A moving average system is used for forecasting weekly demand. $F_{1} (t) and F_{2} (t)$ are sequences of forecasts with parameters $m_{1} and m_{2}, respectively, where m_{1} and m_{2} (m_{1} > m_{2})$ denote the numbers of weeks over which the moving averages are taken. The actual demand shows a step increase from $d_{1} to d_{2}$ at a certain time. Subsequently

neither F_{1} (t) nor F_{2} (t)

will catch up with the value

d_{2}

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F_{2} (t)

will attain the value

d_{2} before F_{1} (t)

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F_{1} (t)

will attain the value

d_{2} before F_{2} (t)

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both sequences

F_{1} (t) and F_{2} (t) will reach d_{2}

in the same period

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Solution

The correct option is B $F_{2} (t)$ will attain the value $d_{2} before F_{1} (t)$
Since, $m_{1} > m_{2}$ the weightage of the latest demand would be more in $F_{2} (t) . Hence F_{2} (t)$ will attain the value $d_{2} before F_{1} (t) .$

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Similar questions

Q. The moving average method is to be used for forecasting demand based on m periods of data. Two values of m are tried,

m_{1}

and

m_{2}

with

m_{1} > m_{2}

, to get two different forecasts, denoted by F(t) and G(t), respectively.
(P) F(t) has less variability than G(t)
(Q) Forecast error of F (t) is less than that of G(t)
Which of the above statements are CORRECT?

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is: