Chapter 7, Problem 8E

a.

Program Plan Intro

Given:

Fraction of work performed by CPU (f) = $60\%=\frac{60}{100}=0.6$

Fraction of work performed by DISK (f) = $40\%=\frac{40}{100}=0.4$

Speedup of CPU (k) = 1.4

Speedup of the DISK (k) = 2.5

To Find: Overall speedup of the CPU and DISK

Solution:

According to Amdahl’s law; $\text{Overall speedup }\left(\text{S}\right)\text{=}\frac{\text{1}}{\text{(1-f)+(}\frac{\text{f}}{\text{k}}\text{)}}$ --- (1)

Substitute the value for “f” and “k” in the above equation. So, the equation (1) becomes,

Overall speedup for CPU:

$\begin{array}{l}{\text{S}}_{\text{CPU}}\text{=}\frac{\text{1}}{\text{(1-0}\text{.6)+(}\frac{\text{0}\text{.6}}{\text{1}\text{.4}}\text{)}}\\ {\text{S}}_{\text{CPU}}\text{=}\frac{\text{1}}{\text{0}\text{.4+0}\text{.43}}\\ {\text{S}}_{\text{CPU}}\text{=}\frac{\text{1}}{0.83}\end{array}$

$\boxed{{\text{S}}_{\text{CPU}}\text{=1}\text{.2}}$

Convert 1.2 to percentage:

=>$1.2\times 100\%=120\%$. Subtract 100% from 120% to get the overall speedup percentage of the new system.

Overall speedup percentage of new system = 120% - 100% = 20%

The overall speedup of the CPU is 20%

Overall speedup for DISK:

$\begin{array}{l}{\text{S}}_{\text{DISK}}\text{=}\frac{\text{1}}{\text{(1-0}\text{.4)+(}\frac{\text{0}\text{.4}}{\text{2}\text{.5}}\text{)}}\\ {\text{S}}_{\text{DISK}}\text{=}\frac{\text{1}}{\text{0}\text{.6+0}\text{.16}}\\ {\text{S}}_{\text{DISK}}\text{=}\frac{\text{1}}{0.76}\end{array}$

$\boxed{{\text{S}}_{\text{DISK}}\text{=1}\text{.31}}$

Convert 1.31 to percentage:

=>$1.31\times 100\%=131\%$. Subtract 100% from 131% to get the overall speedup percentage of the new system.

Overall speedup percentage of new system = 131% - 100% = 31%

The overall speedup of the DISK is 31%

b.

Explanation of Solution

If one does not care about money, then, one would go for the upgrade of disk activity because:

From the above part (a), it is clear that the speed of the DISK activity ${\text{S}}_{\text{DISK}}\text{=31\%}$ is more when compared with CPU

c.

Explanation of Solution

Breakeven points for the upgrade:

From the part (a):

Overall speedup for CPU is: 20%

Money needed to upgrade CPU is: $250

Overall speedup for DISK is: 31%

Money needed to upgrade DISK is: $258.06

The breakeven points for the upgrades is:

$\begin{array}{l}\text{Breakeven point = Money to upgrade CPU × speedup of DISK}\\ \text{Breakeven point = }\$250\text{ × }31\%\\ \boxed{\text{Breakeven point = }7750}\end{array}$

All the things are equal when one pays $7750 for the DISK upgrade