Question

Suppose the daytime processing load consists of 65% CPU activity and 35% disk activity. Your customers are complaining that the system is slow. After doing some research, you learn that you can upgrade your disks for $8,000 to make them 3 times as fast as they are currently. You have also learned that you can upgrade your CPU to make it 1.5 times faster for $6,000. a) Which would you choose to yield the best performance improvement for the least
amount of money?
b) Which option would you choose if you don’t care about the money, but want a
faster system?
c) What is the break-even point for the upgrades? That is, what price would be
charged for the CPU(or the disk--change only one) so the results was the same cost per 1% increase in both.

Answer 1

Answer:

The answer to this question can be described as follows:

Explanation:

Given data:

Performance of the CPU:  

The Fastest Factor Fraction of Work:

$f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65$

Current Feature Speedup:

$K_1=$ 1.5

CPU upgrade=6000

Disk activity:

The quickest part is the proportion of the work performed:

Current Feature Speedup:

$k_2=3$

Disk upgrade=8000

System speedup formula:

$s=\frac{1}{(1-f)+(\frac{f}{k})}$

Finding the CPU activity and disk activity by above formula:

CPU activity:

$S_{CPU}=\frac{1}{(1-f_1)+(\frac{f_1}{k_1})} \\\\=\frac{1}{(1-0.65)+(\frac{0.65}{1.5})} \\\\=1.276 \% ...\rightarrow (1)$

Disk activity:

$S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)$

CPU:

Formula for CPU upgrade:

$= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\ 6,000}{1.276}\\\\= 4702.19....(3)$

DISK:

Formula for DISK upgrade:

$=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\ 8000}{-0.5 \% }\\\\= - 16000....(4)$

equation (3) and (4),

Thus, for the least money the CPU alternative is the best performance upgrade.

b)

From (3) and (4) result,

The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.

c)

The break-event point for the upgrades:

=4702.19 x-0.5

= -2351.095

From (2) and (3))

Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095