c. Management believes when the average weekly demand for beds
will be more than 400, the ER will collapse. When does this
situation expected to occur with 90% confidence, based on the
Holt's forecasting? Use the model you found in question 'b' above.
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200
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240
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300
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360
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400
f(x) = 3.47842401500938 x + 262.192307692308
Series
Column C
Linear (Column C)

a.
The equation appears in the graph below. Here are the calculations for the next 6 weeks:
Week
41
Demand = 3.478*41+262.19
=
404.788
42
Demand = 3.478*42+262.20
=
408.266
43
Demand = 3.478*43+262.21
=
411.744
44
Demand = 3.478*44+262.22
=
415.222
45
Demand = 3.478*45+262.23
=
418.7
b.
To compare performance we need to determine MSE of the two models. For the regression
this value is readily available in the template. MSE(Regression) = 222.25. For the Holt's model
we need to first optimize its parameters alpha and gamma. We use "Data Table" for alpha and gamma.
After solving we get: alpha = .25. and
gamme = .35
I used margins of 0.05.
MSE(Holt's) =
59.67
The Holt's model performs much better.
c.
This situation might occur in 5 weeks. At this time the upper limit of the 95% prediction interval
will exceed 400. Note that the Holt's actual forecast F(t+k) remains under 400 for more than 12 periods ahead
but when adding the confidence level consideration, it becomes 4 week.
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400
450
Chart Title
Column J
Column K
Column E
Column B
Holt's forecasts

Problem 20 from the text pg. 410
Year
Expenses %
1
13.9
a. It appears that a negative trend is present.
2
12.2
3
10.5
4
10.4
5
11.5
6
10
7
8.5
b. The model parameters can be found from the graphical tool directly:
The Expenses percent = -0.7*t + 13.8
c. F(t=8) = -0.7*8+13.8
d. F(t) = -.7*t + 13.8 = 5
t = (5 - 13.8)/(-.7) =
12.57143 years
Additional Analysis
Now let us run the Holt's model on this data, and answewr the same quaestions. We'll p
and then complete the analysis with the template.
Let us use alpha = .3 and gamma
L2 = 12.2.
T2 = 12.2-13.9 =
-1.7
F3 = 12.2+(-1.7) =
10.5
L3= .3*10.5 + 0.7*10.5 =
10.5
T3 = .2*(10.5-12.2)+.8*(-1.7) =
-1.7
F4 = 10.5 +(-1.7) =
8.8
L5 = .3*11.4 +.7*8.8 =
9.58
T5 = .2*(9.58-10.5)+.8*(-1.7) =
-1.544
F6 = 9.58+(-1.544) =
8.306
c. As of period 7 we have (from the termplate)
L7 =
7.444
T7 =
-1.131
F8 = L7 + T7 =
6.313
d. F(t) = 7.444+(-1.131*t) = 5. Now solve for 't'.
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2
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5
7
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15
f(x) = − 0.7 x + 13.8
Chart Title

practice a few stages at the beginning
= .2.

Problem 21 from the textbook (pg. 411)
Year
Cost/Unit
1
20
2
24.5
3
28.2
4
27.5
5
26.6
6
30
7
31
8
36
a. The pattern exhibits positive linear trend
b. The regression prediction equation is: Unit Cost = 1.7738t+19.993
c. The average cost increase is 1.7738 per year.
d. The unit cost next year is expected to be F(9) = 1.7738(9) + 19.993
Additional analysis
(i) You were offerred to use erxponential smoothing for the foreast of this time series.

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- Linear Regression, Regression Analysis, Holt, R Square