Suppose a small community\'s water demand in the future 10 years is estimated to

Question

Suppose a small community's water demand in the future 10 years is estimated to be a random variable following a normal distribution with mean 200 million litres per year (MLPY) and standard deviation 30 MLPY. On the other hand, the supply system has a random capacity due to many factors such as equipment breakdown, watermain breaks and so on. The supply capacity is estimated to be also a normal random variable with a coefficient of variation 0.10. In order to ensure a 99% reliability of supply, what is the mean capacity the system should be designed to?

Explanation / Answer

Let mean capacity of system be m

So, std/mean= coefficient of variation

std/m = 0.1

std = 0.1m

So, capacity is normal with mean m and standard deviation = 0.1m

We want 99% of the time capacity should be more than the demand

So, z = (Cap - demand)/sqrt(var(cap) + var(demand)) = (m-200)/sqrt(900+0.01*m^2)

Pr(z>0) = 0.99

So, (m-200)/sqrt(900+0.01*m^2) = 2.33

(m-200)^2 = 2.33*2.33*(900 + 0.01*m^2)

m^2 - 400m + 40000 = 0.05429m^2 + 4886.01

0.946m^2 - 400m + 35114 = 0

Solving, m = 298.47

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