Percent Growth

Example 1

Population projection and design water demand — Percent Growth

A city of 103,281 grows at 1.5% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 206 L/capita/day.

Given

$P_{0} = 103,281$
$i = 1.5%/yr$
$n = 14 yr$
$Per capita use = 206 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 103281(1 + 0.015)^14 = 127,217 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 103281(1 + 0.015 \times 14) = 124,970 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 127,217(206)/1000 = 26,207 m^{3}/d$
Peak day
$Q_peak = 2.7 \times 26,207 = 70,758 m^{3}/d$

Answer:

P = 127,217 (geometric) vs 124,970 (arithmetic); Q_avg = 26,207 m^{3}/d, Q_peak = 70,758 m^{3}/d

Why the other options are there

21,689 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 2

Population projection and design water demand — Percent Growth (2)

A city of 120,255 grows at 2.3% per year. Project the population in 19 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 487 L/capita/day.

Given

$P_{0} = 120,255$
$i = 2.3%/yr$
$n = 19 yr$
$Per capita use = 487 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 120255(1 + 0.023)^19 = 185,242 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 120255(1 + 0.023 \times 19) = 172,806 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 185,242(487)/1000 = 90,213 m^{3}/d$
Peak day
$Q_peak = 2.4 \times 90,213 = 216,511 m^{3}/d$

Answer:

P = 185,242 (geometric) vs 172,806 (arithmetic); Q_avg = 90,213 m^{3}/d, Q_peak = 216,511 m^{3}/d

Why the other options are there

52,551 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 3

Population projection and design water demand — Percent Growth (3)

A city of 122,118 grows at 1.2% per year. Project the population in 21 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 499 L/capita/day.

Given

$P_{0} = 122,118$
$i = 1.2%/yr$
$n = 21 yr$
$Per capita use = 499 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 122118(1 + 0.012)^21 = 156,881 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 122118(1 + 0.012 \times 21) = 152,892 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 156,881(499)/1000 = 78,284 m^{3}/d$
Peak day
$Q_peak = 2.9 \times 78,284 = 227,023 m^{3}/d$

Answer:

P = 156,881 (geometric) vs 152,892 (arithmetic); Q_avg = 78,284 m^{3}/d, Q_peak = 227,023 m^{3}/d

Why the other options are there

30,774 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 4

Population projection and design water demand — Percent Growth (4)

A city of 43,420 grows at 1.3% per year. Project the population in 10 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 463 L/capita/day.

Given

$P_{0} = 43,420$
$i = 1.3%/yr$
$n = 10 yr$
$Per capita use = 463 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 43420(1 + 0.013)^10 = 49,407 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 43420(1 + 0.013 \times 10) = 49,065 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 49,407(463)/1000 = 22,875 m^{3}/d$
Peak day
$Q_peak = 2.3 \times 22,875 = 52,613 m^{3}/d$

Answer:

P = 49,407 (geometric) vs 49,065 (arithmetic); Q_avg = 22,875 m^{3}/d, Q_peak = 52,613 m^{3}/d

Why the other options are there

5,645 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 5

Population projection and design water demand — Percent Growth (5)

A city of 49,675 grows at 2.9% per year. Project the population in 16 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 250 L/capita/day.

Given

$P_{0} = 49,675$
$i = 2.9%/yr$
$n = 16 yr$
$Per capita use = 250 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 49675(1 + 0.029)^16 = 78,484 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 49675(1 + 0.029 \times 16) = 72,724 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 78,484(250)/1000 = 19,621 m^{3}/d$
Peak day
$Q_peak = 2.1 \times 19,621 = 41,204 m^{3}/d$

Answer:

P = 78,484 (geometric) vs 72,724 (arithmetic); Q_avg = 19,621 m^{3}/d, Q_peak = 41,204 m^{3}/d

Why the other options are there

23,049 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 6

Population projection and design water demand — Percent Growth (6)

A city of 71,715 grows at 0.9% per year. Project the population in 24 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 247 L/capita/day.

Given

$P_{0} = 71,715$
$i = 0.9%/yr$
$n = 24 yr$
$Per capita use = 247 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 71715(1 + 0.009)^24 = 88,920 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 71715(1 + 0.009 \times 24) = 87,205 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 88,920(247)/1000 = 21,963 m^{3}/d$
Peak day
$Q_peak = 2.1 \times 21,963 = 46,123 m^{3}/d$

Answer:

P = 88,920 (geometric) vs 87,205 (arithmetic); Q_avg = 21,963 m^{3}/d, Q_peak = 46,123 m^{3}/d

Why the other options are there

15,490 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 7

Population projection and design water demand — Percent Growth (7)

A city of 138,278 grows at 0.9% per year. Project the population in 20 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 345 L/capita/day.

Given

$P_{0} = 138,278$
$i = 0.9%/yr$
$n = 20 yr$
$Per capita use = 345 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 138278(1 + 0.009)^20 = 165,416 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 138278(1 + 0.009 \times 20) = 163,168 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 165,416(345)/1000 = 57,068 m^{3}/d$
Peak day
$Q_peak = 2.0 \times 57,068 = 114,137 m^{3}/d$

Answer:

P = 165,416 (geometric) vs 163,168 (arithmetic); Q_avg = 57,068 m^{3}/d, Q_peak = 114,137 m^{3}/d

Why the other options are there

24,890 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 8

Population projection and design water demand — Percent Growth (8)

A city of 15,459 grows at 2.0% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 265 L/capita/day.

Given

$P_{0} = 15,459$
$i = 2.0%/yr$
$n = 14 yr$
$Per capita use = 265 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 15459(1 + 0.020)^14 = 20,398 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 15459(1 + 0.020 \times 14) = 19,788 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 20,398(265)/1000 = 5,405 m^{3}/d$
Peak day
$Q_peak = 2.5 \times 5,405 = 13,514 m^{3}/d$

Answer:

P = 20,398 (geometric) vs 19,788 (arithmetic); Q_avg = 5,405 m^{3}/d, Q_peak = 13,514 m^{3}/d

Why the other options are there

4,329 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 9

Population projection and design water demand — Percent Growth (9)

A city of 108,849 grows at 1.0% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 380 L/capita/day.

Given

$P_{0} = 108,849$
$i = 1.0%/yr$
$n = 26 yr$
$Per capita use = 380 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 108849(1 + 0.010)^26 = 140,987 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 108849(1 + 0.010 \times 26) = 137,150 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 140,987(380)/1000 = 53,575 m^{3}/d$
Peak day
$Q_peak = 2.0 \times 53,575 = 107,150 m^{3}/d$

Answer:

P = 140,987 (geometric) vs 137,150 (arithmetic); Q_avg = 53,575 m^{3}/d, Q_peak = 107,150 m^{3}/d

Why the other options are there

28,301 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 10

Population projection and design water demand — Percent Growth (10)

A city of 163,973 grows at 0.9% per year. Project the population in 18 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 296 L/capita/day.

Given

$P_{0} = 163,973$
$i = 0.9%/yr$
$n = 18 yr$
$Per capita use = 296 L/cap/d$

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0}(1 + i)ⁿ$
Substituting
$P = 163973(1 + 0.009)^18 = 192,670 people$
Formula (arithmetic)
$P = P_{0}(1 + i n)$
Substituting
$P = 163973(1 + 0.009 \times 18) = 190,537 people$
Formula
$Q_avg = P \times per capita use$
Substituting
$Q_avg = 192,670(296)/1000 = 57,030 m^{3}/d$
Peak day
$Q_peak = 2.0 \times 57,030 = 114,060 m^{3}/d$

Answer:

P = 192,670 (geometric) vs 190,537 (arithmetic); Q_avg = 57,030 m^{3}/d, Q_peak = 114,060 m^{3}/d

Why the other options are there

26,564 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Core formulas for this FE topic

Worked exam-style examples

Percent Growth

Core formulas for this FE topic

Pt=P0(1+k)nPt = P_{0}(1 + k)nPt=P0​(1+k)n

Pt=populationattimetPt = population at time tPt=populationattimet

P0=populationattimezeroP_{0} = population at time zeroP0​=populationattimezero

k=growthratek = growth ratek=growthrate

n=numberofperiodsn = number of periodsn=numberofperiods

Worked exam-style examples