review midterm wlp
TRANSCRIPT
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2008 Prentice-Hall, Inc.
MIDTERM EXAMINATION DATE Quantitative Method for Business
T est Date: April 09, 2011T
ime: 10:00 AMRoom: A309(Please check this info. with yours)
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MIDTERM EXAMINATION DATE Quantitative Method for Business
O pen BookT heory and Practice:
IntroductionGame T heoryDecision Analysis: T able, T reeForecastingLinear Programing: Graphical Method
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Review for Midterm Exam
Part: Exercises
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C hapter 1Introduction to Quantitative Analysis
Ex ample 1: Break-even Point. A company AB wants to sell a new type of product tomarket. After conducting some surveys, they thinkthat they can sell the new product with price $40/unit.The price of investing some new machines is about$30,000. The total cost to produce one product is$25.
a) Determine the number of sale at which company ABbegins to get profit.
b) If the average demand of this new product is1,000/year. Determine the amount of time thiscompany begins to get profit.
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SolutionEx ample: Break-even Point.
F ixed cost: f = $30,000Selling price per unit: s = $40Variable cost per unit: v = $25
a) Calculate the Break-even point (BEP):
units000,22540
000,30
00 profit
!!!
vv!!
!
!
v s
f BEP
BEP v f BEP s X BEP
vX f sX profit
b) Calculate the payback time:
yea rs2000,1000,2
me payback ti !!! Demand
BEP
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Ex ample 2: C alculate the demandsaccording to production strategies
Ex ample 2: Calculate the demands according toproduction strategies.
A company AB consider 3 strategies to produce the new
product: (X: number of products)1. Cost = $20,000 + $20X2. Cost = $30,000 + $10X3. Cost = $40,000 + $6X
Determine the ranges of potential demands according toeach chosen strategy.
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Solution: C alculate the demands according to production strategies
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
100000
0 500 1000 1500 2000 2500 3000 3500 4000
Strategy 1
Strategy 2
Strategy 3
Choosing:
Strategy 1:
Demand 1, 000
Strategy 2:1,000 Demand 2, 500
Strategy 3:
Demand 2, 500
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Solution: C alculate the demands according to production strategies
H int:
The strategy include fixed cost and variable cost
cost = fixed cost + variable cost x number of products (demand)
We do not need to calculate the intersection point between strategy 1and 3 in this example.
To reduce the calculation, we follow some steps: Arrange the strategies following the order of decreasing of variable cost.
Just determine the intersection between 1 st and 2 nd strategy; 2 nd and 3 rdstrategy, 3 rd and 4 th
Drawing the chart and choose the strategy which has the lowest cost for each range.
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C hapter 2 Game Theories
Ex ample 1:We consider the payoff table for a two-person zero-sum game following:
bb 11 bb33bb 22
Player BPlayer B
7 5 67 5 68 68 6 --66
aa 11
aa 22aa 33
Player APlayer A
10 810 8 --55
Determine the best strategy for each player.
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S olution: Game Theories
Consider the dominated strategy to reduce the size of gameF or player A, strategy a 3 is dominated by strategy a 1. Eliminate strategya 3.F or player B, strategy b 1 is dominated by strategy b 3. Eliminate strategyb1.
bb 11 bb33bb 22
Player BPlayer B
7 5 67 5 68 68 6 --66
aa 11aa 22aa 33
Player APlayer A
10 810 8 --55
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S olution: Game Theories
F ind the minimum value for each row. F ind the maximum of theseminimum values (maximin).F ind the maximum value for each column. F ind the minimum of these maximum values (minimax).
bb 33bb 22
Player BPlayer B
5 65 6
aa 11
aa 22
Player APlayer A
88 --55
row Min.row Min.
--55
55
Column Max.Column Max. 8 68 6
We can see that the minimax is not equal the maximin. We do not
have pure strategy => consider the mixed strategy.
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S olution: Game Theories
bb 33bb 22
Player BPlayer B
5 65 6
aa 11aa 22
Player APlayer A
88 --55
row Min.row Min.
--55
55
Column Max.Column Max. 8 68 6F or player A:
EV = 8p+5(1-p) = -5p+6(1-p)
3p+5 = -11p+6p = 1/14 = 0.0 7 1 => 1-p = 0. 9 29
Player A should choose strategy a 1with probability 0.0 7 1 and strategya 2 with probability 0. 9 29 .
EV = 5.214
p p11-- p p
q 1-q
F or player B:EV = 8q-5(1-q) = 5q+6(1-q)
13q-5 = -q+6q = 11/14 = 0. 7 86 => 1-q = 0.214
Player B should choose strategy b 2with probability 0. 7 86 and strategyb3 with probability 0.214.
EV = 5.214
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C hapter 3Decision Analysis
Ex ample 1: Uncertainty. A company XY who produces the battery wants to sell a new type of battery for electrical bicycles. They consider two possible plan:
Build a new factory to produce this type of battery and can earn about
$200,000 per year if the demand or market is good.H
owever, if the market isnot good, they can lose $150,000.Import the batteries from a foreign company and sell in dominant market withtheir brand name. They can earn only $100,000 per year with this planbecause of some limitations if the market is good. H owever, they will loseonly $30,000 if the market is not good.
In general, company can do nothing also. Assume that the management board does not know anything about the
market conditions (states of nature). Choose the best alternative for thiscompany according to 5 criteria (Maximax, maximin, H urwicz with =0. 7 ,Laplace, minimax regret)
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S olutionExample 1: Uncertainty
State of natureMinimumin a row
Max imumin a row
Hurwicz = 0.7 LaplaceAlternativ
esGood
MarketBad
Market
Build afactory 200,000 -150,000
-150,000 200,000 95,000
=0.7*200,000+0.3*
( -150,000 )
25,000 =( 200,000 -
150,000 )/ 2
Import 100,000 -30,000 -30,000 100,000 61,000 35,000
Donothing 0 0
0 0 0 0
Maximin Maximax Maximum in the columnBest decision:
Maximax: Build a factory
Maximin: Do nothing.
H urwicz: Build a factory ( =0. 7 )
Laplace: Import from a foreign company
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S olutionExample 1: Uncertainty
State of nature
Alternatives
GoodMarket
BadMarket
Build afactory 200,000 -150,000
Import 100,000 -30,000
Do nothing 0 0
MinimaxBest decision: Import from a foreign company
Opportunity Loss TableState of nature Max imum
in arow
Alternatives Good Market Bad Market
Build afactory
0= 200,000 -200,000
150,000=0- (-150,000) 150,000
Import 100,000= 200,000 -100,00030,000
=0-(-30,000)100,000
Do nothing 200,000=200,000 -00
=0-0200,000
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C hapter 3Decision Analysis
Ex ample 2: Risk.
Consider same company XY in the example 1.
Now, we assume that the management board know believe the good market
conditions (states of nature) will occur with the probability of 60%.Choose the best alternative for this company.Construct the decision tree for this company.Calculate EMV, EOL, EVPI
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S olutionExample 2: Risk
State of nature
E MVAlternatives
GoodMarket
(0.6)
BadMarket
(0.4)
Build afactory 200,000 -150,000 60,000
Import 100,000 -30,000 48 ,000
Do nothing 0 0 0
Opportunity Loss TableState of nature
EO LAlternatives
GoodMarket
(0.6)
BadMarket
(0.4)Build a
factory 0 150,000 60,000
Import 100,000 30,000 72,000
Do nothing 200,000 0 120,000
Import
Good market (0.6)
Good market (0.6)
Bad market (0.4)
Bad market (0.4)
200,000
-150,000100,000
-30,000
0
E MV=60,000
E MV=48,000
EVPI=EVwPI - max EMV
=0.6*200,000+0.4*0 60,000
= 60,000
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C hapter 3Decision Analysis
Ex ample 3: Risk with further information.
Consider same company XY in the example 2 with more information.
Now, the management board want to conduct a survey to revise their belief
about the condition of market. The cost of this survey is $30,000. According to their experiences in past, the probability that the survey willgive exactly the positive result when the market is good is 7 5%. Andwhen the market is not good, the probability that the survey can detectwith the negative result is 9 0%.
Construct the decision tree for this company.Determine the conditional probabilities.Choose the best alternative for this company.Calculate EMV, EVSI
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S olutionExample 3: Risk with further information
First DecisionPoint
Second DecisionPoint
Good Market (?)Bad Market (?)Good Market (?)Bad Market (?)
Good Market (?)Bad Market (?)Good Market (?)Bad Market (?)
Good Market (0.60)Bad Market (0.4)Good Market (0.60)Bad Market (0.40)
Im-port
Do nothing
6
7
Im-port
Do nothing
2
3
Im-port
Do nothing
4
5
1
Payoffs
$180,000$170,000
$70,000 $60,000
$30,000
$150,000$200,000
$100,000 $30,000
$0
$180,000$170,000
$70,000 $60,000
$30,000
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S olutionExample 3: Risk with further information
4.0)( ;6.0)(10.0)|(90.0)|(
25.0)|(75.0)|(
!!!!
!!
BM P GM P
BM PS P BM NS P
GM NS P GM PS P
GM: good market; BM: bad market;
PS: surveys positive result; NS: surveys negative result
51.0)(1)(
49.04.010.06.075.0
)()|()()|( )()()(
!!
!vv!
!!
PS P NS P
BM P BM PS P GM P GM PS P BM PS P GM PS P PS P
The total probabilities:
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S olutionExample 3: Risk with further
informationFirst DecisionPoint Second DecisionPointGood Market (0.92)Bad Market (0.08)Good Market (0.92)Bad Market (0.08)
Good Market (0.29)Bad Market (0.71)Good Market (0.29)Bad Market (0.71)
Good Market (0.60)Bad Market (0.4)Good Market (0.60)Bad Market (0.40)
Im-port
Do nothing
6
7
Im-port
Do nothing
2
3
Im-port
Do nothing
4
5
1
Payoffs
$180,000$170,000
$70,000 $60,000
$30,000
$150,000$200,000
$100,000 $30,000
$0
$180,000$170,000
$70,000 $60,000
$30,000
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S olutionExample 3: Risk with further information
Calculate the EMVs.
000,48)000,30(4.0000,1006.0
000,60)000,150(4.0000,2006.0
300,22)000,60(71.0000,7029.0
500,78)000,180(71.0000,17029.0
600,59)000,60(08.0000,7092.0
000,142)000,180(08.0000,17092.0
7
6
5
4
3
2
!vv!
!vv!
!vv!
!vv!
!vv!
!vv!
EMV
EMV
EMV
EMV
EMV
EMV
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S olutionExample 3: Risk with further information
207,58)300,22(51.0000,14249.01 !vv! EMV
Calculate the EMV 1.
Calculate the EVSI.
207,28000,60000,30207,58
info. sampl ewithoutcostinfo. sampl ewith!!
! EV EV EV SI
The best choice is not to conduct the survey and build the factory.
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S olutionExample 3: Risk with further information
First DecisionPoint Second DecisionPoint
Good Market (0.92)Bad Market (0.08)Good Market (0.92)Bad Market (0.08)
Good Market (0.29)Bad Market (0.71)Good Market (0.29)Bad Market (0.71)
Good Market (0.60)Bad Market (0.4)Good Market (0.60)Bad Market (0.40)
Im-port
Do nothing
6
7
Im-port
Do nothing
2
3
Im-port
Do nothing
4
5
1
Payoffs
$180,000$170,000
$70,000 $60,000
$30,000
$150,000$200,000
$100,000 $30,000
$0
$180,000$170,000
$70,000 $60,000
$30,000
$142,000
$59,600$58,207
$-78,500
$-22,300$-22,300
$60,000
$48,000$60,000
$142,000
$60,000
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C hapter 4Forecasting
Ex ample: Forecasting. A car dealer want to forecast the sale of a car model in next quarter. H e has some data of salefrom 3 years 2008, 200 9 and 2010 as the besidetable.
H e know some forecasting methods such asMoving average, weighted moving average,Exponential smoothing, Exponential smoothingwith trend and Decomposition models includingmultiplicative and additive methods. H owever, hedoes not know which method is the best for hissituation.Develop some forecasting models listed aboveand help him choose the best.
N ote: moving average , choosing n= 2.Ex ponential smoothing: = 0.3; = 0.7
Periods Sales
1. Q1/2008 236
2. Q2/2008 189
3. Q3/2008 2454. Q4/2008 208
5. Q1/2009 245
6. Q2/2009 199
7. Q3/2009 253
8. Q4/2009 2139. Q1/2010 267
10. Q2/2010 210
11. Q3/2010 273
12. Q4/2010 234
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Ex ample
Periods Sales CMAs Seasonal ratios Seasonal Indices
Q1/2008 236
Q2/2008 189
Q3/2008 245220.6= 0.5 *236+189+245
+208+ 0.5 *245 1.111= 245/220.61.105= (1.111+0.933+1.087+
0.877)/4Q4/2008 208 223 0.933 0.921
Q1/2009 245 225.3 1.087 1.1035
Q2/2009 199 226.9 0.877 0.87
Q3/2009 253 230.3 1.099 1.105
Q4/2009 213 234.4 0.909 0.921
Q1/2010 267 238.3 1.12 1.1035
Q2/2010 210 243.4 0.863 0.87
Q3/2010 273 1.105
Q4/2010 234 0.921
Q1/2011 1.1035
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Ex ample
Periods Sales
Seasonal
Indices
Deseasonalized Sales
ForecastedBase Values
(Y=3.351X+209)
ForecastedSales
AdditiveDecom.Model
(Y=232.8+3.313X1-53.313X2+1.042X3-
40.938X4)
1.Q1/2008 236 1.1035 213.9= 236/1.1035 212.4= 3.351*1+209 234.4= 212.4*1.1035 236.1
2.Q2/2008 189 0.87 217.2 215.7= 3.351*2+209 187.7 186.1
3.Q3/2008 245 1.105 221.7 219.1 242.1 243.8
4.Q4/2008 208 0.921 225.8 222.4 204.8 205.1
5.Q1/2009 245 1.1035 222 225.8 249.2 249.4
6.Q2/2009 199 0.87 228.7 229.1 199.3 199.4
7.Q3/2009 253 1.105 229 232.5 256.9 257
8.Q4/2009 213 0.921 231.3 235.8 217.2 218.4
9.Q1/2010 267 1.1035 242 239.2 264 262.6
10.Q2/2010 210 0.87 241.4 242.5 211 212.6
11.Q3/2010 273 1.105 247.1 245.9 271.7 270.3
12.Q4/2010 234 0.921 254.1 249.2 229.5 231.6
13.Q1/2011 1.1035 252.6 278.7 275.9
MAD = 2.62 2.78
MSE = 9 10
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C hapter 5 Linear programming problem
Ex ample: Forecasting. A car dealer want to forecast the sale of a car model in next quarter. H e has some data of salefrom 3 years 2008, 200 9 and 2010 as the besidetable.
H e know some forecasting methods such asMoving average, weighted moving average,Exponential smoothing, Exponential smoothingwith trend and Decomposition models includingmultiplicative and additive methods. H owever, hedoes not know which method is the best for hissituation.Develop some forecasting models listed aboveand help him choose the best.
N ote: moving average , choosing n= 2.Ex ponential smoothing: = 0.3; = 0.7
Periods Sales
1. Q1/2008 236
2. Q2/2008 189
3. Q3/2008 2454. Q4/2008 208
5. Q1/2009 245
6. Q2/2009 199
7. Q3/2009 253
8. Q4/2009 2139. Q1/2010 267
10. Q2/2010 210
11. Q3/2010 273
12. Q4/2010 234
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LP Model: Example
L abor L abor Clay Clay RevenueRevenuePRODUCTPRODUCT (hr/unit)(hr/unit) (lb/unit)(lb/unit) ($/unit)($/unit)BowlBowl 11 44 4040MugMug 22 33 5050
There are 40 hours of labor and 120 pounds of clay availableThere are 40 hours of labor and 120 pounds of clay available
each dayeach day
RESOURCE REQUIREMENTSRESOURCE REQUIREMENTS
Pottery Example
Beaver Creek Pottery Company is located on a Native Americanreservation. Each day, the company has available 40 hours of labor and 120pounds of clay. The firm makes two products, bowls and mugs. A bowl requires 1hour of labor and 4 pounds of clay. A mug requires 2 hours of labor and 3 poundsof clay. The firm's profit is $40 per bowl and $50 per mug. The company wants tomaximize profit
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LP Formulation: Example
MaximizeMaximize Z Z = $40= $40 x x 11 + 50+ 50 x x 22
Subject toSubject to x x 11 ++ 22 x x 22 ee 40 hr 40 hr (labor constraint)(labor constraint)
44 x x 11 ++ 33 x x 22 ee 120 lb120 lb (clay constraint)(clay constraint) x x 11 ,, x x 22 uu 00
Decision variablesDecision variables x x 1 = number of bowls to produce1 = number of bowls to produce x x 2 = number of mugs to produce2 = number of mugs to produce
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Graphical S olution: Example
44 x x 11 + 3+ 3 x x 22 ee 120 lb120 lb
x x 11 + 2+ 2 x x 22 ee 40 hr 40 hr
Area common to Area common to
both constraintsboth constraints
5050
4040
3030
2020
1010
00 |1010
|6060
|5050
|2020
|3030
|4040 x x 11
x x 2 2
O bjective lineO bjective line
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C omputing Optimal Values
x x 11 ++ 22 x x 22 == 404044 x x 11 ++ 33 x x 22 == 120120
44 x x 11 ++ 88 x x 22 == 160160
--44 x x 11 -- 33 x x 22 == --12012055 x x 22 == 4040 x x 22 == 88
x x 11 ++ 2(8)2(8) == 4040 x x 11 == 2424
44 x x 11 + 3+ 3 x x 22 ee 120 lb120 lb
x x 11 + 2+ 2 x x 22 ee 40 hr 40 hr
4040
3030
2020
1010
00 |1010
|2020
|3030
|4040
x x 11
x x 2 2
Z Z = $50(24) + $50(8) = $1,360= $50(24) + $50(8) = $1,360
24
8
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