# CMIS 310 Amdahl

## Basic law

• s = final speed up
• f = fraction of time spent in part sped up
• k = amount of speed up
• f' = 1-f
• k' = 1 - k

$s = \frac\left\{1\right\}\left\{f\text{'} +f / k\right\}$

$Example:$

• f = 70% = 0.7
• k = 1.5 = 50% faster
• f' = 0.3

$s = \frac\left\{1\right\}\left\{0.3 +\frac\left\{0.7\right\}\left\{1.5\right\}\right\} = 1 + 7 / 23 \cong 1.304347826086957...$

## Percent

After some algebra, we find the following:

• P = cost per fraction of improvement
• if improvement ratio is 1.30, P = 0.30 = 30%
• M = cost of improvement
• so cost per % is M/30

$100 * P = \frac\left\{M\right\}\left\{s - 1\right\} = M * \frac\left\{f\text{'}k + f\right\}\left\{fk\text{'}\right\}$

• So, if M = \$10,000, and s, f, and k as above

$P = \frac\left\{\10,000\right\}\left\{100\right\} * \frac\left\{23\right\}\left\{7\right\} = \328.571428571428571...$

## Ratio

To compute the relative percentages, again after some algebra, we get:

$\frac\left\{P1\right\}\left\{P2\right\} = \frac\left\{M1\right\}\left\{M2\right\} * \frac\left\{k2\text{'}\right\}\left\{k1\text{'}\right\} * \frac\left\{\frac\left\{f\text{'}\right\}\left\{f\right\}k1 + 1\right\}\left\{\frac\left\{f\right\}\left\{f\text{'}\right\}k2 + 1\right\}$