Population 2055 AD by four methods

The last census is 2011, so for 2055, $n = (2055-2011)/10 = 4.4$ decades. Four methods are used.

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{57{,}000-32{,}000}{4} = 6{,}250.0$

$$P_{2055} = P_n + n\bar x = 57{,}000 + 4.4\times 6{,}250.0 = 84{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{57{,}000}{32{,}000}\right)^{1/4}-1 = 15.526\%$

$$P_{2055} = P_n(1+r)^n = 57{,}000\times(1+0.15526)^{4.4} = 107{,}565$$

Incremental increase method

Increase per decade: 4,000, 8,000, 8,000, 5,000; incremental increase: 4,000, 0, -3,000. $\bar x = 6{,}250.00$, $\bar y = 333.33$

$$P_{2055} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 57{,}000 + 4.4(6{,}250.00) + \frac{4.4(5.4)}{2}(333.33) = 88{,}460$$

Decreasing rate of growth method

Average decrease in rate $= 0.962\%$ per decade.

Population at 2055 is obtained by linear interpolation between the decade values: $P_{2055} = 76{,}741$.

Summary

Design population and justification

The census increases are 4000, 8000, 8000 and 5000 per decade, so the growth is irregular and slowing in the last decade. The geometrical method gives an extreme value (it assumes the growth rate stays constant) and the decreasing rate method gives the lowest value. A mean of the four methods balances them, so the mean is adopted:

$$P_{2055} = \frac{84{,}500+107{,}565+88{,}460+76{,}741}{4} = 89{,}317$$

Daily water requirement

For a rural area, assume 45 lpcd and 15% losses and wastage.

$$Q = 89{,}317\times45\times1.15 = 4{,}622{,}130\ \text{l/day} = 4{,}622\ \mathrm{m^3/day}$$

Answer: Population in 2055 = 89,317 (adopted mean); daily water requirement = 4,622 m$^3$/day (53.5 l/s).