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postnew [5]
3 years ago
14

Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent server

s in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal?
Computers and Technology
1 answer:
sattari [20]3 years ago
6 0

Answer:

The smallest number of servers required is 3 servers

<em />

Explanation:

Given

Reliability = 99.98\%

Individual Servers = 95\%

Required

Minimum number of servers needed

Let p represent the probability that a server is reliable and the probability that it wont be reliable be represented with q

Such that

p = 95\%\\

It should be noted that probabilities always add up to 1;

So,

p + q = 1'

Subtract p from both sides

p - p + q = 1 - p

q = 1 - p

Substitute p = 95\%\\

q = 1 - 95\%

Convert % to fraction

q = 1 - \frac{95}{100}

Convert fraction to decimal

q = 1 - 0.95

q = 0.05

------------------------------------------------------------------------------------------------

<em>To get an expression for one server</em>

The probabilities of 1 servers having 99.98% reliability is as follows;

p = 99.98\%

Recall that probabilities always add up to 1;

So,

p + q = 1

Subtract q from both sides

p + q - q = 1 - q

p = 1 - q

So,

p = 1 - q = 99.98\%

1 - q = 99.98\%

Let the number of servers be represented with n

The above expression becomes

1 - q^n = 99.98\%

Convert percent to fraction

1 - q^n = \frac{9998}{10000}

Convert fraction to decimal

1 - q^n = 0.9998

Add q^n to both sides

1 - q^n + q^n= 0.9998 + q^n

1 = 0.9998 + q^n

Subtract 0.9998 from both sides

1 - 0.9998 = 0.9998 - 0.9998 + q^n

1 - 0.9998 = q^n

0.0002 = q^n

Recall that q = 0.05

So, the expression becomes

0.0002 = 0.05^n

Take Log of both sides

Log(0.0002) = Log(0.05^n)

From laws of logarithm Loga^b = bLoga

So,

Log(0.0002) = Log(0.05^n) becomes

Log(0.0002) = nLog(0.05)

Divide both sides by Log0.05

\frac{Log(0.0002)}{Log(0.05)} = \frac{nLog(0.05)}{Log(0.05)}

\frac{Log(0.0002)}{Log(0.05)} = n

n = \frac{Log(0.0002)}{Log(0.05)}

n = \frac{-3.69897000434}{-1.30102999566}

n = 2.84310893421

n = 3 (Approximated)

<em>Hence, the smallest number of servers required is 3 servers</em>

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