The number of fishing rods selling each day is given below. Perform analyses of the time series to determine which model should be used for forecasting.

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a. **3 day moving average **analysis

b. **4 day moving average **analysis

c. **3 day weighted moving average **analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data

d. **exponential smoothing **analysis with **a = 0.3. **

e. Which model provides a better fit of the data?

f. Forecast day 13 sales of fishing rods using the model chosen in part (e).

Day

Rods sold

1 60

2 70

3 110

4 80

5 70

6 85

7 115

8 105

9 65

10 75

11 95

12 85

- Analysis of methods:

1.1 **3 day moving average**

“\def\arraystretch{1.5}n \begin{array}{c:c:c:c:c}n day & sales & Model & Error & Cummulated \space Err\\ \hlinen 1 & 60 & 60 &0& 0 \\ \hlinen 2 & 70 & 70 &0& 0 \\ \hlinen 3 & 110 & \frac{60+70+110}{3}=80 &110-80=30& 0+30=30 \\ \hlinen 4 & 80 & \frac{70+110+80}{3}=88.6 &86.6-80=6.6& 30+6.6=36.6 \\ \hlinen5 & 70 & \frac{110+80+70}{3}=88.6 &86.6-70=16.6& 36.6 +16.6=53.2\\ \hlinen6 & 85 & \frac{80+70+85}{3}=78.3 &85-78.3=6.6& 53.2+6.6=60\\ \hlinen7 & 115 & \frac{70+85+115}{3}=90 &115-90=25& 60+25=85\\ \hlinen8 & 105 & \frac{85+115+105}{3}=101.6 &105-101.6=3.4& 85+3.4=88.4\\ \hlinen9 & 65 & \frac{115+105+65}{3}=95 &95-65=30& 88.4+30=118.4\\ \hlinen10 & 75 & \frac{105+65+75}{3}=81.6 &81.6-75=6.6&118.4+6.6=125\\ \hlinen11 & 95 & \frac{65+75+95}{3}=78.3 &95-78.3=16.6&125+6.6=141.6\\ \hlinen12 & 85 & \frac{75+95+85}{3}=85 &85-85=0&141.6+0=141.6\\ \hlinen\end{array}”

MEAN ERROR =“141.6/12=11.8”

1.1 4** day moving average**

“\def\arraystretch{1.5}n \begin{array}{c:c:c:c:c}n day & sales & Model & Error & Cummulated \space Err\\ \hlinen 1 & 60 & 60 &0& 0 \\ \hlinen 2 & 70 & 70 &0& 0 \\ \hlinen 3 & 110 & 110 &0& 0+30=30 \\ \hlinen 4 & 80 & \frac{60+70+110+80}{4}=80 &80-80=0& 0+0=0 \\ \hlinen5 & 70 & \frac{70+110+80+70}{4}=82.5 &82.5-70=12.5& 0 +12.5=12.5\\ \hlinen6 & 85 & \frac{110+80+70+85}{4}=86.25 &86.25-85=1.25& 12.5.+1.25=13.75\\ \hlinen7 & 115 & \frac{80+70+85+115}{4}=87.5 &115-87.5=1.27.5& 13.75+27.5=41.25\\ \hlinen8 & 105 & \frac{70+85+115+105}{4}=93.75 &105-93.75=11.25.5& 41.25+11.25=52.5\\ \hlinen9 & 65 & \frac{85+115+105+65}{4}=92.5 &92.5-65=27.5& 52.5+27.5=80\\ \hlinen10 & 75 & \frac{115+105+65+75}{4}=90 &90-75=15& 80+15=95\\ \hlinen11 & 95 & \frac{105+65+75+95}{4}=85 &95-85=10& 95+10=105\\ \hlinen12 & 85 & \frac{65+75+95+85}{4}=80 &85-80=5& 105+5=110\\ \hlinenn\end{array}”

MEAN ERROR =“110/12=9.17”

Thus method of 4** **day moving average more exactly correspond to data than method of 3 day moving average.

1,3** **3 day weighted moving average** **analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data

“\def\arraystretch{1.5}nn \begin{array}{c:c:c:c:c}nn day & sales & Model & Error & Cummulated \space Err\\ \hlinenn 1 & 60 & 60 &0& 0 \\ \hlinen 2 & 70 & 70 &0& 0 \\ \hlinen 3 & 110 & 60\cdot 0.2+70\cdot0.3+110\cdot 0.5=88 &110-88=22& 0+22=22 \\ \hlinen 4 & 80 & 70\cdot 0.2+110\cdot0.3+80\cdot 0.5=87 &87-80=7& 22+7=29 \\ \hlinen 5 & 70 & 110\cdot 0.2+80\cdot 0.3+70\cdot 0.5=81 &81-70=11& 29+11=40 \\ \hlinen 6 & 85 & 80\cdot 0.2+70\cdot 0.3+85\cdot 0.5=79.5 &85-79.5=5.5& 40 +5.5=45.5\\ \hlinen 7 & 115 & 70\cdot 0.2+85\cdot 0.3+115\cdot 0.5=97 &115-97=18& 45.5+18=63.5\\ \hlinen 8 & 105 & 85\cdot 0.2+115\cdot 0.3+105\cdot 0.5=104 &105-104=1& 63.5=1=64.5\\ \hlinen 9 & 65 & 115\cdot 0.2+105\cdot 0.3+65\cdot 0.5=87 &87-65=22&64.5+22=86.5\\ \hlinen 10 & 75 & 105\cdot 0.2+65\cdot 0.3+75\cdot 0.5=78 &78-75=3&86.5+3=89.5\\ \hlinen11 & 95 & 65\cdot 0.2+75\cdot 0.3+95\cdot 0.5=83 &95-83=12&89.5+12=101.5\\ \hlinen12 & 85 & 75\cdot 0.2+95\cdot 0.3+85\cdot 0.5=88 &86-85=1&101.5+1=102.5\\ \hlinen\end{array}”

MEAN ERROR =“102.5/12=8.5”

Thus last merhod is a best from three considered.

1.4 exponential smoothing** **analysis with **a = 0.3**

“\def\arraystretch{1.5}nn \begin{array}{c:c:c:c:c}nn day & sales & Model & Error & Cummulated \space Err\\ \hlinenn 1 & 60 & 60 &0& 0 \\ \hlinen 2 & 70 & 0.7\cdot 60 +0.3\cdot 70=63&70-63=7& 0+7=7 \\ \hlinen 3 & 110 & 0.7\cdot 63 +0.3\cdot 110=77.1&110-77.1=32.9& 7+32.9=39.9 \\ \hlinen4 & 80 & 0.7\cdot 77.1 +0.3\cdot 80=77.97&80-77.97=2.03& 39.9 +2.03=41.93\\ \hlinen5 & 70 & 0.7\cdot 77.97 +0.3\cdot 70=75.579&75.579-70=5.579& 41.93+5.579=47.509\\ \hlinen6 & 85 & 0.7\cdot 75.579 +0.3\cdot 85=78.4053&85-78.4053=6.5947&47.509+6.5947=54.1037\\ \hlinen7 & 115 & 0.7\cdot 78.4053 +0.3\cdot 115=89.383&115-89.383=25.616&54.1037+25.616=79.720\\ \hlinen8 & 105 & 0.7\cdot 89.383 +0.3\cdot 105=94.069&105-94.069=10.931&79.720+9.931=90.651\\ \hlinen9 & 65 & 0.7\cdot 94.069 +0.3\cdot 65=85.348&85.348-65=20.348&90.651+20.348=110.999\\ \hlinen10 & 75 & 0.7\cdot 85.348 +0.3\cdot 75=82.244&82.244-75=7.244&110.999+7.244=118.243\\ \hlinen11 & 95 & 0.7\cdot 82.244 +0.3\cdot 95=86.07&95-86.07=9.929&118.243+8,929=127.17\\ \hlinen12 & 85 & 0.7\cdot 86.07 +0.3\cdot 85=85.749&85.749-85=0.749&127.17+0.749=127.92\\ \hlinen\end{array}”

MEAN ERROR =“127.9/12=10.7”

Thus the method of exponential smoothing** **analysis with **a = 0.3 **gives the bigger mean error than best method of day weighted moving average** **analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data and finally we take lass method as a best at whole.

f. Forecast day 13 sales of fishing rods using the model chosen in part (e)

we apply /the best method of * ‘day weighted moving average** **analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data ‘,*

use data of last three days and calculate forecast value:

“75\cdot 0.2+95\cdot 0.3+85\cdot 0.5=88”

Thus in 13 day we expect 88 sales.

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