now suppose on a new

hardware, $[$X$]\%$ of instructions can be run parallel by a factor of $9$ times. Calculate the

overall speed up that has been achieved. Also what is the new execution time.

Note: For value of $[$X$],$ use last two digits of your student number.

Note: For all questions, submit your work with detailed calculations.

Formulas:

$1.$ Average CPI Calculation:

Average CPI$=($CPI$\_$i $\backslash$times Fraction of Instructions$\_$i$).$

$2.$ Execution Time Calculation:

Execution Time$=($Instruction Count$)\backslash$times $($CPI$)\backslash$times $1/$Clock Rate

$3.$ MIPS $($Million Instructions Per Second$)$ Calculation:

MIPS$=$Clock Rate $\backslash$times $106$

$/$ CPI Or MIPS$=$Execution Time $\backslash$times $106$

$/$ Instruction Count

$4.$ speedup $($S$)$

$$ S$=1/((1$P$)+$ P $/$ Sp$)$

$$ P is the proportion of the program that can be parallelized. P $=$ X$/100$;

$$ Sp is the speedup factor of the parallel portion.