12.6.1 Free-Hand Method

The free-hand method is the simplest method of ascertaining trend. On the graph paper, time t is measured horizontally, whereas the values of the variable y are measured vertically. The points (t₁, y₁), (t₂, y₂), …, (t_n, y_n) are plotted on a graph paper by taking t_i on the x-axis and y_i on the y-axis. The plotted points are then joined by straight lines and the trend line is fitted with the help of a transparent ruler or a smooth curve by hand. The trend line is drawn in such way that the following are satisfied as far as possible:

A free-hand curve removes the short-term variations and exhibits a general trend. Free-hand method is one of the simplest and most flexible methods and sometimes yield good results. It is too subjective. In this method the trend line varies from person to person, as it depends on individual’s judgment. Hence it cannot be used as a basis of prediction.

Example: Fit a straight line trend to the following data by using free-hand graph method:

12.6.2 Semiaverage Method

In this method the series is divided into two halves. Then the average is found out for each half of the series. The average values are plotted on the graph paper against the mid periods of the corresponding each half series. The line joining these two plotted points gives the trend line. The direction of the line indicates about rising, falling, or constant trend of business movements.

If the number of years in a series is odd, the middle year is excluded at the time of dividing the series into two halves, and then either excluded in both series or excluded totally depending on whether the leftover series contains an even or odd number of years respectively.

The semiaverage method is not subjective and for one series, there is only one trend line. But it is affected by extreme values. This method does not ensure elimination of short-term and cyclic variations. Semiaverage method is superior to free-hand method. It may be used when the data are given for a longer period.

Example 12.1: Fit a straight line trend by using the following data:

Use semiaverage method. Also estimate the profit for the year 1988.

Solution: Trend line by semiaverage method number of years is 7, i.e., odd. We leave the middle most value, i.e., the value corresponding to the year 1984.

$\begin{array}{ll}\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{series}\hfill & =\mathrm{Mean}\hfill \\ \hfill & =\frac{20+22+27}{3}=\frac{69}{3}=23\hfill \end{array}$

$\begin{array}{ll}\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{last}\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{series}\hfill & =\mathrm{Mean}\hfill \\ \hfill & =\frac{30+29+40}{3}=\frac{99}{3}=33\hfill \end{array}$

Plot 23 against 1982 and 33 against 1986. Join these points as shown in figure below.

The profit in the year 1988 is Rs. 37 lakhs.

Example 12.2: Draw the trend by semiaverage method from the following data:

Solution: The data correspond to 14 years

$\mathrm{No}.\mathrm{of}\mathrm{years}\mathrm{in}\mathrm{each}\mathrm{half}=\frac{14}{2}=7$

We break the data into two equal parts of 7 years each.

$\mathrm{Average}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{half}\mathrm{series}=\frac{729}{7}=104.14=104(\mathrm{approx}.)$

$\mathrm{Average}\mathrm{of}\mathrm{the}\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{series}=\frac{854}{7}=122$

Plot 104 against 1830 and 122 against 1900. Join these points (Fig. 12.1).

12.6.3 Moving Average Method

The moving average method is an improvement over the semiaverage method and short-term fluctuations are eliminated by it. A moving average is defined as an average of fixed number of items in the time series which move through the series by dropping the top items of the previous averaged group and adding the next in each successive average.

Let (t₁, y₁), (t₂, y₂), …, (t_n, y_n) denote given time series y₁, y₂, …, y_n are the values of the variable y; corresponding to time periods t₁, t₂, …, t_n, respectively.

The moving averages of order m are defined as

$\frac{y_1+y_2+\dots +y_m}{m};\frac{y_2+y_3+\dots +y_{m+1}}{m};$

Here y₁+y₂+…+y_m, y₂+y₃+…+y_m+1,… are called moving totals of m.

In using moving averages in estimating the trend, we shall have to decide as what should be the order of the moving averages. The order of the moving average should be equal to the length of the cycles in the time series. In case the order of the moving averages is given in the problem itself, then we shall use that order for computing the moving average. The order of the moving averages may either be odd or even.

The moving averages of order 3 are

$\frac{y_1+y_2+y_3}{3};\frac{y_2+y_3+y_4}{3};\dots ;\frac{y_{n-2}+y_{n-1}+y_n}{3}$

These moving averages are called the trend values. They are considered to correspond 2nd, 3rd, …, (n−1)th years, respectively. Calculation of trend values by using moving averages of even order is slightly complicated. The following steps are involved in the method:

12.6.4 Method of Least Square

The method of least squares is a widely used method of fitting curve for a given data. It is the most popular method used to determine the position of the trend line of a given time series. The trend line is technically called the best fit. In this method a mathematical relationship is established between the time factor and the variable given. Let (t₁, y₁), (t₂, y₂), …, (t_n, y_n) denote the given time series. In this method the trend value y_c of the variable y are computed so as to satisfy the conditions:

1. The sum of the deviations of y from their corresponding trend values is zero.
i.e.,

$\sum (y-y_c)=0$

2. The sum of the square of the deviations of the values of y from their corresponding trend values is the least.
i.e.,

$\sum {(y-y_c)}^2$

is least.

The equation of the trend line can be expressed as

$y_c=a+bx$

where a and b are constants and the trend line satisfies the conditions:

The values of a and b determined such that they satisfy the equations.

$\sum y=na+b\sum x$ (12.1)

(12.1)

$\sum xy=b\sum x+a\sum x^2$ (12.2)

(12.2)

Eqs. (12.1) and (12.3) are called normal equations.

Solving Eqs. (12.1) and (12.2) we get

$a=\frac{\sum y·\sum x^2-\sum x\sum xy}{n\sum x^2-{\left(\sum x\right)}^2}$

and

$b=\frac{n·\sum xy-\sum x\sum y}{n\sum x^2-{\left(\sum x\right)}^2}$

In the equation, y_c=a+bx, of the trend, a represents the trend of the variable when x=0 and b represents the slope of the trend line. If b is positive, the trend line will be upward and if b is negative the trend line will be downward.

When the origin is mentioned and the deviations from the origin is denoted by x, we get

$a=\frac{\sum y}{n},b=\frac{\sum xy}{\sum x^2}$

$\because$(The sum of derivation from the origin=$\sum x=0$).

Note: If n is odd, we take the middle value (Middle year) as the origin. If n is even, there will be two middle values. In this case we take the mean of the two middle values as the origin.

Merits

Demerits

12.6.4.1 Solved Examples

Example 12.6: Find the least square line y=a+bx for the data:

x	–2	–1	0	1	2
y	1	2	3	3	4

Solution:

x	y	x2	xy
–2	1	4	–2
–1	2	1	–2
0	3	0	0
1	3	1	3
2	4	4	8
$\sum x=0$	$\sum y=13$	$\sum x^2=10$	$\sum xy=7$

∴ $\sum x=0$, $\sum y=13$, $\sum x^2=10$, $\sum xy=7$, and x=5

The normal equations are

$na+b\sum x=\sum y$

$a\sum x+b\sum x^2=\sum xy$

Putting the values of n, $\sum x,\sum y,\sum x^2,\sum xy$ in the above equation, we get

$5a+b·0=13\Rightarrow 5a=13$ (12.3)

(12.3)

$a·0+b(10)=7\Rightarrow 10b=7$ (12.4)

(12.4)

From Eqs. (12.3) and (12.4) we get

$a=\frac{13}{5}=2.6,b=\frac{7}{10}=0.7$

The required of least square line is

$y=2.6+(0.7)x$

Example 12.7: Fit a straight line trend by the method of least square from the following data and find the trend values.

Year	1958	1959	1960	1961	1962
Sales (in lakhs of units)	65	95	80	115	105

Solution: We have n=5

∴ n is odd.

Taking middle year, i.e., 1960 as the origin. We get,

Year	Sales	x	x2	xy
1958	65	–2	4	–130
1959	95	–1	1	–95
1960	80	0	0	0
1961	115	1	1	115
1962	105	2	4	210
Total	$\sum y=460$	$\sum x=0$	$\sum x^2=10$	$\sum xy=100$

∴ n=5, $\sum x=0$, $\sum x^2=10$, $\sum y=460$, and $\sum xy=100$

$a=\frac{\sum y}{n}\Rightarrow a=\frac{460}{5}=92$

$b=\frac{\sum xy}{\sum x^2}=\frac{100}{10}=10$

∴ The equation of the straight line trend is

$y_c=a+bx\Rightarrow y_c=92+10x$

For the year 1958, x=–2

$\begin{array}{ll}\Rightarrow y_c\hfill & =y_{1958}=92+10(-2)\hfill \\ \hfill & =92-20=72\hfill \end{array}$

For the year 1959, x=–1

$\begin{array}{ll}\Rightarrow y_c\hfill & =y_{1959}=92+10(-1)\hfill \\ \hfill & =92-10=82\hfill \end{array}$

For the year 1960, x=0

$\begin{array}{ll}\Rightarrow y_c\hfill & =y_{1960}=92+10(0)\hfill \\ \hfill & =92+0=92\hfill \end{array}$

For the year 1961, x=1

$\begin{array}{ll}\Rightarrow y_c\hfill & =y_{1961}=92+10(1)\hfill \\ \hfill & =92+10=102\hfill \end{array}$

For the year 1962, x=2

$\begin{array}{cc}\Rightarrow y_c& =y_{1962}=92+10(2)\\ & =92+20=112\end{array}$

We have

Year	Trend value
1958	72
1959	82
1960	92
1961	102
1962	112

and the straight line trend is y_c=92+10x or simply y=92+10x.

Example 12.8: Determine the trend by the method of least squares. Also find the trend values.

Year	1950	1951	1952	1953	1954	1955	1956	1957
Value	346	411	392	512	626	640	611	796

Solution: Here n=9

∴ n is even.

1953 and 1954 are the middle years.

The origin is $\frac{1953+1954}{2}=1953.5$

We take x=year−1953.5

Year	x	y	x2	xy
1950	–3.5	346	12.25	–1211.00
1951	–2.5	411	6.25	–1027.50
1952	–1.5	392	2.25	–588.00
1953	–0.5	512	0.25	–256.00
1954	0.5	626	0.25	313.00
1955	1.5	640	2.25	960.00
1956	2.5	611	6.25	1527.50
1957	3.5	796	12.25	2786.00
Total	0	4334	42.00	2504.00

∴ We have n=8, $\sum x=0,$ $\sum y=4334,$ $\sum x^2=42$, and $\sum xy=2504$

$a=\frac{\sum y}{n}\Rightarrow a=\frac{4334}{8}=541.75$

$b=\frac{\sum xy}{\sum x^2}=\frac{2504}{42}=59.60$

∴ The equation of the trend line is

$y=541.75+(59.60)x$

The trend values are:

Year	x	Trend: y=541.75+(59.60)x
1950	–3.5	333.15
1951	–2.5	392.75
1952	–1.5	452.35
1953	–0.5	511.95
1954	0.5	571.95
1955	1.5	631.15
1956	2.5	690.75
1957	3.5	750.35

12.6.5 Nonlinear Trend

So far the discussion about trend is confined mostly to the linear trend. There are situations where linear trend is not found suitable. For instance consider the population growth. In the beginning of a period population growth will be slow but will start to multiply at a faster rate, in the later period. For such a type of time series data, a nonlinear trend would depict a better trend than the linear one. Some well-known methods of determining nonlinear trends are:

The above methods have been discussed in this chapter. If the trend is changing frequently, a curve will give a picture of trend. Which curve is best suited to the data may be guessed by plotting time series data on a graph. When the shape of the curve is known, the mathematical equation for it may be given. Once the equation is decided, it can be fitted to the available time series data. Some of the frequently used curve are:

We shall restrict ourselves only to the study of parabolic curves.

Parabola: The parabolic curve is mathematically represented by a second-degree polynomial of the form

$y=a+bx+c{x}^2$

It can be fitted by the method of least square, using the method of least squares we get the three normal equations.

$na+b\sum x+c\sum x^2=\sum y$

$a\sum x+b\sum x^2+c\sum x^3=\sum xy$

$a\sum x^2+b\sum x^3+c\sum x^4=\sum x^{2}y$

Solving the above equations we obtain the values for a, b, and c.

Substituting the values of a, b, and c is

$y=a+bx+c{x}^2$

We get the parabola. The curve is a quadratic curve and is also known as a response curve. The shape of the curve depends on the values of b and c. Its geometric shapes are as shown in Fig. 12.1 (Fig. 12.2).

Example 12.9: The prices to the commodities during 1978–83 are given below. Fit a parabola y=a+bx+cx² to the data. Estimate the price of the commodity for the year 1984.

Also find the trend values.

Solution: No. of periods=n=6

Trend by least square method.

The normal equation are

$na+b\sum x+c\sum x^2=\sum y$

$a\sum x+b\sum x^2+c\sum x^3=\sum xy$

$a\sum x^2+b\sum x^3+c\sum x^4=\sum x^{2}y$

i.e.,

$6a+b.0+70c=848\Rightarrow 6a+70c=848$ (12.5)

(12.5)

$a.0+70b+c.b=694\Rightarrow 70b=694$ (12.6)

(12.6)

$70.a+b.0+1414c=10160\Rightarrow 70a+1414c=10160$ (12.7)

(12.7)

From Eq. (12.6) we get

$70b=694\Rightarrow b=\frac{694}{70}=9.914$

Consider Eqs. (12.5) and (12.7)

$\begin{array}{l}35\times (\text{6a}+\text{70c}=\text{848})\hfill \\ 3\times \frac{(70a+1414c=10160)}{\begin{array}{l}210a+2450c=29680\\ \frac{210a+4242c=30480}{-1792c=-800}\end{array}}\hfill \end{array}$ (12.8)

(12.8)

$\Rightarrow c=\frac{800}{1792}=0.446$

$\therefore c=0.446$

From Eq. (12.7) $6a+70(0.446)=848$

$\Rightarrow 6a+848–70(0.446)\Rightarrow a=136.13$

$\therefore a=136.13,b=9.14x+0.4462x^2$

where 1980.5 is the origin and x unit is $1/2$ years.

Estimated value of price for the year 1984:

We have x=(1984−1980.5) × 2=7

Estimated price for the year 1984:

$\begin{array}{ll}{y}_c=y_{1984}\hfill & =136.13+(9.914)(7)+(0.446)(49)\hfill \\ \hfill & =227.382\mathrm{units}\hfill \end{array}$

12.6.6 Conversions of Trend Equations

A time series has three characteristics namely

We can shift the origin and change the trend equation or we can change unit of variable to find the new trend equation. In this section we discuss the following conversions:

$y_c=f(x+k)$

If the origin is shifted back by k periods then the new trend equation is of the form

$y_c=f(x-k)$

(where k>0)

Example 12.10: The parabolic trend of the equation of a time series is given by

$y_c=1200+4x–2x^2$

with origin=1980, and x unit=1 year shift the origin to (a) 1984, (b) 1979 and find the new trend equations.

Solution: We have

$y_c=1200+4x–2x^2$

The origin under reference=1980

a. When the origin is shifted to 1984, x is to be replaced by x+(1984−1980)=x+4 in the given equation.
i.e.,

$k=1984–1980=4$

The new trend equation with new origin at 1984 is

$y_c=1200+4(x+4)–2{(x+4)}^2$

$\Rightarrow y_c=1200+4x+16–2x^2–16x–32$

$\Rightarrow y_c=1184–12x–2x^2$

b. When the origin is shifted to 1975, then

$k=1979–1980=–1$

The new trend equation with new origin at 1984 is

$y_c=1200+4(x–1)–2{(x–1)}^2$

$\Rightarrow y_c=1200+4x–4–2(x^2–2x+1)$

$\Rightarrow y_c=1194+8x–2x^2$

Let

$y_c=a+bx\mathrm{is}\mathrm{the}\mathrm{trend}\mathrm{line}.$

with

$\begin{array}{l}x\mathrm{unit}=1\mathrm{year}\hfill \\ y\mathrm{unit}=\mathrm{annual}\mathrm{total}\hfill \end{array}$

If we convert the trend equation, so that x unit is 1 month and y unit is monthly average then the equation of new trend line will be

$y_c=\frac{a}{12}+\frac{b}{12\times 12}x$

i.e.,

$y_c=\frac{a}{12}+\frac{b}{144}x$

If the annual trend equation y_c=a+bx is converted to half yearly trend equation, then the new trend equation will be of the form

$y_c=\frac{a}{\left(\frac{12}{6}\right)}+\frac{b}{\left(\frac{144}{6}\right)}x=\frac{a}{2}+\frac{b}{24}x$

If the annual trend equation is converted into quarterly trend equation, then it is of the form

$y_c=\frac{a}{\left(\frac{12}{3}\right)}+\frac{b}{\left(\frac{144}{3}\right)}x=\frac{a}{4}+\frac{b}{48}x$

where x unit: one quarter; y unit: quarterly values.

Example 12.11: Convert the following annual trend equation for cloth production in a factory to a monthly trend equation.

$y=108+7.2x$

(origin 1976, time limit 1 year, y=cloth production in ′000 m).

Solution: Annual trend equation is y=108+7.2x

∴ Monthly trend equation is

$y=\frac{108}{12}+\frac{7.2}{144}x$

$y=9+0.5x$

Note: When the data are given monthly averages per year, the value of a remains unchanged in the conversion process, in which case we take new trend lines as

$y=a+\frac{b}{12}x$

Example 12.12: Convert the following annual trend equation of ABC corporation to a monthly level: y=41+7.2x (Origin: 1976, x units are year, y=average monthly sales).

Solution: y is average monthly sales.

∴ The monthly trend equation can be written as

$y=41+\frac{7.2}{12}x=41+(0.6)x$

(Origin July 1, 1976, time unit 1 month, y=monthly cloth production in ′000 m).

Example 12.13: The parabolic trend equation of a time series is given by y_c=5+x−x² with origin 1980, and x limit=1 year, y=annual production. Convert the trend equation so that

Solution: The trend equation is y_c=5+x−x² with origin 1980, x unit=1 year, and y unit=annual production.

1. When x limit=1 year and y unit=average monthly production.
The new trend equation is

$y_c=\frac{15}{12}+\frac{x}{12}-\frac{x^2}{12}$

2. When x unit=1 month; y unit=average monthly production.

The new trend equation is

$y_c=\frac{15}{12}+\frac{x}{{12}^2}-\frac{x^2}{{12}^3}$

i.e.,

$y_c=\frac{15}{12}+\frac{x}{144}-\frac{x^2}{1728}$

Exercise 12.1

Find the average production, rate of growth, and trend ordinates (apply least square method)?