13 Trend and Seasonal Methods of Forecasting
Vikas Singla
13.1 OBJECTIVES
To make students understand:
- Trend method of variation in data and its quantitative forecasting method.
- Seasonal method of data variation.
- Multiplicative method of data prediction if it is varying in seasonal manner
13.2 INTRODUCTION
Previous module discussed time series method of forecasting when data variation occurred over a short term period. Also the fluctuation behavior of data was random and unknown. This module would focus on methods of forecasting when variation data is occurring from short to mid-term period. Quarterly sales, sales of products depending on weather conditions, traffic congestion on daily basis, movie ticket sales during weekends are some of the examples that fall in the category of such time period. Moreover, such data show a particular behavior in its variation. Sale of smart phones shows an increasing trend whereas sales of desktops show a decreasing trend over a monthly or quarterly time period. Whereas, sale of milk products, demand of winter clothing, demand of electricity, rush hours show cyclical trend occurring over regular time period. In all such cases data does follow a particular pattern which was missing in random variations. Thus, this module focuses on two forecasting techniques namely trend and seasonal used when data variation shows either of the following pattern over a specified time interval:
- either increasing or decreasing (discussed under category of trend method)
- regularly repeating upward or downward movements at regular intervals (discussed under seasonal method)
13.3 TREND METHOD OF FORECASTING
The trend component in the data may be either increasing or decreasing. This increase or decrease variation may be linear or not. But this chapter would limit itself only to linear variation in the data showing any particular trend. Analysis of trend involves developing a mathematical equation that would appropriately describe a relationship between two variables. The relationship can be positive or negative. Positive relationship would imply that with increase in one variable other variable would also show an increasing trend. Whereas, negative relationship would imply that with increase in one variable other variable would tend to show a decreasing trend. For example, cases of lung cancer would show an increasing trend with increase in smoking habits. However, with increase in awareness about harmful effects of smoking sales of tobacco products might decrease.
Thus, in trend forecasting there are two variables. One of the variable which is under control like time, money spend on advertisement, media time devoted to advertisements related to tobacco etc. is called as independent variable. The effect of these controlled variables on some other variable such as sales, awareness level etc. is called as dependent variable. When independent variable is time and dependent variable changes with change in time periods then it is time series analysis. If independent variable is other than time such as money spend on advertisement and effect of this is studied on amount of sales then it is studied under causal relationship. For both cases, one aspect is common i.e. there should be relationship between dependent and independent variable.
For forecasting purposes, this linear relationship between two variables linear regression analysis is used.
Regression analysis is the process of building a model involving two variables that can be used to predict one variable by another variable. The most elementary regression model is called simple regression in which the variable to be predicted is called the dependent variable and variable which predicts is called independent variable. The dependent variable is denoted by ‘y’ and independent variable is denoted by ‘x’. The model is denoted by following equation.
Dependent or predicted variable (y) = intercept (a) + slope (b) * independent or predictor variable (x)
Thus, a basic regression model where there is only one independent variable and the regression function is linear can be expressed as y = a + bx …..(1)
where y = dependent variable
x = independent variable
a and b are two parameters where a = intercept and b = slope intercept
The above stated model in (1) is said to be simple, because there is only one independent variable linear in parameters, because no parameter (a and b) appears as an exponent or is multiplied or divided by another parameter, and linear in independent variable, as it appears only in first power.
For instance in the above example if sales is considered as dependent and advertisement expenditure as independent variable then the model would become as:
Sales = a + b * (advertisement expenditure)
The parameters ‘a’ and ‘b’ in equation 1 are called as regression coefficients. ‘b’ is the slope of regression equation which indicates the change in dependent variable ‘Y’ with unit increase in independent variable ‘X’. The parameter ‘a’ is the Y intercept of the regression line. If the scope of regression model includes value of X as zero then the mean of Y is given by parameter ‘a’ otherwise it does not have any particular meaning. For example, figure 1 indicates following regression equation where Y is the rate in percentage and X is the time in years.
Y = -814.99 + 0.426*(X)
Where Y intercept ‘a’ = -814.99 indicates value of regression function if X=0. But as regression line does not start from zero so it does not carry any meaning. The slope parameter ‘b’ = 0.426 indicates that unit increase in X would lead to 0.426 increase in Y.
Thus, in this process entire aim would be calculate values of ‘a’ and ‘b’ which can be done by using following formulas.
Slope intercept ‘b’ can be calculated by using the following equation:
b = (∑XiYi – (∑Xi)( ∑Yi)/n ) / (∑X 2 – (∑Xi)2/n ) ………(2)
y intercept of the regression line can be calculated as:
a = (∑Yi – b∑Xi) / n ……….(3)
By using above discussed formulas we would explain following two examples. In the first example time would be taken as independent variable. This example would be considered as an application of trend method of forecasting. Second example would take time as dependent variable.
Example 13.2.1: A firm’s sales for a product line during 13 quarters of past three years are shown in Table 13.2.1.
Forecast sales for each quarter of fourth year.
| Table 13.2.1 | ||||||||||||
| Quarter
(X) |
1. | 2. | 3. | 4. | 5. | 6. | 7. | 8. | 9. | 10. | 11. | 13. |
| Sales (Y) | 600 | 1550 | 1500 | 1500 | 2400 | 3100 | 2600 | 2900 | 3800 | 4500 | 4000 | 4900 |
Solution:
| Table 13.2.2 | |||
| Quarter (X) | Sales (Y) | XY | X square |
| 1 | 600 | 600 | 1 |
| 2 | 1550 | 3100 | 4 |
| 3 | 1500 | 4500 | 9 |
| 4 | 1500 | 6000 | 16 |
| 5 | 2400 | 13000 | 25 |
| 6 | 3300 | 19800 | 36 |
| 7 | 2600 | 18200 | 49 |
| 8 | 2900 | 23200 | 64 |
| 9 | 3800 | 34200 | 81 |
| 10 | 4500 | 45000 | 100 |
| 11 | 4000 | 44000 | 131 |
| 13 | 4900 | 58800 | 144 |
| Total = 78 | 33550 | 269400 | 650 |
By using the formula for slope intercept ‘b’ as given in equation (2) b
= (∑XiYi – (∑Xi)( ∑Yi)/n ) / (∑Xi2 – (∑Xi)2/n )
= (269400 – 78*33550/13) / (650 – 6084/13)
= 358.91
Similarly ‘a’ is calculated by using formula given in equation (3) a
= (∑Yi – b∑Xi) / n
= (33550 – 358.91*78) / 13
= 462.87
Thus equation becomes:
Y = 462.87 + 358.91*(X)
For all quarters of fourth year i.e. for period 13, 14, 15 and 16 sales was forecasted as:
| Forecast (for period 13) | = | 462.87 + 358.91*(X) |
| = | 462.87 + 358.91*(13) | |
| = | 5128.7 | |
| Forecast (for period 14) | = | 462.87 + 358.91*(X) |
| = | 462.87 + 358.91*(14) | |
| = | 5487.61 | |
| Forecast (for period 15) | = | 462.87 + 358.91*(X) |
| = | 462.87 + 358.91*(15) | |
| = | 5846.52 | |
| Forecast (for period 16) | = | 462.87 + 358.91*(X) |
| = | 462.87 + 358.91*(16) | |
| = | 6205.43 |
Example 13.2.2: Example: A manufacturing company produces a spare part in batches which vary in size as demand fluctuates. Table 13.2.3 contains data on batch size and number of man hours required to produce a particular batch. It was suggested that larger batch size would require more effort in terms of time taken to produce. Also estimate mean number of hours required to produce a batch of 55 units
| Table 13.2.3 | ||||||||||
| Batch size (X) | 30 | 20 | 60 | 80 | 40 | 50 | 60 | 30 | 70 | 60 |
| Man Hours (Y) | 73 | 50 | 138 | 170 | 87 | 108 | 135 | 69 | 148 | 132 |
Solution: As amount of time taken to produce a particular batch depends on size of that batch so, batch size is considered as independent variable and man hours as dependent variable.
By using equation (2) and using the values of Table 13.2.3 we can calculate values of ‘a’ and ‘b’ as shown in Table 13.2.4:
| Table 13.2.4 | |||
| Batch Size (X) | Man hours (Y) | XY | X2 |
| 30 | 73 | 2190 | 900 |
| 20 | 50 | 1000 | 400 |
| 60 | 138 | 7680 | 3600 |
| 80 | 170 | 13600 | 6400 |
| 40 | 87 | 3480 | 1600 |
| 50 | 108 | 5400 | 2500 |
| 60 | 135 | 8100 | 3600 |
| 30 | 69 | 2070 | 900 |
| 70 | 148 | 10360 | 4900 |
| 60 | 132 | 7920 | 3600 |
| Total = 1100 | 500 | 61800 | 28400 |
By using calculations in Table 13.2.4 regression parameters are:
b = (61800 – (500)(1100)/10 ) / (28400 – (500)2/10 )
= 2.0
a = (1100 – 2(500)) / 10
= 10.0
Thus equation becomes:
Y = 10 + 2*X
And estimated man hours for batch size of 55 would be
Y = 10 + 2*(55)
= 130
13.4 SEASONAL METHOD OF FORECASTING
Seasonal variations in time series data represent upward or downward movement occurring repeatedly at regular intervals. For instance, sales of winter clothes increases during winter months and decreases during summer months. Same cycle gets repeated after same time period. Two essential characteristics of seasonal variation is that:
- Every cycle of upward and downward movement repeats itself after similar time period.
- Secondly, regular cycles can be used to infer the time for which there would be an increase and time for which there would be decrease in demand.
For example, rush hours happen twice a day. For illustration purposes suppose there is increase in traffic during two hours in morning and for two hours in evening every day. So, in this cycle there would be increase during morning and then traffic would start decreasing which would again increase during evening and then again would drop. This cycle repeats itself every day regularly making it a seasonal variation where each season is of one day. So, if naïve forecast is used then demand for Tuesday can be estimated as demand for Monday during peak hours. Similarly, in case of barber shop there is increase in demand of hair cuts during Sundays and then drops. This cycle gets repeated every week. In this case each season is of one week. Demand of hair cuts for one Saturday can be estimated as demand for previous Saturday.
Models of seasonality:
To estimate seasonal forecast there are two different models: additive and multiplicative.
In the additive model seasonality is expressed as a quantity which is added or subtracted to average of the data. In the multiplicative model seasonality is expressed as percentage of the average amount which is then used to multiply the value of a series to incorporate seasonality. These seasonal percentages are termed as seasonal indexes. For instance, seasonal index of 1.20 regarding sales of toys for a particular time period suggests that sales of toys for that quarter are 20% percent more than the average sales. Similarly, seasonal index of 0.90 would imply that sales for that period are 10% below the average.
This chapter discusses only multiplicative model as this model is more widely used and it tends to be more representative of actual forecast.
Example 13.2.3: The following data clearly depicts seasonal pattern of a particular business. It was tentatively estimated by looking at increasing trend of demand per year that (as shown in Total row) that total demand in fifth year would be 2600. Predict quarterly demand for fifth year.
| Table 13.2.5 | ||||
| Quarter | Year 1 | Year 2 | Year 3 | Year 4 |
| 1. | 45 | 70 | 100 | 100 |
| 2. | 335 | 370 | 585 | 725 |
| 3. | 520 | 590 | 830 | 1160 |
| 4. | 100 | 170 | 285 | 215 |
| Total | 1000 | 1300 | 1800 | 2200 |
Solution: Some important points that should be considered by looking at the data
- The data shows that there is increase in demand as we move from first quarter which peaks in third quarter and then shows a decline in fourth quarter. This cycle gets repeated for every year in four year data shown.
- Here, data of each year is divided into four seasons pertaining to quarterly data. If we use naïve forecast then demand for next season i.e. for first quarter of fifth year would be more than 100.
- So, quarter wise demand comparison follows a trend pattern, whereas year wise comparison indicates seasonal pattern with increase and decrease components.
Step 1: Firstly, calculate average demand per season:
Year 1: 1000 / 4 = 250,
Year 2: 1300 / 4 = 300
Year 3: 1800 / 4 = 450,
Year 4: 2200 / 4 = 550
Step 2: Calculation of seasonal indices: Divide the actual demand for a season by the average demand of that season.
| Quarter | Year 1 | Year 2 | Year 3 | Year 4 |
| 1. | 45/250 | 70/300 | 100/450 | 100/550 |
| 2. | 335/250 | 370/300 | 585/450 | 725/550 |
| 3. | 520/250 | 590/300 | 830/450 | 1160/550 |
| 4. | 100/250 | 170/300 | 285/450 | 215/550 |
Step 3: Calculate the average of seasonal indices of each season.
| Quarter | Year 1 | Year 2 | Year 3 | Year 4 | Average of seasonal indices |
| 1. | 0.18 | 0.23 | 0.22 | 0.18 | (0.18 + 0.23 + 0.22 + 0.18)/4 = 0.20 |
| 2. | 1.34 | 1.23 | 1.30 | 1.32 | 1.30 |
| 3. | 2.08 | 1.97 | 1.84 | 2.11 | 2.00 |
| 4. | 0.40 | 0.57 | 0.63 | 0.39 | 0.50 |
Step 4: As it has been estimated that fifth year demand would be 2600, so it can be suggested that average demand per quarter in fifth year would be 2600/4 = 650 units.
Thus, quarterly forecast can be made by multiplying the seasonal index of each quarter with average demand of that quarter.
| Forecast demand of first quarter of fifth year | = 650*0.20 | = 130 |
| Forecast demand of second quarter of fifth year | = 650*1.30 | = 845 |
| Forecast demand of third quarter of fifth year | = 650*2.00 | = 1300 |
| Forecast demand of fourth quarter of fifth year | = 650*0.50 | = 325 |
13.5 SUMMARY
This chapter discusses forecasting techniques when data shows a certain pattern over short to medium term. Specifically, trend and seasonal methods have been discussed with some illustrations. Trend method pertains to either increasing or decreasing fluctuation of data over selected time period. Such time period can be daily, weekly or monthly but preferably should be less than one year. Trend method involves time series data and uses method of linear regression for prediction purposes. Causal method also uses same methodology but does not involve time series data. Seasonal method of forecasting is used where data shows both increasing and decreasing pattern over a certain time period. In this method pattern is repeated over regular interval of time period. Multiplicative method of prediction has been illustrated as it is more representative of forecast.
13.6 GLOSSARY
- Trend Forecasting: A forecasting technique in which time series data shows either increasing or decreasing pattern.
- Causal Forecasting: A situation in which one variable causes another.
- Linear Regression: A least squares method which assumes that past data and future projections fall around a straight line.
- Regression parameters: Intercept ‘a’ and slope intercept ‘b’ are used to construct regression line with minimum error.
- Seasonal method: is forecasting technique where past data shows increasing and decreasing pattern regularly repeatedly.
13.7 REFERENCES/ SUGGESTED READINGS
- Chase, B.R., Shankar, R., Jacobs, F.R. and Aquilano, N.J., Operations & Supply Chain Management, 13th Edition, McGraw Hill.
- Stevenson, W.J., Operations Management, 9th Edition, Tata McGraw Hill.
- Lee J. Krajewski, Operations Management, Prentice-Hall of India, New Delhi, 8th Edition.
13.8 SHORT ANSWER QUESTIONS
1. Using some fictitious data we wish to predict the musical ability for a person who scores 8 on a test for mathematical ability. We know the relationship is positive. We know that the slope is 1.63 and the intercept is 8.41. What is their predicted score on musical ability?
(a) 80.32 (b) 21.45 (c) 68.91 (d) -4.63
Answer: b
2. If b = 0 the line of best fit will conventionally be drawn
.
(a) through the middle of the datapoints
(b) as horizontal
(c) as provides the best fit to the scores.
Answer: b
3. If you read that the slope of a scatter diagram is 2.00, what does this mean?
(a) For every increase of 1 on the x-axis there is an increase of 2.00 on the y-axis.
(b) For every increase of 2.00 on the x-axis there is an increase of twice as much on the y-axis.
(c) For every increase of 2.00 on the x-axis there is an increase of 2.00 on the y-axis.
(d) Very little.
Answer: a
4. The slope of the line is called:
(a) a correlation coefficient and indicates the variability of the points around the regression line in the scatter diagram.
(b) b and gives us a measure of how much y changes as x changes.
(c) a and is the point where the regression line cuts the vertical axis.
(d) none of the above.
Answer: b
5. In seasonal variation time period for increase in demand during one cycle can be used to predict the increase in next cycle.
(a) True (b) False
Answer: a
6. The time period for increase and decrease is similar in each cycle of seasonal variation.
(a) True (b) False
Answer: a
13.9 MODEL QUESTIONS
1. The complete demand data for years 1, 2, and 3 are given below. Compute the forecast for each of quarters in year 4, given that the forecast for the total demand in year 4 is 2980 gallons.
| Quarterly demand (in gallons) | ||||
| Year | Quarter 1 | Quarter 2 | Quarter 3 | Quarter 4 |
| 1 | 350 | 710 | 950 | 420 |
| 2 | 370 | 680 | 1060 | 500 |
| 3 | 450 | 750 | 1020 | 570 |
2. Show how data is fluctuating graphically and use appropriate method to predict for next period.
| X | 1. | 2. | 3. | 4. | 5. | 6. | 7. | 8. | 9. | 10. | 11. | 13. |
| y | 600 | 1550 | 1500 | 1500 | 2400 | 3100 | 2600 | 2900 | 3800 | 4500 | 4000 | 4900 |