Om- Sandeep Raina b54!27!06 OM-I Week1 s10-11 Conceptual Exercise

om- sandeep raina b54-27-06_OM-I_Week1_s10-11_Conceptual Exercise

Question No 7 : Chasewood Apartments…………

Semester

1 7.2 2912 6.3 2283 6.7 2524 7.0 2655 6.9 2706 6.4 2407 7.1 2888 6.7 246

FORECASTING MODEL

Next semester enrolments expected 6.6

a. y=67.238x-196.38 247.3908 247

b.R2=0.927 92.7%

University Enrolments (thousands)

No. of units Leased

Using Regression Equation we can find the number of apartments leased "y" for 6.6k expected enrolments

R2 explains the variation in y w.r.t. to variation in y w.r.t. variation in x. Pvalue <0.05; hence the model is accurate and not random

6.2 6.4 6.6 6.8 7.0 7.2 7.40

50

100

150

200

250

300

350

f(x) = 67.2384219554x - 196.3807890223R² = 0.9270350804

No. of units Leased


Variation% in the Model 7.30%Variation in No of units 18.031 +-18 unitsActual units leased can vary from 229-265 units

c.

Hence the forecasting model will predict upto 92.7% the output

The model is quite accurate with Rsqaure=0.927 and Pvalue <0.05. He is also now aware that he has sufficient 300 units, more than max 265 required in the range and there will not be any shortage.Hence the model can be useful for the Manager to estimate the no of units to be kept ready for lease and associated works involved.


SUMMARY OUTPUT

Regression StatisticsMultiple RR SquareAdjusted RStandard ErObservatio

ANOVA

RegressionResidualTotal

CoefficientsInterceptUniversity

say

6.2 6.4 6.6 6.8 7.0 7.2 7.40

50

100

150

200

250

300

350

f(x) = 67.2384219554x - 196.3807890223R² = 0.9270350804

No. of units Leased


SUMMARY OUTPUT

Regression Statistics0.9628270.9270350.9148746.574158

8

df SS MS F Significance F1 3294.683 3294.683 76.2313 0.0001256 259.3173 43.219557 3554

CoefficientsStandard Error t Stat P-value Lower 95%Upper 95%Lower 95.0%Upper 95.0%-196.3808 52.32264 -3.753266 0.009473 -324.4097 -68.3519 -324.4097 -68.351967.23842 7.701066 8.731054 0.000125 48.39459 86.08225 48.39459 86.08225


Upper 95.0%


Q9 a large health maint organisation (HMO)…

a) Develop a moving average forecast for the past 10 months (Month 15-24) for AP=2,4,6 and 8

forecastsMonth No of Lawsuits AP=2months1 162 253 164 245 386 467 548 529 5110 5611 6712 4513 5314 6115 55 57.016 69 58.017 63 62.018 57 66.019 48 60.020 55 53.021 61 52.022 51 58.023 56 56.024 53 54.0

b Which AP results in the mean absolute forecasting error……

AP=2monthsMonth No of Lawsuits forecast15 55 57.016 69 58.017 63 62.018 57 66.019 48 60.020 55 52.521 61 51.522 51 58.0


23 56 56.024 53 53.5

total absolute deviationmean absolute deviation

AP=6months results in lowest forecasting error as MAD is least compared to other Aps. (2,4,8)Hence this AP=2 is recommended as forecasting will be more accurate, deviation will be least from the actual outcome, so that Cindy is better prepared

c Using your recommended forecast….FORECASTING FOR MONTH 25 WILL BE AS BELOW: F25 =AVERAGE(C56:C61) 54.0


Develop a moving average forecast for the past 10 months (Month 15-24) for AP=2,4,6 and 8

forecastsAP=4months AP=6months AP=8months

57.0 56.0 55.054.0 57.0 55.060.0 59.0 58.062.0 58.0 59.061.0 60.0 59.060.0 59.0 57.056.0 58.0 58.056.0 59.0 59.054.0 56.0 58.056.0 55.0 58.0

Which AP results in the mean absolute forecasting error……

AP=4months AP=6months AP=8monthsabs deviation forecast abs deviation forecast abs deviati forecast2.0 56.5 1.5 55.5 0.5 54.911.0 53.5 15.5 56.2 12.8 55.01.0 59.5 3.5 58.3 4.7 57.19.0 62.0 5.0 57.7 0.7 58.612.0 61.0 13.0 59.7 11.7 58.82.5 59.3 4.3 58.8 3.8 56.49.5 55.8 5.3 57.8 3.2 57.67.0 55.3 4.3 58.8 7.8 58.6

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240

10

20

30

40

50

60

70

80

Chart Title

No of Lawsuits AP=2months AP=4months AP=6months AP=8months


0.0 53.8 2.3 55.8 0.2 57.40.5 55.8 2.8 54.7 1.7 57.554.5 57.3 47.05.5 5.7 4.7

AP=6months results in lowest forecasting error as MAD is least compared to other Aps. (2,4,8)Hence this AP=2 is recommended as forecasting will be more accurate, deviation will be least from the actual outcome, so that Cindy is better prepared


abs deviation0.114.05.91.610.81.43.47.6

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240

10

20

30

40

50

60

70

80

Chart Title

No of Lawsuits AP=2months AP=4months AP=6months AP=8months


1.44.550.65.1

Hence this AP=2 is recommended as forecasting will be more accurate, deviation will be least from the actual outcome, so that Cindy is better prepared


Ques no 13A toy company ..

forecasts

a) Month α=0.1 α=0.3 α=0.5

1 0.39 0.390 0.390 0.3902 0.41 0.390 0.390 0.3903 0.45 0.392 0.396 0.4004 0.44 0.398 0.412 0.4255 0.4 0.402 0.421 0.4336 0.41 0.402 0.414 0.4167 0.38 0.403 0.413 0.4138 0.36 0.400 0.403 0.3979 0.35 0.396 0.390 0.37810 0.38 0.392 0.378 0.36411 0.39 0.391 0.379 0.37212 0.43 0.390 0.382 0.38113 0.37 0.394 0.396 0.40614 0.38 0.392 0.389 0.38815 0.36 0.391 0.386 0.38416 0.39 0.388 0.378 0.372

forecasts

b)Month α=0.1 α=0.3

6 0.41 0.402 0.417 0.38 0.403 0.02 0.418 0.36 0.400 0.04 0.409 0.35 0.396 0.05 0.39

10 0.38 0.392 0.01 0.3811 0.39 0.391 0.00 0.3812 0.43 0.390 0.04 0.3813 0.37 0.394 0.02 0.4014 0.38 0.392 0.01 0.3915 0.36 0.391 0.03 0.3916 0.39 0.388 0.00 0.38

Total Absolute deviation 0.23Mean Absolute dev 0.023

c) For Computing F17 we will use the best alpha as 0.1F17=

0.39

Platic PelletsPrice/Pount

Platic PelletsPrice/Pount

abs deviation

So α=0.1 leads to leads to least mean absolute deviation of 0.023 from weeks 7-16 better than alpha of 0.3 &0.5

F16+α(A16-F16)


α F2=

0.1 F1+α(A1-F1) 0.39

0.3 F1+α(A1-F1) 0.390.5 F1+α(A1-F1) 0.39

forecasts

α=0.5

0.420.03 0.41 0.030.04 0.40 0.040.04 0.38 0.030.00 0.36 0.020.01 0.37 0.020.05 0.38 0.050.03 0.41 0.040.01 0.39 0.010.03 0.38 0.020.01 0.37 0.020.25 0.270.025 0.027

abs deviation

abs deviation

α=0.1 leads to leads to least mean absolute deviation of 0.023 from weeks 7-16 better than alpha of 0.3 &0.5

Om- Sandeep Raina b54!27!06 OM-I Week1 s10-11 Conceptual Exercise

Documents

Transcript of Om- Sandeep Raina b54!27!06 OM-I Week1 s10-11 Conceptual Exercise