Forecast bake-off: naive, average or smoothing?
A budget airline in Madrid needs a quick forecast of daily seat sales. The analysts compare three simple methods on past data and use the winner to forecast tomorrow.
- Naive: prediction for day
tis the value of dayt-1. - SMA3: the mean of the three days before day
t. - EMA:
ema[0] = x[0],ema[i] = alpha * x[i] + (1 - alpha) * ema[i-1]; the prediction for daytisema[t-1].
Task: line 1 is alpha (a decimal between 0 and 1). Line 2 is the daily series (at least 4 numbers). Days are numbered from 0. Score each method on days t = 3, 4, ..., n-1 with the mean absolute error (MAE). Print four lines: Naive MAE: a, SMA3 MAE: b, EMA MAE: c (3 decimals each) and Best: NAME. The best method has the lowest MAE; when two MAEs are equal after rounding to 6 decimals, prefer in the order Naive, SMA3, EMA. Then print Forecast for day n: F with 2 decimals, where F is the best method's prediction for the next day (naive: last value, SMA3: mean of the last three values, EMA: the final ema value).
Sample tests
Input
0.5 120 135 128 150 160 155 170 182 176 190
Expected output
Naive MAE: 12.000 SMA3 MAE: 15.714 EMA MAE: 15.468 Best: Naive Forecast for day 10: 190.00
Input
0.3 50 52 51 53
Expected output
Naive MAE: 2.000 SMA3 MAE: 2.000 EMA MAE: 2.280 Best: Naive Forecast for day 4: 53.00
Your answer is also checked against 3 hidden tests.
Need a hint?
Write one helper that returns the prediction of a method for day t, then reuse it for scoring and for the final forecast.
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