Population 2070 by three methods, demand 2080

The census interval is 10 years (B.S.) and the last census is 2050. For 2070, $n = 2$ decades.

Population in 2070

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{60{,}000-25{,}000}{4} = 8{,}750.0$

$$P_{2070} = P_n + n\bar x = 60{,}000 + 2\times 8{,}750.0 = 77{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{60{,}000}{25{,}000}\right)^{1/4}-1 = 24.467\%$

$$P_{2070} = P_n(1+r)^n = 60{,}000\times(1+0.24467)^{2} = 92{,}952$$

Decreasing rate of growth method

Average decrease in rate $= 10.303\%$ per decade.

$P_{2070} = 52{,}447$

Because the growth rate falls sharply in the last decade, this method predicts a falling population, which should be treated with caution for a growing city.

Population in 2080 (n = 3)

Water demand in 2080

The per capita demand is not given, so 110 lpcd is assumed for a city. The last increment (5,000) is much smaller than the earlier ones, so growth is slowing; the methods give very different results, and the average of the three is adopted as the design population.

$$P_{2080} = \frac{86{,}250+115{,}694+41{,}004}{3} = 80{,}983$$ $$Q = 80{,}983\times110 = 8{,}908{,}094\ \mathrm{l/day} = 8{,}908\ \mathrm{m^3/day}$$

Answer: Population in 2070: arithmetical 77,500, geometrical 92,952, decreasing rate 52,447. Water demand in 2080 = 8,908 m$^3$/day (103.1 l/s).