Question

A hospital records the number of floral deliveries its patients receive each day. For a two week
period, the records show

15, 27, 26, 24, 18, 21, 26, 19, 15, 28, 25, 26, 17, 23

Use exponential smoothing with a smoothing constant of .4 to forecast the number of deliveries.

Answer 1

Answer on Question #44592 – Math – Other

A hospital records the number of floral deliveries its patients receive each day. For a two week period, the records show

15, 27, 26, 24, 18, 21, 26, 19, 15, 28, 25, 26, 17, 23

Use exponential smoothing with a smoothing constant of 0.4 to forecast the number of deliveries.

Solution

Unlike moving average models, which use a fixed number of the most recent values in the time series for smoothing and forecasting, exponential smoothing incorporates all values time series, placing the heaviest weight on the current data, and weights on older observations that diminish exponentially over time. Because of the emphasis on all previous periods in the data set, the exponential smoothing model is recursive. When a time series exhibits no strong or discernible seasonality or trend, the simplest form of exponential smoothing – single exponential smoothing – can be applied. The formula for single exponential smoothing is given by

$$\widehat{Y}_{t+1} = \alpha Y_{t} + (1 - \alpha) \widehat{Y}_{t}$$

In this equation, $\widehat{Y}_{t+1}$ represents the forecast value for period $t + 1$; $Y_t$ is the actual value of the current period, $t$; $\widehat{Y}_t$ is the forecast value for the current period, $t$; and $\alpha$ is the smoothing constant, or alpha, a number between 0 and 1. Alpha is the weight we assign to the most recent observation in our time series. Essentially, we are basing our forecast for the next period on the actual value for this period, and the value we forecasted for this period, which in turn was based on forecasts for periods before that.

We apply the noted above formula to our problem.

$$\widehat{Y}_{2} = (1 - \alpha) \widehat{Y}_{1} + \alpha Y_{1} = (1 - 0.4) \cdot 15 + 0.4 \cdot 15 = 15$$ $$\widehat{Y}_{3} = (1 - \alpha) \widehat{Y}_{2} + \alpha Y_{2} = (1 - 0.4) \cdot 15 + 0.4 \cdot 27 = 19.8$$ $$\widehat{Y}_{4} = (1 - \alpha) \widehat{Y}_{3} + \alpha Y_{3} = (1 - 0.4) \cdot 19.8 + 0.4 \cdot 26 = 22.28$$ $$\widehat{Y}_{5} = (1 - \alpha) \widehat{Y}_{4} + \alpha Y_{4} = (1 - 0.4) \cdot 22.28 + 0.4 \cdot 24 = 22.968$$ $$\widehat{Y}_{6} = (1 - \alpha) \widehat{Y}_{5} + \alpha Y_{5} = (1 - 0.4) \cdot 22.968 + 0.4 \cdot 18 = 20.9808$$ $$\widehat{Y}_{7} = (1 - \alpha) \widehat{Y}_{6} + \alpha Y_{6} = (1 - 0.4) \cdot 20.9808 + 0.4 \cdot 21 = 20.98848$$ $$\widehat{Y}_{8} = (1 - \alpha) \widehat{Y}_{7} + \alpha Y_{7} = (1 - 0.4) \cdot 20.98848 + 0.4 \cdot 26 = 22.993088$$ $$\widehat{Y}_{9} = (1 - \alpha) \widehat{Y}_{8} + \alpha Y_{8} = (1 - 0.4) \cdot 22.993088 + 0.4 \cdot 19 = 21.3958528$$ $$\widehat{Y}_{10} = (1 - \alpha) \widehat{Y}_{9} + \alpha Y_{9} = (1 - 0.4) \cdot 21.3958528 + 0.4 \cdot 15 = 18.83751168$$

First, we provide the obtained results in Table 1.

Table 1 Actual and forecasting the number of deliveries with the smoothing constant $\alpha = 0.4$ .

We can represent the results of calculations in Figure 1.

Figure 1 Actual and forecasting the number of deliveries with the smoothing constant $\alpha = 0.4$ .

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