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3500 1000 -1500 7000 2000 -9000 0.4 0.3 0.3 De ci si on Al te rn at iv es Heavy, s1 Moderate, s2 Light, s3 Bl ad e At ta ch me nt , d1 Ne w Sn ow pl ow , d2 Martin's Se season. Mar new heavy-d alternative if the snow profits hav for additi

Linear Programme - Solution ToMartin Service Station

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Page 1: Linear Programme - Solution ToMartin Service Station

3500 1000 -1500

7000 2000 -9000

0.4 0.3 0.3

heavy Moderate Light

Normal 0.7 0.8 0.9

Decision Alternatives

Heavy, s1

Moderate, s2 Light, s3

Blade Attachment, d1

New Snowplow, d2

Conditional Probability For A Given State of Nature

Test Results 

Martin's Service Station is considering entering the snowplowing business for the coming winter season. Martin can purchase either a snowplow blade attachment for the station's pick-up truck or a new heavy-duty snowplow truck. Martin has analyzed the situation and believes that either alternative would be a profitable investment if the snowfall is heavy. Smaller profits would result if the snowfall is moderate, and losses would result if the snowfall is light. The following profits have been determined. Martin is willing to pay $50 for additional information on the weather.

Page 2: Linear Programme - Solution ToMartin Service Station

Unreasonable 0.3 0.2 0.1

Answerpart a - bayes probability- 5 marks

we use bayes theorem see text for steps

S1=normal

State of nature (1) prior prob(2)

H 0.4 0.7M 0.3 0.8L 0.3 0.9

S2 =unreasonable

State of nature (1) prior prob(2)

H 0.4 0.3M 0.3 0.2L 0.3 0.1

part a

981.65

1,455 0.79

1,455 forecast 0.21 3235.714285714

The probabilities for the states of nature are P(s1) = 0.4, P(s2) = 0.3, and P(s3) = 0.3. Suppose that Martin decides to wait until September before making a final decision. Estimates of the probabilities associate with a normal (

condit prob(S1/h,), S1/m,S1/l (3)

condit prob(S2/H,), S2/M,S2/L (3)

Page 3: Linear Programme - Solution ToMartin Service Station

No test

no FORECAST

12501250

New plow

700

part bGet the forecast, If normal get the blade, if unreasonable get the plow. The expected value is 1455

part cEVSI EWSI+ COST OF SURVEY - EWOSI

1,455 50 1250 255part dEVPI EWPI -EWOPI REVENUE

2950

EWOPI 1,250

EVPI 1,700part eEfficiency evsi/evpi 0.15

EWPI ( assume the best of all situation)

=0.4*7000+0.3*2000+0.3*-1500

blade

Page 4: Linear Programme - Solution ToMartin Service Station

Martin's Service Station is considering entering the snowplowing business for the coming winter season. Martin can purchase either a snowplow blade attachment for the station's pick-up truck or a new heavy-duty snowplow truck. Martin has analyzed the situation and believes that either alternative would be a profitable investment if the snowfall is heavy. Smaller profits would result if the snowfall is moderate, and losses would result if the snowfall is light. The following profits have been determined. Martin is willing to pay $50 for additional information on the weather.

Page 5: Linear Programme - Solution ToMartin Service Station

0.28 0.350.24 0.300.27 0.34

0.79 1.00

S2 =unreasonable

0.12 0.570.06 0.290.03 0.140.21 1.00

Payoffsadjust for SOP

982 H/N 0.35 3,450

M/N 0.30 950

L/N 0.34 -1,550

0new plow

-37H/N 0.35443 6,950M/N 0.303797 1950L/N 0.341772 -9,050

2,021

H/U 0.57 3,450

M/U 0.29 950

0 L/U 0.14 -1,550

) = 0.3. Suppose that Martin decides to wait until September before making a final decision. Estimates of the probabilities associate with a normal (

Joint prob(4)

posterior prob=col(4)/.79

Joint prob(4)

posterior prob=(4)/.21

Page 6: Linear Programme - Solution ToMartin Service Station

New plow

3,236

H/U 0.57 6,950M/U 0.29 1950L/U 0.14 -9,050

HM 0.4 3,500 1400L 0.3 1000 300

0.3 -1,500 -4501250 expected value

0 expected valueH 0.4 7,000 2800M 0.3 2000 600L 0.3 -9,000 -2700

Get the forecast, If normal get the blade, if unreasonable get the plow. The expected value is 1455 700 expected value

Page 7: Linear Programme - Solution ToMartin Service Station

) = 0.3. Suppose that Martin decides to wait until September before making a final decision. Estimates of the probabilities associate with a normal (N) or unreasonably cold (U) September are as follows

Page 8: Linear Programme - Solution ToMartin Service Station

) September are as follows

Page 9: Linear Programme - Solution ToMartin Service Station

payoffs states of nature maximax maximin

alternatives stable moderate high max payoff choice min payoff

Houses 70000 30000 10000 70000 10000

Shopping center80000 20000 5000 80000 best 5000

Leasing 40000 30000 20000 40000 20000

payoffs states of nature equally likely hurwicz

alternatives stable moderate high max payoff choice realism

Houses 70000 30000 10000 36666.666667 best 40000

Shopping center 80000 20000 5000 35000 42500

Leasing 40000 30000 20000 30000 30000

alpha

regrets states of nature equally likely

alternatives good fair poor max payoff choice

Houses 10000 0 10000 10000 best

Shopping center 0 10000 15000 15000

Leasing 40000 0 0 40000

f)probablities 0.30 0.40 0.30

Decsion good fair poor

Houses 21,000 12,000 3,000 36,000 best

Shopping center 24,000 8,000 1,500 33,500

Leasing 12,000 12,000 6,000 30,000

0 moveEWOPI

g most to be paid for additional informationuse regret table

0.30 0.40 0.30

Decsion market down

Houses 3,000 0 3,000 6,000 best

Shopping center 0 4,000 4,500 8,500

Leasing 12,000 0 0 12,000

Expected value

Expected value

market up

market stable

Expected regret

Min expeted regret

Page 10: Linear Programme - Solution ToMartin Service Station

or EVPI =ewpi -ewopi

ewopi=382part f 36,000

ewpi =0.3*80000+0.4*30000+0.3*20000 42,000

EVPI =ewpi -ewopi (42000-36000)= 6,000

Page 11: Linear Programme - Solution ToMartin Service Station

The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company's equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:

Material Costschoice Decision Stable

Houses $70,000 105,000

best Leasing 40,000

Determine the best decision, using the following decision criteria.choice

best

0.5

Alternative stable moderate high Equal likelihood

Houses 70000 30000 10000

80000 20000 5000

Leasing 40000 30000 20000

Shopping center

a.       Maximax

b.      Maximin

c.       Minimax regret

d.      Hurwicz (a = .2)

Shopping center

Page 12: Linear Programme - Solution ToMartin Service Station

The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company's equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:

Material CostsIncrease$30,000 20,000

40,000

Determine the best decision, using the following decision criteria.

Minimax regret

Hurwicz (a = .2)

Page 13: Linear Programme - Solution ToMartin Service Station

The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company's equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:

Page 14: Linear Programme - Solution ToMartin Service Station

The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company's equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:

Page 15: Linear Programme - Solution ToMartin Service Station

The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company's equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:

Page 16: Linear Programme - Solution ToMartin Service Station

The owner of the Columbia Construction Company must decide between building a housing development, constructing a shopping center, and leasing all the company's equipment to another company. The profit that will result from each alternative will be determined by whether material costs remain stable or increase. The profit from each alternative, given the two possibilities for material costs, is shown in the following payoff table:

Page 17: Linear Programme - Solution ToMartin Service Station

3 semester MA

Month Fund Priceforecast error

1 63 1/42 60 1/83 61 3/44 64 1/4 61.71 2.54 2.54 6.46 3.96%

5 59 3/8 62.04 (2.67) 2.67 7.11 4.49%

6 57 7/8 61.79 (3.92) 3.92 15.34 6.77%

7 62 1/4 60.50 1.75 1.75 3.06 2.81%

8 65 1/8 59.83 5.29 5.29 28.00 8.13%

9 68 1/4 61.75 6.50 6.50 42.25 9.52%

10 65 1/2 65.21 0.29 0.29 0.09 0.45%

11 68 1/8 66.29 1.83 1.83 3.36 2.69%

12 63 1/4 67.29 (4.04) 4.04 16.34 6.39%

13 64 3/8 65.63 (1.25) 1.25 1.56 1.94%

14 68 5/8 65.25 3.38 3.38 11.39 4.92%

15 70 1/8 65.42 4.71 4.71 22.17 6.71%

16 72 3/4 67.71 5.04 5.04 25.42 6.93%

17 74 1/8 70.50 3.63 3.63 13.14 4.89%

18 71 3/4 72.33 (0.58) 0.58 0.34 0.81%

19 75 1/2 72.88 2.63 2.63 6.89 3.48%

20 76 3/4 73.79 2.96 2.96 8.75 3.85%

21 74.67 AVERAGE 3.12 12.45 4.63%

MAD mse MAPE

weighted three-quarter

Month Fund Priceforecast error

1 63 1/4 0.1

2 60 1/8 0.3

3 61 3/4 0.6

4 64 1/4 61.4125 2.84 2.84 8.05 4.42%

5 59 3/8 63.0875 (3.71) 3.71 13.78 6.25%

6 57 7/8 61.075 (3.20) 3.20 10.24 5.53%

7 62 1/4 58.9625 3.29 3.29 10.81 5.28%

8 65 1/8 60.65 4.47 4.47 20.03 6.87%

9 68 1/4 63.5375 4.71 4.71 22.21 6.90%

10 65 1/2 66.7125 (1.21) 1.21 1.47 1.85%

11 68 1/8 66.2875 1.84 1.84 3.38 2.70%

12 63 1/4 67.35 (4.10) 4.10 16.81 6.48%

absolute error

Squared error

Absolute % error

absolute error

Squared error

Absolute % error

Page 18: Linear Programme - Solution ToMartin Service Station

13 64 3/8 64.9375 (0.56) 0.56 0.32 0.87%

14 68 5/8 64.4125 4.21 4.21 17.75 6.14%

15 70 1/8 66.8125 3.31 3.31 10.97 4.72%

16 72 3/4 69.1 3.65 3.65 13.32 5.02%

17 74 1/8 71.55 2.58 2.58 6.63 3.47%

18 71 3/4 73.3125 (1.56) 1.56 2.44 2.18%

19 75 1/2 72.5625 2.94 2.94 8.63 3.89%

20 76 3/4 74.2375 2.51 2.51 6.31 3.27%

21 75.875AVERAGE 2.98 10.18 4.46%

next period MAD mse MAPE

exponential

Month Fund Priceforecast error

1 63 1/4 63.25

2 60 1/8 63.25 (3.13) 3.13 9.77 5.20%

3 61 3/4 62.00 (0.25) 0.25 0.06 0.40%

4 64 1/4 61.90 2.35 2.35 5.52 3.66%

5 59 3/8 62.84 (3.46) 3.46 12.01 5.84%

6 57 7/8 61.45 (3.58) 3.58 12.81 6.18%

7 62 1/4 60.02 2.23 2.23 4.96 3.58%

8 65 1/8 60.91 4.21 4.21 17.74 6.47%

9 68 1/4 62.60 5.65 5.65 31.94 8.28%

10 65 1/2 64.86 0.64 0.64 0.41 0.98%

11 68 1/8 65.12 3.01 3.01 9.06 4.42%

12 63 1/4 66.32 (3.07) 3.07 9.42 4.85%

13 64 3/8 65.09 (0.72) 0.72 0.51 1.11%

14 68 5/8 64.80 3.82 3.82 14.59 5.57%

15 70 1/8 66.33 3.79 3.79 14.38 5.41%

16 72 3/4 67.85 4.90 4.90 24.01 6.74%

17 74 1/8 69.81 4.32 4.32 18.62 5.82%

18 71 3/4 71.54 0.21 0.21 0.05 0.30%

19 75 1/2 71.62 3.88 3.88 15.04 5.14%

20 76 3/4 73.17 3.58 3.58 12.80 4.66%

21 74.60 average 2.99 11.25 4.45%

alpha 0.4 MAD mse MAPE

MAD mse MAPE3 MA 3.117647 12.45118 4.63%weighted average 2.982353 10.18482 4.46%

absolute error

Squared error

Absolute % error

Page 19: Linear Programme - Solution ToMartin Service Station

exponential 2.989146 11.24746 4.45%

Using MAD the weighted average is most accurateUsing MSE the weighted average is most accurateusing MAPE the exponential is most accurate

Page 20: Linear Programme - Solution ToMartin Service Station

(a) F4 = 400.000, F5 = 406.67, F6 = 423.33, F7 = 498.33, F8 = 521.67, F9 = 571.67; (b) F2 = 400.00, F3 = 410.00, F4 = 398.00, F5 = 402.40, F6 = 421.92, F7 = 452.53, F8 = 460.00, F9 = 498.02; (c) 3-sem. MAD = 80.33, exp. smooth. MAD = 87.16

Page 21: Linear Programme - Solution ToMartin Service Station

= 521.67, F9 = 571.67; (b) F2 = 400.00, F3 = 410.00, F4 = 398.00, F5 = 402.40, F6 = 421.92, F7 = 452.53, F8 = 460.00, F9 = 498.02; (c) 3-sem. MAD = 80.33, exp. smooth. MAD = 87.16

Page 22: Linear Programme - Solution ToMartin Service Station

= 460.00, F9 = 498.02; (c) 3-sem. MAD = 80.33, exp. smooth. MAD = 87.16