Respuesta :
Answer:
The answer to this question can be described as follows:
Explanation:
Given data:
Performance of the CPU:
The Fastest Factor Fraction of Work:
[tex]f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65[/tex]
Current Feature Speedup:
[tex]K_1=[/tex] 1.5
CPU upgrade=6000
Disk activity:
The quickest part is the proportion of the work performed:
Current Feature Speedup:
[tex]k_2=3[/tex]
Disk upgrade=8000
System speedup formula:
[tex]s=\frac{1}{(1-f)+(\frac{f}{k})}[/tex]
Finding the CPU activity and disk activity by above formula:
CPU activity:
[tex]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) \\[/tex]
Disk activity:
[tex]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)[/tex]
CPU:
Formula for CPU upgrade:
[tex]= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\$ 6,000}{1.276}\\\\= 4702.19....(3)[/tex]
DISK:
Formula for DISK upgrade:
[tex]=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\$ 8000}{-0.5 \% }\\\\= - 16000....(4)[/tex]
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
