APSEd

# Population Forecasting Methods | Formulas | Example Problems | Practice Problem

Population forecasting is a method to predict/forecast the future population of an area. Usually, the population at the __design period__ of water supply systems is predicted to find the water demand at that time, as the systems are required to fulfill their purposes till the end of the design period. Methods to predict the population are discussed further.

## Arithmetical Increase or Arithmetical Mean Method

### Where Arithmetical Increase Method is used

The arithmetical Increase Method is mainly adopted for **old and developed **towns**, **where the rate of population growth is nearly constant. Therefore, it is assumed that the rate of growth of the population is constant. It is similar to simple interest calculations. The population predicted by this method is the lowest of all.

### Arithmetical Increase Method Derivation

* dP/dt = K* (say), where, dP/dt represents rate of growth of population.

Integrating the above equation over P1 to P2 over a time period of t1 to t2,

*∫dP = K∫dt*

*[P2 - P1] = k * [t2 - t1]*

**P2 = P1 + K * Δt**

**P2 = P1 + x̄ * n**

* P2 = P1 + n*x̄

### Arithmetical Increase Method Formula

**Pn = Po + nx̄,**

where,

Po - last known population

Pn - population (predicted) after 'n' number of decades,

n - number of decades between Po and Pn and,

x̄ - the rate of population growth.

### Arithmetical Increase Method Example Problem

The following data (common data) will be used in the example problems for all other methods to be discussed.

Year | Population |

1930 | 25000 |

1940 | 28000 |

1950 | 34000 |

1960 | 42000 |

1970 | 47000 |

**Question: **With the help of the common data find the population for the year 2020 using the arithmetic increase method.

**Solution:**

Step 1: Find the increase in population each decade.

Year | Population | Increase |

1930 | 25000 | - |

1940 | 28000 | 3000 |

1950 | 34000 | 6000 |

1960 | 42000 | 8000 |

1970 | 47000 | 5000 |

Step 2: Find the average rate of increase of population (x̄)

x̄ = (3000+6000+8000+5000)/4

x̄ = 22000/4

x̄ = 5500

Step 3: Find the number of decades (n) between the last known year and the required year

n = 5 (5 decades elapsed between 1970 and 2020)

Step 4: Apply the formula **Pn = Po + nx̄,**

P[2020] = P[1970] + (5 * 5500)

P[2020] = 47000 + 27500

** P[2020] = 74,500**. Therefore, population at 2020 will be 74,500.

## Geometrical Increase Method

### Where Geometrical Increase Method is used

This method is adopted for **young and developing towns**, where the rate of growth of population is proportional to the population at present (i.e., dP/dt ∝ P). Therefore, it is assumed that the percentage increase in population is constant. It is similar to compound interest calculations. The population predicted by this method is the highest of all.

### Geometrical Increase Method Derivation

Let's say, for the 0th-year population is **P**

For 1st year/decade, according to this method, the population would become,

* P + (r/100)P*, where r is the growth rate.

For 2nd year/decade, according to this method, population would become,

*[P + (r/100)P] + (r/100)[P + (r/100)P]*

*= P[1+(r/100)]^2*

Generalizing the above equation, we get,

**Pn = Po[1 + (r/100)]^n**

### Geometrical Increase Method Formula

** Pn = Po[1 + (r/100)]^n**,

where,

Po - last known population,

Pn - population (predicted) after 'n' number of decades,

n - number of decades between Po and Pn and,

**r - growth rate =****(increase in population/initial population) * 100 (%).**

r could be found as arithmetic mean (i.e., (r1 + r2 + r3 ... rn)/n) or as a geometric mean (i.e., nth root of (r1 * r2 * r3 ... rn)), for the given data. According to Indian standards r should be calculates using geometric mean method.

### Geometrical Increase Method Example Problem

**Question: **With the help of the common data find the population for the year 2020 using the Geometrical increase method.

**Solution:**

Step 1: Find the increase in population each decade.

Step 2: Find the growth rate.

Year | Population | Increase in population | Growth rate |

1930 | 25000 | - | - |

1940 | 28000 | 3000 | (3000/25000) * 100 = 12% |

1950 | 34000 | 6000 | (6000/28000) * 100 = 21.4% |

1960 | 42000 | 8000 | (8000/34000) * 100 = 23.5% |

1970 | 47000 | 5000 | (5000/42000) * 100 = 11.9% |

Step 3: Find the average growth rate (r) using geometrical mean.

r = ∜(12 * 21.4 * 23.5 * 11.9)

r = 16.37 %

Step 4: Find the number of decades (n) between the last known year and the required year

n = 5 (5 decades elapsed between 1970 and 2020)

Step 5: Apply the formula **Pn = Po[1 + (r/100)]^n**

P[2020] = P[1970][1 + (16.37/100)]^5

P[2020] = 47000[1.1637]^5

** P[2020] = 1,00,300. **Therefore, population at 2020 will be 1,00,300.

## Incremental Increase Method

### Where Incremental Increase Method is used

This method is adopted for **average-sized**** ****towns**** under normal conditions**, where the rate of population growth is not constant i.e., either increasing or decreasing. It is a combination of the arithmetic increase method and geometrical increase method. Population predicted by this method lies between the arithmetical increase method and the geometrical increase method.

### Incremental Increase Method Formula

** Pn = (Po + nx̄) + ((n(n+1))/2)* ȳ**,

where,

Po - last known population,

Pn - population (predicted) after 'n' number of decades,

n - number of decades between Po and Pn,

x̄ - mean or average of increase in population and,

ȳ - algebraic mean of incremental increase (an increase of increase) of population.

### Incremental Increase Method Example Problem

**Question: **With the help of the common data find the population for the year 2020 using the Incremental Increase method.

**Solution:**

Step 1: Find the increase in population in each decade.

Step 2: Find the incremental increase i.e., increase of increase.

Year | Population | Increase in population | Incremental Increase |

1930 | 25000 | - | - |

1940 | 28000 | 3000 | - |

1950 | 34000 | 6000 | 6000 - 3000 = 3000 |

1960 | 42000 | 8000 | 8000 - 6000 = 2000 |

1970 | 47000 | 5000 | 5000 - 8000 = -3000 |

Step 3: Find x̄ and ȳ as average of Increase in population and Incremental increase values respectively.

x̄ = (3000+6000+8000+5000)/4

x̄ = 5500

ȳ = (3000+2000-3000)/3

ȳ = 2000/3

Step 4: Find the number of decades (n) between the last known year and the required year

n = 5 (5 decades elapsed between 1970 and 2020)

Step 5: Apply the formula *Pn = (Po + n**x̄) + ((n(n+1))/2)* *** ȳ**,

P[2020] = (P[1970] + nx̄) + ((n(n+1))/2)* ȳ

P[2020] = 47000 + (5 * 5500) + (((5 * 6)/2) * (2000/3))

** P[2020] = 84,500. **Therefore, population at 2020 will be 84,500.

## Decreasing Rate of Growth Method

### Where Decreasing Rate of Growth Method is used

This method is adopted for a town which is reaching **saturation population**, where the rate of population growth is decreasing. In this method, an average decrease in growth rate (S) is considered.

### Decreasing Rate of Growth Method Formula

*Pn = P(n-1) + ((r(n-1) - S)/100) * P(n-1),*

where,

Pn - population at required decade,

P(n-1) - population at previous decade (predicted or available),

r(n-1) - growth rate at previous decade and,

S - average decrease in growth rate.

Due to the very nature of the formula, which requires population data at the previous decade i.e., P(n-1), this method requires the calculation of population at each successive decade (from the last known decade) instead of directly calculating population at the required decade.

### Decreasing Rate of Growth Method Example Problem

**Question: **With the help of the common data find the population for the year 2020 using the decreasing rate of growth method.

**Solution:**

Step 1: Find the increase in population.

Step 2: Find the growth rate (r) as in the geometrical increase method.

Step 3: Find the decrease in the growth rate.

Year | Population | Increase in population | Growth rate (r) | Decrease in growth rate |

1930 | 25000 | - | - | - |

1940 | 28000 | 3000 | 12% | - |

1950 | 34000 | 6000 | 21.4% | 12 - 21.4 = -9.4% |

1960 | 42000 | 8000 | 23.5% | 21.4 - 23.5 = -2.1% |

1970 | 47000 | 5000 | 11.9% | 23.5 - 11.9 = 11.6% |

Step 4: Find the average of decrease in growth rate(s).

S = (-9.4-2.1+11.6)/3

S = 0.1/3

S = 0.03%

Step 5: Apply the formula ** Pn = P(n-1) + ((r(n-1) - S)/100) * P(n-1), **and find the population at successive decade till the population at required data is arrived.

P[1980] = P[1970] + ((r[1970] - S)/100) * P[1970]

P[1980] = 47000 + ((11.9 - 0.03)/100) * 47000

P[1980] = 52579

P[1990] = P[1980] + ((r[1980] - S)/100) * P[1980]

P[1990] = 52579 + ((11.87 - 0.03)/100) * 52579, here r[1980] is directly found as 11.9 - 0.03 i.e., r[1970] - S, which equals to 11.87.

P[1990] = 58,804

Similarly, P[2020] could be found.

## Graphical Method

In this method, the population vs time graph is plotted and is extended accordingly to find the future population. It is to be done by an experienced person and is almost always prone to error.

## Comparative Graphical Method

In this method, the population of a town is predicted by comparing it with a similar town.

## Master Plan Method

This method is used for a completely planned city that is not meant to be developed in a haphazard manner.

## Practice Problem

For more insights please refer to the video lecture below on population forecasting methods.

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