Question

What is this year's forecast using exponential smoothing with alpha = .4, if last year's smoothed forecast was 2600?

The dean of a school of business is forecasting total student enrollment for this year's summer session classes based on the following historical data:

Four years ago 2000

There years ago 2200

Two years ago 2800

Last year 3000


What is this year's forecast using exponential smoothing with alpha = .4, if last year's smoothed forecast was 2600?
A. 2,600
B. 2,760
C. 2,800
D. 3,840
E. 3,000


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Answer #1
Concepts and reason

The Time-Series analysis is used here to evaluate the required forecast because time series data can be smoothed by the exponential smoothing forecasting technique. In fitting the smoothed version of the time series data, there can be some window functions. These can be either the simple moving average or the weighed moving average.

Fundamentals

The simplest of all of the smoothing function can be:

st=α.xt+(1α).st1{s_t} = \alpha .{x_t} + \left( {1 - \alpha } \right).{s_{t - 1}}

Here,

st=theforescastoftheperiodst1=thepreviousforcastedmeasureα=smoothingfactorxt=actualpreviousvalue\begin{array}{l}\\{s_t} = {\rm{the forescast of the period}}\\\\{{\rm{s}}_{t - 1}} = {\rm{the previous forcasted measure}}\\\\\alpha = {\rm{smoothing factor}}\\\\{{\rm{x}}_t} = {\rm{actual previous value}}\\\end{array}

Use the following formula, to obtain the required forecasted value:

st=α.xt+(1α).st1{s_t} = \alpha .{x_t} + \left( {1 - \alpha } \right).{s_{t - 1}}

Here,

st1=2600α=0.4xt=3000\begin{array}{l}\\{s_{t - 1}} = 2600\\\\\alpha = 0.4\\\\{x_t} = 3000\\\end{array}

The required forecast value is:

st=0.4×3000+(10.4)×2600=2760\begin{array}{c}\\{s_t} = 0.4 \times 3000 + \left( {1 - 0.4} \right) \times 2600\\\\ = 2760\\\end{array}

Ans:

Hence, the required forecast is obtained as 27602760 .

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