Population 2090 by three methods, 110 lpcd

Census interval 10 years (B.S.), last census 2070; for 2090, $n = 2$ decades.

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{70{,}000-28{,}000}{4} = 10{,}500.0$

$$P_{2090} = P_n + n\bar x = 70{,}000 + 2\times 10{,}500.0 = 91{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{70{,}000}{28{,}000}\right)^{1/4}-1 = 25.743\%$

$$P_{2090} = P_n(1+r)^n = 70{,}000\times(1+0.25743)^{2} = 110{,}680$$

Decreasing rate of growth method

Average decrease in rate $= 5.008\%$ per decade.

$P_{2090} = 89{,}620$

Summary

Water demand in 2090

The decreasing rate of growth method follows the falling trend of the growth rate and is adopted for design. Per capita demand 110 lpcd.

$$Q = 89{,}620\times110 = 9{,}858{,}187\ \mathrm{l/day} = 9{,}858.2\ \mathrm{m^3/day}$$

Answer: Population in 2090: arithmetical 91,000, geometrical 110,680, decreasing rate 89,620. Total water demand = 9,858 m$^3$/day (114.1 l/s) on the adopted population.