Decision tree and decision analysis

7/28/2019 Decision tree and decision analysis

1/20

ASIGNMENT NO: 1

Name: Mohammad Abdullah

Enrollment: 2010-E-37

Class: MBA (5th quarter)

Course: Quantitative Analysis.

Topic: Decision Analysis...

Submitted To: Mr. Taimoor Shah

Submission Date: 21-06-2012

INSTITUTE OF MANAGEMENT SCIENCES UNIVERSITY OF

BALUSHISTAN QUETTA


2/20

Question No: 2

Finance manger's optimal policy would be Drill upto 20 meters because it has the lowest expected cost

No water P= 0.8$ 275 000

$ 125 000

1

2

Water Struck P= 0.2

Do not Drill

Drill upto 25

Drill upto 20 meter

Do not drill

No water P= 0.30

Water struck P= 0.70

$100,000

2n

1st

Cost = 245 000$ 150,000

$ 250,000

$ 245,000

Cost = $ 77,500

$ 77,500


3/20

Problem No: 3-27

a. Decision Table.

State of nature/ activity

Size of first stationGoodmarket($)

Fairmarket($)

PoorMarket($)

Small$

50,000.00$

200,000.00$

(10,000.00)

medium$

80,000.00$

30,000.00$

(20,000.00)

large$

100,000.00$

30,000.00$

(40,000.00)

very large$

300,000.00$

25,000.00$

(160,000.00)

b. Optimistic decision criterion Maximax.

Maximax = {Max in 1st row,Max in 2nd row,Max in 3rd row,Max in 4th row} = {Max}

Maximax = {200000,80000,100000,300000} = {300000}

Conclusion: Optimal decision is to build a very large size first station.

c. Passimistic decision criterion (Maximin).

Maximin = {-10000, -20000, -40000, -160000} = {-10000}

Conclusion: Passimistic decision Is to build a small size of first station.

d. Equallly likely decision criterion (Laplace).

Small 50000+20000+(-10000)/3=60000/3=20000

medium 80000+30000+(-20000)/3=90000/3=30000

large 100000+30000+(-40000)/3=90000/30=30000

very large 300000+25000+(-160000)/3=165000/3=55000

Laplace={20000,30000,30000,55000}={55000}

Conclusion: Equally likely decision is to build a Very large size of f irst station.

e. Realism decision criterion (Hurwicz). a = 0.8

Hurwicz={Max in row x + Min in row (1-)}

Small = {200000 x 0.8 = (1- 0.8) (-10000)} = 158000

Medium = {80000 x 0.8 = (1- 0.8) (-20000)} = 60000

Large = {100000 x 0.8 = (1- 0.8) (-40000)} = 72000

Very large = {300000 x 0.8 = (1- 0.8) (-160000)} = 208000

Hurwicz={158000,60000,72000,208000} = {208000}

Conclusion: Laplace decision Is to build a Very large size of first station.

f. Loss Table

State of nature/ activity

Size of first stationGoodmarket($)

Fairmarket($)

PoorMarket($)

Small$

250,000.00$

10,000.00$

10,000.00

medium$

220,000.00 $ -$

20,000.00

large

$

200,000.00 $ -

$

40,000.00

very large $ - $ $


4/20

5,000.00 160,000.00

g. Minimax Regret decision criterion.

Minimax={250000,220000,200000,160000} = {160000}

Conclusion: Minimax regret decision Is to build a Very large size of first station.

Question No: 2

1st decision pay off tableDecision alternatives state of nature /activity

highdemand

lowdemand

Expand small plant 2.5 m (-3.1 m)

Do not expand small plant 1.5 m 1.1 m

Probability 0.85 0.15

Last decision pay off table

Decision alternatives state of nature /activityhighdemand

lowdemand

construct full size pland 5 m (-2 m)construct small size plant 1.66 m 1 m

Probability 0.85 0.15

Full size plant high demand

cost = 5 million

payoff =1million/annum

period = 10 years

Net pay off = {(1 x 10)} - {5}

10 - 5 = $ 5 million

Full size plant low demand

cost = 5 million

payoff = 0.3 million/annum

period = 10 years

Net pay off ={(0.3 x 10)} -{5}

3 - 5 = $ -2 million

Expanded small plant high demand

cost = $ 1 million

expansion cost = $ 4.2 million

High demand yeild = $ 0.25 million/annumHigh demad period = 2 years


5/20

expanded high demand yeild = $ 0.9 million/ annum

expanded high demand Period = 8 years

Net pay off ={(0.25 x 2 + 0.9 x 8)} - {1+4.2}

7.7 - 5.2 = $ 2.5 million

Expanded small plant low demand

cost = $ 1 million

expansion cost = $ 4.2 million

High demand yeild = $ 0.25 million/annum

High demad period = 2 years

expanded low demand yeild = $ 0.2 million/ annum

expanded low demand Period = 8 years

Net pay off = {(0.25 x 2 + 0.2 x 8)} - {1 + 4.2}

7.7 - 5.2 = $ -3.1 million

Do not expand small plant high demand

cost = 1 million

High demand yeild = 0.25 million/annumHigh demad period = 10 years

Net pay off = {(0.25 x 2 + 0.25 x 8)} - {1}

2.5 - 1 = $ 1.5 million

Do not expand small plant low demand

cost = $ 1 million

High demand yeild = $ 0.25 million/annum


Low demad yield = $ 0.2 million/ annum

Low demad Period = 8 years

Net pay off = {(0.25 x 2 + 0.2 x 8)} - {1}

2.1 - 1 = $ 1.1 million

Small plant low demand

cost = $ 1 million

Payoff = $ 0.2 million/annum


Period = 10 years

Net pay off ={(0.2 x 10)} -{1}

2 - 1 = $ 1 million

Decision Table

EMV AT Nod- 4 = {(0.85 x 2.5 ) + (0.15 x -3.1)} = {$ 1.66 million}

EMV AT Nod- 3 = {(0.85 x 1.5 ) + (0.15 x 1.1)} = {$ 1.44 million}

EMV AT Nod- 2 = {(0.85 x 5 ) + (0.15 x -2)} = {$ 3.95 million}

EMV AT Nod- 1 = {(0.85 x 1.66 ) + (0.15 x 1)} = {$ 1.56 million}


6/20

Question No: 2

No water struck

Out source cost = Rs. 150000

Drilling cost = Rs. 5000/meter

Drill upto 25 meters Cost=(25 merters x Rs. 5000) + 150,000 = Rs.275,000

Water struck

Drilling cost =Rs.5000/meter

Drill upto 25 meters Cost= 25 merters x Rs. 5000 = Rs. 125,000

Do not drill further

Outsource cost= Rs.150,000


Drill upto 20 meters= (Rs. 5000/meter x 20 Meters) + Rs. 150,000 =Rs. 250,000

Water struck at 20 meters P= 0.70


Drill upto 20 meters= (Rs. 5000/meter x 20 Meters) = Rs. 100,000

Do not drill any well

Out source cost= Rs. 150,000

Analysis Table

Expected cost AT Node 2 = (275,000 x 0.8) + (125,000 x 0.2) = Rs. 245,000

Decision at the decision point 2= Is to drill upto 25 meters i.e Rs. 245,000

Expected cost AT Node 1 = (25,000 x 0.3) + (100,000 x 0.7) = Rs. 77,500Decision at the decision point 1= Is to drill upto 20 meters i.e Rs. 77,500


7/20

The optimal policy would be to construct full size plant because it has the highiest ecpected monitory value

High demand P=$ 2.5 Million

$ -3.1 Million

2

1

3

4

$ 1.5 Million

$ 1.1 Million

Low demand P= 0.15

High demand P=

Low demand P= 0.15

Do not Expand Small Plant

Expand Small Plant

Construct Small size

Construct Full size

High demand P=

High demand P= 0.85

Low demand P= 0.15

Low demand P= 0.15

$ 5 Million

$ -2 Million

$1 Million

$ 1.66 Million

2n

1st

EMV = $ 1.66 Million

EMV = $ 1.44 Million

$ 3.95 Million

EMV= $ 1.56 Million

EMV= $ 3.95 Million


8/20

Problem 1

page 225

D= 2500 units

C = $ 110/unit

F = $16/ orderr = 0.20/$/year

Cr = $ 22/unit/year

a.

Q * = 2 D f/Cr

Q * = 2 (2500) (16) / 22

Q * = 60.3022 units / order

b. Tc = Toc + Tcc

Tc = F(D/Q) + Cr(Q/2)

Tc = 16(2500/60.3022) + 22(60.3022/2)

Tc = 663.347 + 663.242

Tc = $ 1326.6321/year

c. Q * = 60 units rounded of to the nearest

Tc = F(D/Q) + Cr(Q/2)

Tc = 16(2500/60) + 22(60/2)

Tc = 666.67 + 660

Tc = $ 1326.67/year

%age increase = 1326.67 / 1326.6321 x 100 = 100.0028568 % = 0.0028568 %


9/20

`

Problem 2


50 % above Q * = 90.4533 units / order

Tc prior = $ 1326.6321/year

Tc = F(D/Q) + Cr(Q/2)

Tc = 16(2500/90.4533) + 22(90.4533/2)

Tc = 442.2171 + 994.9863

Tc = $ 1437.2034/year

%age increase = 1437.2034 / 1326.6321 x 100 = 108.3347 % = 8.3347 %

Problem 3

(a) Ordering 10 % above than EOQ = 66.3324 units / order

Tc = F(D/Q) + Cr(Q/2)

Tc = 16(2500/66.3324) + 22(66.3324/2)

Tc = 603.0233 + 729.6566

Tc = $ 1332.6799/year

%age increase = 1332.6799 / 1326.6321 x 100 = 100.45587% = 0.45587 %


10/20

(b) Ordering 10 % less than EOQ = 54.2719 units / order

Tc = F(D/Q) + Cr(Q/2)

Tc = 16(2500/54.2719) + 22(54.2719/2)

Tc = 737.0285 + 596.9900

Tc = $ 1334.020353/year

%age increase = 1334.020353 / 1326.6321 x 100 = 100.55968 % = 0.55968 %


11/20

Problem 6-37

N * = 5 orders/year

F = $6/order

Cr = $ 10 / grill / year

Stockout Cost = $ 50 unit

Current ROP = 650 units

L = 12 days

Probabilities = 0.3 0.2 0.1 0.1 0.05 0.05 0.05 0.05 0.05 0.03 0.02

Alternatives Demand during lead time600 650 700 750 800 850 900 950 1000 1050 1100 EMV

600 0 12500 25000 37500 50000 62500 75000 87500 100000 112500 125000 33375

650 500 0 12500 25000 37500 50000 62500 75000 87500 100000 112500 24775

700 1000 500 0 12500 25000 37500 50000 62500 75000 87500 100000 18775

750 1500 1000 500 0 12500 25000 37500 50000 62500 75000 87500 17075

800 2000 1500 1000 500 0 12500 25000 37500 50000 62500 75000 10675

850 2500 2000 1500 1000 500 0 12500 25000 37500 50000 62500 7925

900 3000 2500 2000 1500 1000 500 0 12500 25000 37500 62500 6075

950 3500 3000 2500 2000 1500 1000 500 0 12500 25000 37500 4375

1000 4000 3500 3000 2500 2000 1500 1000 500 0 12500 25000 3575

1050 4500 4000 3500 3000 2500 2000 1500 1000 500 0 12500 3425

1100 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 0 3665

Home Inc... should maintain safety stock level of inventory of 1050 units


12/20

Exercise :3

P = 10,000 units / year

C = $ 1.5/unit

F = $ 500/ order

r = 0.25/$/year

Cr = $ 0.375/unit/year

(a) D = 1800 units / year

Q * = 2 D f/Cr(1-D/P)

Q * = 2 (1800) (500) /(0.375)(1-1800/10000)

Q * = 2419.4335 units / order

Tc = Toc + Tcc

Tc = F(D/Q) + Cr(Q/2)(1-D/P)

Tc = 500 (1800/2419.43) + 0.375 (2419.43/2)(1-1800 / 10000)

Tc = 371.98 + 371.98

Tc = $ 737.96/year

(b) D = 7200 units / year

Q * = 2 D f/Cr(1-D/P)

Q * = 2 (7200) (500) /(0.375)(1-7200/10000)

Q * = 8280.7867 units / order

Tc = Toc + Tcc

Tc = F(D/Q) + Cr(Q/2)(1-D/P)

Tc = 500 (7200/8280.7867) + 0.375 (8280.7867/2)(1-7200 / 10000)

Tc = 434.7413 + 434.7413

Tc = $ 869.4826 /year

If ordered four fold than the last year's Q * = 9677.734 units / order

Tc = Toc + Tcc

Tc = F(D/Q) + Cr(Q/2)(1-D/P)

Tc = 500 (7200/9677.734) + 0.375 (9677.734/2)(1-7200 / 10000)

Tc = 371.9879 + 1487.9516

Tc = $ 1859.9395 /year

Q * 9677.734 Total Cost = $ 1859.9395

Q * 8280.786 Total Cost = $ 869.4826

Difference due to four fold = $ 990.4569

Note: This difference is due to increasing the order size exactly by the factor of 4.


13/20

Insufficient order will lead to additional holding cost.

Production manager should increase her order size by the factor of 3 and

she should order 8280.7867 units / order.

Exercise: 6

Page 241

D = 16,000 Ounce

r = 0.10 / $ / year

(a)Table6.7

Vendeor Unit Cost/Ounce $ OrderingCost $ Term of Sales

1 6.1 30 Q >= 1,600

2 6.3 25 Q >= 1,200

3 5.9 27 Q >= 1,800

EOQ for 3rd region:

Q * = 2 D F / Cnr

Q * = 2 (16000) (27) /5.90 x 0.10

Q * = 1210.1267 units

Since Q * do not luy in the 3rd region there fore we will take the lowe limitof

the 3rd region for the calculation of Total cost.

Tc = F(D/Q) + Cnr(Q/2) + DCn

Tc = 27 (16000 / 1800) + 0.59 (1800/2) + (16000 x 5.90)

Tc = 240 + 537 + 94400

Tc = $ 95177 / year

EOQ for 2nd region:

Q * = 2 D F / Cnr

Q * = 2 (16000) (25) / 6.30 x 0.10

Q * = 1126.8723 units


14/20

Since Q * do not luy in the 2nd region there fore we will take the lowe limitof

the 2nd region for the calculation of Total cost.


Tc = 25 (16000 / 1200) + 0.63 (1200/2) + (16000 x 6.30)

Tc = 333.3333 + 378 + 100800

Tc = $ 101511.3333 / year

EOQ for 1st region:

Q * = 2 D F / Cnr

Q * = 2 (16000) (30) / 6.1 x 0.10

Q * = 1254.50 units

Since Q * is not luying in the 1st region there fore we will take lower

limit of 1st region For calculation of Tc.


Tc = 30 (16000 / 1600) + 0.61 (1600/2) + (16000 x 6.10)

Tc = 300 + 488 + 97600

Tc = $ 98,388 / year

VendeorTerm ofSales

Minimumcost

3 Q >= 1,800$

95,177

2 Q >= 1,200$

101,511

1 Q >= 1,600$

98,388

Sandia should purchase from 3rd Vendor because it has the lowest Tc.

(b) Optimal inventory policy

L = 5 days

w.d = 250 days

T * = Q / D x w.d

T * = 1600 / 16000 x 250

T * = 25 days

N * = D / Q


15/20

N * = 16000 / 1600

N * = 10 orders / year

ROP = L x Daily demand

ROP = 5 x 64

ROP = 320 units

Problem: 1

page 225

C = $ 200/unit

F = $ 11 / order

r = 0.13/$/year

Cr = $ 26/unit/year

D = 50 units / month = 600 units / year

L = 7days (1 + 1 + 3 + 2)

w.d = 350 days / year(a) Q * = 2 D f/Cr

Q * = 2 (600) (11) / 26


(b,c,d)

Tc = Toc + Tcc

Tc = F(D/Q) + Cr(Q/2)

Tc = 11 (600/22.5320) + 26 (22.5320/2)

Tc = 292.9167 + 292.9160

Tc = $ 585.8327/year

(e) Optimal inventory policy


T * = Q * / D x w.d

T * = 22.5320 / 600 x 350

T * = 13.14 or 13 days

N * = D / Q*

N * = 600 / 22.5320

N * = 26.63 of 27 orders

ROP = L x Daily demand

ROP = 7 x 600 / 350 = 12 days


16/20

(Inventory level units)

Q * = 23 units

Average Inventory

ROP = 12 units

(Continue = 27 Orders)

(Time = 300 days)

7 days 13 days

page 219

Problem 1

(a) C = $ 110 /unit

F = $ 5 / order

r = 0.12/$/year

Cr = $ 13.2/unit/yearD = 6 units / month = 72 units / year

w.d = 300 days / year

(b)

Q TOC TCC TC

1 360 6.6 366.6

2 180 13.2 193.2

3 120 19.8 139.8

4 90 26.4 116.4

5 72 33 105

6 60 39.6 99.6

7 51 46.2 97.28 45 52.8 97.8

9 40 59.4 99.4

10 36 66 102

page 215

Problem 1 Q * = 2 D f/Cr

Q * = 2 (72) (5) / 13.2


Problem 2 Tc = Toc + Tcc

Tc = F(D/Q) + Cr(Q/2)Tc = 5 (72/7.3854) + 13.2 (7.3854/2)


17/20

Tc = 48.7448 + 48.7436

Tc = $ 97.4884/year

Yes our answer matching with previous answer.

Problem 3

D = 24 units / month = 288 units / yearQ * = 2 D f/Cr

Q * = 2 (288) (5) / 13.2


Selly should not increase her order quantity by factor of 4.

She should increase her order quantity be factor of 2.

Optimal inventory policy


T * = Q * / D x w.d

T * = 14.7709 / 288 x 300

T * = 15.3864 or 15 days

N * = D / Q*

N * = 288 / 14.7709

N * = 16.2062 of 16 orders / year

Exercise 4

C = $ 450 / unit

F = $ 50 / order

r = 0.25 / $ / year

Cr = $ 112.5 / unit / year

D = 90 units / year

L = 10 days


(a) Q * = 2 D F / Cr

Q * = 2 (90) (50) / 112.5

Q * =8.9442 units

M * = 2 D F / Cr

M * = 2 (90) (50) / 112.5

M * =8.9442

M * = 2.0902 units / order

m = Q * - M *

m = 8.9442 - 8.9442

m = 0 units%age of EOQ met by M * = M * / Q* x 100


18/20

%age of EOQ met by M * = 8.9442 / 8.9442 x 100

%age of EOQ met by M * = 100 %

%age of EOQ met by m = m / Q* x 100


T * = Q * / d x w.d

T * = 38.5695 / 90 x 270

T * = 115.7085 or 116 days

N * = D / Q *

N * = 90 / 38.2695

N * = 2.3517 or 2 orders

ROP = L x daily demand

ROP = 10 x 90 /270

ROP = 3.3333 units

Tc = F(D/Q) + CrM2 / 2Q + b (Q - M ) 2 /2Q

Tc = 50(90/38.2695) +112.5(2.0902) 2/ 2 (38.2695) +6.5 (38.2695 - 2.0902 ) 2 /2(38.2695)

Tc = 117.5871 + 6.4216 + 111.1606

Tc = 117.5871 +117.5822

Tc = $ 235.1693

Tc = 117.5871 +117.5822

Tc = $ 235.1693

Tc = 117.5871 +117.5822

Tc = $ 235.1693

(b) C = $ 450 / unit

F = $ 50 / order

r = 0.25 / $ / year

Cr = $ 112.5 / unit / year

D = 90 units / year

L = 10 days

w.d = 270 days / yearb = $ 5 + Promotional $ 135 / 90 = $ 6.5 / unit

Q * = 2 D F / Cr x b + Cr / b

Q * = 2 (90) (50) / 112.5 x 6.5 + 112.5 / 6.5

Q * =8.9442 units x 4.2787 units


M * = 2 D F / Cr x b / b + Cr

M * = 2 (90) (50) / 112.5 x 6.5 / 6.5 + 112.5

M * =8.9442 x 0.2337



19/20

m = Q * - M *

m = 38.2695 - 2.0902

m = 36.1793 units

%age of EOQ met by M * = M * / Q* x 100

%age of EOQ met by M * = 2.08 / 38.25 x 100%age of EOQ met by M * = 5.4617 %


%age of EOQ met by m = 36.1793 / 38.5695 x 100

%age of EOQ met by m = 94.5382 %

T * = Q * / d x w.d

T * = 38.5695 / 90 x 270

T * = 115.7085 or 116 days

N * = D / Q *

N * = 90 / 38.2695

N * = 2.3517 or 2 orders


ROP = 10 x 90 /270

ROP = 3.3333 units

Tc = F(D/Q) + CrM2 / 2Q + b (Q - M ) 2 /2Q

Tc = 50(90/38.2695) +112.5(2.0902)2/ 2 (38.2695)+6.5 (38.2695 - 2.0902 )

2/2(38.2695)

Tc = 117.5871 + 6.4216 + 111.1606

Tc = 117.5871 +117.5822

Tc = $ 235.1693

(c..) C = $ 450 / unit

F = $ 50 / order

r = 0.25 / $ / year

Cr = $ 112.5 / unit / yearD = 90 units / year

L = 10 days


b = $ 5

Q * = 2 D F / Cr x b + Cr / b

Q * = 2 (90) (50) / 112.5 x 5 + 112.5 / 5

Q * =8.9442 units x 4.8476 units


M * = 2 D F / Cr x b / b + Cr

M * = 2 (90) (50) / 112.5 x 5 / 5 + 112.5


20/20

M * =8.9442 x 0.205684


m = Q * - M *

m = 43.3586 - 1.8450

m = 41.5135 units%age of EOQ met by M * = M * / Q* x 100

%age of EOQ met by M * = 1.8450 / 43.3856 x 100

%age of EOQ met by M * = 4.2525 %


%age of EOQ met by m = 41.5135 / 43.3856 x 100

%age of EOQ met by m = 95.6849 %

T * = Q * / d x w.d

T * = 43.3856 / 90 x 270

T * = 130.1568 or 130 days

N * = D / Q *

N * = 90 / 43.3856

N * = 2.0744 or 2 orders


ROP = 10 x 90 /270

ROP = 3.3333 units

Tc = F(D/Q) + CrM2 / 2Q + b (Q - M ) 2 /2Q

Tc = 50(90/43.3856) +112.5(1.8450) 2 / 2 (43.3856) + 5 (43.3856 - 1.8450 ) 2 /2(43.3856)

Tc = 103.7210 + 4.4133 + 99.4351

Tc = 103.7210 +103.8484

Tc = $ 207.5694 / year

Tabulated Comparison

S # Particularsa. No

Backorder

b. withadditional

$ 6.5

c. $5without

additional

1 Q * 8.9442 38.2695 93.3586

2 M * 8.9442 2.0902 1.8453 m 0 36.1793 41.5135

4 M * %age 100% 5.46% 4.25%

5 m %age 0% 94.54% 95.68%

6 Toc 117.5871 117.5871 103.721

7 Tcc 117.5822 117.5822 103.8484

8 Tc 235.1693 235.1693 207.5694

9 N * 2.3517 2.3517 2.0744

10 T * 115.7085 115.7085 130.1568

11 ROP 3.3333 3.3333 3.3333

Documents

Decision tree and decision analysis