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astra-53 [7]
3 years ago
15

What does firewall do

Computers and Technology
1 answer:
Bogdan [553]3 years ago
4 0

Answer:

help

Explanation: to make people blocked on school computer

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Suppose the daytime processing load consists of 65% CPU activity and 35% disk activity. Your customers are complaining that the
bekas [8.4K]

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

4 0
3 years ago
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