--- title: "Population 2055 AD by four methods" subject: "Water Supply Engineering" chapter: 3 key: "semester_5/water_supply_engineering:pop-c70o" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-c70o site: BCT Engineering --- # Population 2055 AD by four methods Water Supply Engineering · Chapter 3: Quantity of Water - Asked in: 2070 Chaitra (old course) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pop-c70o - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3 ## Question Determine the population of a rural area for the year 2055 AD (any four methods). Compute the expected daily water quantity requirement for one proposed population with justification. | Year | 1971 | 1981 | 1991 | 2001 | 2011 | |---|---|---|---|---|---| | Population | 32000 | 36000 | 44000 | 52000 | 57000 | ## Short answer Four methods for 2055: arithmetic 84500, geometric 107565, incremental 88460, decreasing rate 76741; adopt mean 89317. Rural demand at 45 lpcd plus 15% losses is 4622 m3/d (53.5 l/s). ## Answer 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** | Decade | Increase | Growth rate (%) | Decrease in rate (%) | |---|---|---|---| | 1971-1981 | 4,000 | 12.50 | - | | 1981-1991 | 8,000 | 22.22 | -9.72 | | 1991-2001 | 8,000 | 18.18 | 4.04 | | 2001-2011 | 5,000 | 9.62 | 8.57 | Average decrease in rate $= 0.962\%$ per decade. | Year | Growth rate (%) | Population | |---|---|---| | 2021 | 8.65 | 61,933 | | 2031 | 7.69 | 66,697 | | 2041 | 6.73 | 71,186 | | 2051 | 5.77 | 75,293 | | 2061 | 4.81 | 78,913 | Population at 2055 is obtained by linear interpolation between the decade values: $P_{2055} = 76{,}741$. #### Summary | Method | Population (2055) | |---|---| | Arithmetical increase | 84,500 | | Geometrical increase | 107,565 | | Incremental increase | 88,460 | | Decreasing rate of growth | 76,741 | #### 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 2055 with last census 2011, what n applies? A. 44 decades B. 4.4 decades C. 4.0 decades D. 5.5 decades Answer: B. (2055-2011)/10 = 4.4. (Key point: Value of n) 2. Which formula is the incremental increase method? A. P = Pn + n x B. P = Pn (1+r)^n C. P = Pn + n x + n(n+1)/2 y D. P = Pn + n x + n y Answer: C. Adds the average increase and the average incremental increase (y) with the n(n+1)/2 factor. (Key point: Incremental increase formula) 3. Why is the mean of the four methods adopted? A. The geometric method is always correct B. Census data are exact so any method works C. The decreasing rate value is the safest highest value D. Growth is irregular and slowing; geometric is extreme high and decreasing rate lowest, so the mean balances them Answer: D. Averaging reduces the bias of the extreme methods when trends are irregular. (Key point: Why take the mean) 4. How is the daily water quantity computed? A. 89,317 x 45 lpcd x 1.15 (15% losses) B. 89,317 x 110 lpcd C. 89,317 x 45 lpcd without losses D. 89,317 x 135 x 1.15 Answer: A. Rural area uses 45 lpcd, plus 15% losses and wastage. 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