Performance and execution time have an inverse relationship.
Consider the following relation:

Performance = \frac{1}{Execution Time}

Thus, if the execution time of a program decreases (i.e. the program gets faster), then the performance increases.

The following relations can be used to determine the performance of a program:

$E T = C C \cdot C T = \frac{C C}{C R}$
$C C = E T \cdot C R = C P I \cdot I C$

$E T = Execution Time$
$C C = Clock Tycles$
$C T = Clock cycle Time$
$C P I = average Clock cycles Per Instruciton$
$I C = Instruction Count$

Examples

The following examples are unchecked and may be wrong.

(a) If computer A runs a program in 10 seconds and computer B runs the same program in 15 seconds, how much faster is A than B?

$P_{A} = \frac{1}{E T _{A}} = 0.1 P_{B} = \frac{1}{E T _{B}} = 0.067 \frac{P _{A}}{P _{B}} = \frac{0.1}{0.067} = 1.5 Thus, computer A is 1.5 times faster than computer B.$

(b) Our favorite program runs in 10 seconds on computer A, which has a 2 GHz clock. We are trying to help a compute rdesigned build a computer, B, which will run this program in 6 seconds. THe designed has determined that if computer B has 1.2 times as many clock cycles as computer A. > What clock rate shoud we tell the deigner to target?

$E T_{A} = 10 s C R_{A} = 2 G H z C C_{A} = ? E T_{B} = 6 s C R_{B} = ? C C_{B} = 1.2 C C_{A} E T_{A} = \frac{C C _{A}}{C R _{A}} ⟹ C C_{A} = E T_{A} \cdot C R_{A} = 10 \cdot 2 \cdot 1 0^{9} = 20 \cdot 1 0^{9} cy c l es C C_{B} = 1.2 \cdot C C_{A} = 1.2 \cdot 20 \cdot 1 0^{9} = 24 \cdot 1 0^{9} cy c l es E T_{B} = \frac{C C _{B}}{C R _{B}} ⟹ C R_{B} = \frac{C C _{B}}{E T _{B}} = \frac{24 \cdot 1 0 ^{9}}{6} = 4 G H z Thus, we are expecting a 40% im p r o v e m e n t .$

(c)

$C T_{A} = 250 p s C P I_{A} = 2.0 E T_{A} = C C_{A} \cdot C T_{A} = I C_{A} \cdot C P I_{A} \cdot C T_{A} = \frac{I C _{A} \cdot C P I _{A}}{C R _{A}} = 500 p s C T_{B} = 500 p s S P I_{B} = 1.2 E T_{B} = I C_{B} \cdot C P I_{B} \cdot C T_{B} = 1.2 \cdot 500 = 600 p s Thus, A is faster.$

(d)

Program 1 has 5 instructions
Program 2 has 6 instructions
$E T = \frac{I C \cdot C P I}{C R} We can ignore CR since they both have the same C C_{1} = (2 \cdot 1) + (1 \cdot 2) + (2 \cdot 3) = 2 + 2 + 6 = 10 cy c l es C C_{2} = (4 \cdot 1) + (1 \cdot 2) + (1 \cdot 3) = 4 + 2 + 3 = 9 cy c l es$

(e)
a.

P = \frac{1}{E T} E T_{P 1} = \frac{C P I _{P 1}}{C R _{P 1}} = \frac{1.5}{3 \cdot 1 0 ^{9}} = \frac{1}{2 \cdot 1 0 ^{9}} = 0.5 nan oseco n d s

Explorer

Calculating Performance

Examples