optimization i. © the mcgraw-hill companies, inc., 2004 operations management -- prof. juran2...
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Optimization I
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Outline
• Basic Optimization: Linear programming– Graphical method– Spreadsheet Method
• Extension: Nonlinear programming– Portfolio optimization
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What is Optimization?
• A model with a “best” solution• Strict mathematical definition of
“optimal”• Usually unrealistic assumptions• Useful for managerial intuition
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Elements of an Optimization Model
• Formulation– Decision Variables– Objective– Constraints
• Solution – Algorithm or Heuristic
• Interpretation
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Optimization Example:
Extreme Downhill Co.
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1. Managerial Problem Definition
Michele Taggart needs to decide how many sets of skis and how many snowboards to make this week.
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2. Formulation
a. Define the choices to be made by the manager (decision variables).
b. Find a mathematical expression for the manager's goal (objective function).
c. Find expressions for the things that restrict the manager's range of choices (constraints).
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Variable Name Symbol UnitsSkis X 100s of pairs of skis
Snowboards Y 100s of snowboards
2a: Decision Variables
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Extreme Downhill
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Skis (X 100)
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Extreme Downhill
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Skis (X 100)
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2b: Objective Function
Find a mathematical expression for the manager's goal (objective function).
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EDC makes $40 for every snowboard it sells, and $60 for every pair of skis. Michele wants to make sure she chooses the right mix of the two products so as to make the most money for her company.
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YX 40006000Profit
What Is the Objective?
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Extreme Downhill
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Skis (X 100)
profit = $40,000
profit = $80,000
profit = $120,000
profit = $160,000
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Extreme Downhill
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Skis (X 100)
profit = $40,000
profit = $80,000
profit = $120,000
profit = $160,000
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2c: Constraints
Find expressions for the things that restrict the manager's range of choices (constraints).
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Molding Machine Constraint
The molding machine takes three hours to make 100 pairs of skis, or it can make 100 snowboards in two hours, and the molding machine is only running 115.5 hours every week.
The total number of hours spent molding skis and snowboards cannot exceed 115.5.
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Molding Machine Constraint
511523 .YX
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Extreme Downhill
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Skis (x 100)
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Cutting Machine Constraint
Michele only gets to use the cutting machine 51 hours per week. The cutting machine can process 100 pairs of skis in an hour, or it can do 100 snowboards in three hours.
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Cutting Machine Constraint
5131 YX
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Extreme Downhill
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Skis (x 100)
Cutting
Molding
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Delivery Van Constraint
There isn't any point in making more products in a week than can fit into the van The van has a capacity of 48 cubic meters. 100 snowboards take up one cubic meter, and 100 sets of skis take up two cubic meters.
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Delivery Van Constraint
4812 YX
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Extreme Downhill
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Van
Cutting
Molding
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Demand Constraint
Michele has decided that she will never make more than 1,600 snowboards per week, because she won't be able to sell any more than that.
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Demand Constraint
16Y
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Extreme Downhill
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Demand
Van
Cutting
Molding
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Non-negativity Constraints
Michele can't make a negative number of either product.
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Non-negativity Constraints
0X0Y
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Extreme Downhill
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Skis (x 100)
Nonnegativity - Snowboards
Nonnegativity - Skis
Demand
Van
Cutting
Molding
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Extreme Downhill
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Solution MethodologyUse algebra to find the best solution.
(Simplex algorithm)
George B. Dantzig
1914 - 2005
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Extreme Downhill
Point A Point E
Point CPoint B
Point D
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Point X YA 0 0B 0 16C 3 16D 18.6 10.8E 24 0
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Calculating Profits
Point X Y Objective function ProfitA 0 0 6000(0)+4000(0)=$0.00B 0 16 6000(0)+4000(16)=$64,000.00C 3 16 6000(3)+4000(16)=$82,000.00D 18.6 10.8 6000(18.6)+4000(10.8)=$154,800.00E 24 0 6000(24)+4000(0)=$144,000.00
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The Optimal Solution
• Make 1,860 sets of skis and 1,080 snowboards.
• Earn $154,800 profit.
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Spreadsheet Optimization
1
23
4
56789
101112
A B C D E F G H I Jobjective: 6000 4000 10,000$ = profit
decision variables: (in 100s) skis snowboards
1 1
constraints:Molding 3 2 5 <= 115.5Cutting 1 3 4 <= 51Van 2 1 3 <= 48Demand 0 1 1 <= 16Nonnegativity (skis) 1 0 1 >= 0Nonnegativity (snowboards) 0 1 1 >= 0
=SUMPRODUCT(C1:D1,C4:D4)
=SUMPRODUCT($C$4:$D$4,C7:D7)
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1
234
56789
101112
A B C D E F Gobjective: 6000 4000 154,800$ = profit
decision variables: (in 100s) skis snowboards18.6 10.8
constraints:Molding 3 2 77.4 <= 115.5Cutting 1 3 51 <= 51Van 2 1 48 <= 48Demand 0 1 10.8 <= 16Nonnegativity (skis) 1 0 18.6 >= 0Nonnegativity (snowboards) 0 1 10.8 >= 0
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14151617181920212223242526272829303132
A B C D E F GObjective Cell (Max)
Cell Name Original Value Final Value$E$1 objective: 154,800$ 154,800$
Variable CellsCell Name Original Value Final Value Integer
$C$4 skis 18.6 18.6 Contin$D$4 snowboards 10.8 10.8 Contin
ConstraintsCell Name Cell Value Formula Status Slack
$E$11 Nonnegativity (skis) 18.6 $E$11>=$G$11 Not Binding 18.6$E$12 Nonnegativity (snowboards) 10.8 $E$12>=$G$12 Not Binding 10.8$E$7 Molding 77.4 $E$7<=$G$7 Not Binding 38.1$E$8 Cutting 51 $E$8<=$G$8 Binding 0$E$9 Van 48 $E$9<=$G$9 Binding 0$E$10 Demand 10.8 $E$10<=$G$10 Not Binding 5.2
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Most important number: Shadow PriceThe change in the objective function that would result from a one-unit increase in the right-hand side of a constraint
67891011121314151617181920
A B C D E F G HVariable Cells
Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease
$C$4 skis 18.6 0 6000 2000 4666.666667$D$4 snowboards 10.8 0 4000 14000 1000
ConstraintsFinal Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease$E$11 Nonnegativity (skis) 18.6 0 0 18.6 1E+30$E$12 Nonnegativity (snowboards) 10.8 0 0 10.8 1E+30$E$7 Molding 77.4 0 115.5 1E+30 38.1$E$8 Cutting 51 400 51 13 27$E$9 Van 48 2800 48 27.2 26$E$10 Demand 10.8 0 16 1E+30 5.2
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Nonlinear Example: Scenario Approach to Portfolio
OptimizationYear Ford Lilly Kellogg Merck HP 1983 14.13 14.47 8.09 5.02 42.38 1984 15.21 16.50 10.00 5.22 33.88 1985 19.33 27.88 17.38 7.61 36.75 1986 28.13 37.13 25.88 13.76 41.88 1987 37.69 39.00 26.19 17.61 58.25 1988 50.50 42.75 32.13 19.25 47.25 1989 43.63 68.50 33.81 25.83 31.88 1990 26.63 73.25 37.94 29.96 57.00 1991 28.13 83.50 65.38 55.50 69.88
Use the scenario approach to determine the minimum-risk portfolio of these stocks that yields an expected return of at least 22%, without shorting.
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The percent return on the portfolio is represented by the random variable R.
In this model, xi is the proportion of the portfolio (i.e. a number between zero and one) allocated to investment i.
Each investment i has a percent return under each scenario j, which we represent with the symbol rij.
5
1iiixrR
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W e c a l c u l a t e t h e p e r c e n t r e t u r n o n e a c h o f t h e s t o c k s i n e a c h y e a r : Y e a r F o r d L i l l y K e l l o g g M e r c k H P 1 9 8 4 0 . 0 7 6 0 . 1 4 0 0 . 2 3 6 0 . 0 4 0 - 0 . 2 0 1 1 9 8 5 0 . 2 7 1 0 . 6 9 0 0 . 7 3 8 0 . 4 5 8 0 . 0 8 5 1 9 8 6 0 . 4 5 5 0 . 3 3 2 0 . 4 8 9 0 . 8 0 8 0 . 1 4 0 1 9 8 7 0 . 3 4 0 0 . 0 5 0 0 . 0 1 2 0 . 2 8 0 0 . 3 9 1 1 9 8 8 0 . 3 4 0 0 . 0 9 6 0 . 2 2 7 0 . 0 9 3 - 0 . 1 8 9 1 9 8 9 - 0 . 1 3 6 0 . 6 0 2 0 . 0 5 2 0 . 3 4 2 - 0 . 3 2 5 1 9 9 0 - 0 . 3 9 0 0 . 0 6 9 0 . 1 2 2 0 . 1 6 0 0 . 7 8 8 1 9 9 1 0 . 0 5 6 0 . 1 4 0 0 . 7 2 3 0 . 8 5 2 0 . 2 2 6
F o r e x a m p l e , F o r d w e n t f r o m $ 1 4 . 3 1 t o $ 1 5 . 2 1 i n 1 9 8 4 , s o t h e r e t u r n o n F o r d s t o c k i n 1 9 8 4 w a s :
076.013.14
13.1421.15
0
011
S
SSr j
j
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The portfolio return under any scenario j is given by:
5
1iiijj xrR
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Let Pj represent the probability of scenario j occurring.
The expected value of R is given by:
8
1jjjR PR
8
1
2
jjRjR PR
The standard deviation of R is given by:
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In this model, each scenario is considered to have an equal probability of occurring, so we can simplify the two expressions:
8
8
1 j
j
R
R
8
8
1
2
j
Rj
R
R
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Decision VariablesWe need to determine the proportion of our portfolio to invest in each of the five stocks.
ObjectiveMinimize risk.
ConstraintsAll of the money must be invested. (1)The expected return must be at least 22%.(2)No shorting. (3)
Managerial Formulation
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Mathematical FormulationDecision Variablesx1, x2, x3, x4, and x5 (corresponding to Ford, Lilly, Kellogg, Merck, and HP). ObjectiveMinimize Z = Constraints
(1)
(2)
For all i, xi ≥ 0 (3)
8
8
1
2
j
Rj
R
R
0.15
1
i
ix
22.08
8
1 j
j
R
R
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123
45678910111213141516171819202122232425262728
A B C D E F G H I JTotal Ford Lilly Kellogg Merck HP
Exp Return = 0.127 1 1.000 0.000 0.000 0.000 0.000StDev = 0.265
req. return = 0.220 Historical dataYear Ford Lilly Kellogg Merck HP1983 14.13 14.47 8.09 5.02 42.381984 15.21 16.50 10.00 5.22 33.881985 19.33 27.88 17.38 7.61 36.751986 28.13 37.13 25.88 13.76 41.881987 37.69 39.00 26.19 17.61 58.251988 50.50 42.75 32.13 19.25 47.251989 43.63 68.50 33.81 25.83 31.881990 26.63 73.25 37.94 29.96 57.001991 28.13 83.50 65.38 55.50 69.88
Historical data on returnsreturn deviation^2 Year Ford Lilly Kellogg Merck HP0.076 0.003 1984 0.076 0.140 0.236 0.040 -0.2010.271 0.021 1985 0.271 0.690 0.738 0.458 0.0850.455 0.108 1986 0.455 0.332 0.489 0.808 0.1400.340 0.045 1987 0.340 0.050 0.012 0.280 0.3910.340 0.045 1988 0.340 0.096 0.227 0.093 -0.189-0.136 0.069 1989 -0.136 0.602 0.052 0.342 -0.325-0.390 0.267 1990 -0.390 0.069 0.122 0.160 0.7880.056 0.005 1991 0.056 0.140 0.723 0.852 0.226
mean 0.127 0.265 0.325 0.379 0.114stdevp 0.265 0.235 0.271 0.290 0.341
=AVERAGE(B19:B26)
=SQRT(AVERAGE(C19:C26))
=SUM(G2:K2)
=SUMPRODUCT($F$2:$J$2,F19:J19)
=(B19-$C$2)^2
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The decision variables are in F2:J2.
The objective function is in C3.
Cell E2 keeps track of constraint (1).
Cells C2 and C5 keep track of constraint (2).
Constraint (3) can be handled by checking the “Unconstrained Variables Non-negative” box.
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123
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A B C D E F G H I JTotal Ford Lilly Kellogg Merck HP
Exp Return = 0.220 1 0.173 0.426 0.054 0.105 0.241StDev = 0.128
req. return = 0.220 Historical dataYear Ford Lilly Kellogg Merck HP1983 14.13 14.47 8.09 5.02 42.381984 15.21 16.50 10.00 5.22 33.881985 19.33 27.88 17.38 7.61 36.751986 28.13 37.13 25.88 13.76 41.881987 37.69 39.00 26.19 17.61 58.251988 50.50 42.75 32.13 19.25 47.251989 43.63 68.50 33.81 25.83 31.881990 26.63 73.25 37.94 29.96 57.001991 28.13 83.50 65.38 55.50 69.88
Historical data on returnsreturn deviation^2 Year Ford Lilly Kellogg Merck HP0.042 0.032 1984 0.076 0.140 0.236 0.040 -0.2010.450 0.053 1985 0.271 0.690 0.738 0.458 0.0850.366 0.021 1986 0.455 0.332 0.489 0.808 0.1400.205 0.000 1987 0.340 0.050 0.012 0.280 0.3910.076 0.021 1988 0.340 0.096 0.227 0.093 -0.1890.194 0.001 1989 -0.136 0.602 0.052 0.342 -0.3250.175 0.002 1990 -0.390 0.069 0.122 0.160 0.7880.253 0.001 1991 0.056 0.140 0.723 0.852 0.226
mean 0.127 0.265 0.325 0.379 0.114stdevp 0.265 0.235 0.271 0.290 0.341
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Invest 17.3% in Ford, 42.6% in Lilly, 5.4% in Kellogg, 10.5% in Merck, and 24.1% in HP.
The expected return will be 22%, and the standard deviation will be 12.8%.
Conclusions
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2. Show how the optimal portfolio changes as the required return varies.
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3031323334353637383940414243444546474849505152535455
A B C D E F G HRequired Return Risk Return Ford Lilly Kellogg Merck HP
0.000 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.010 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.020 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.030 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.040 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.050 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.060 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.070 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.080 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.090 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.100 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.110 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.120 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.130 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.140 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.150 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.160 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.170 0.115 0.179 0.289 0.407 0.000 0.000 0.3040.180 0.115 0.180 0.285 0.413 0.000 0.000 0.3020.190 0.116 0.190 0.249 0.430 0.029 0.007 0.2860.200 0.119 0.200 0.224 0.429 0.038 0.039 0.2710.210 0.123 0.210 0.198 0.428 0.046 0.072 0.2560.220 0.128 0.220 0.173 0.426 0.054 0.105 0.2410.230 0.133 0.230 0.148 0.425 0.063 0.138 0.2260.240 0.139 0.240 0.122 0.424 0.071 0.171 0.211
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Optimal Portfolio
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
15% 20% 25% 30% 35%
Required Return
Pro
po
rtio
n o
f P
ort
folio
Lilly
HP
Merck
Kellogg
Ford
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3. Draw the efficient frontier for portfolios composed of these five stocks.
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Efficient Frontier
0%
5%
10%
15%
20%
25%
30%
35%
40%
45%
50%
0% 5% 10% 15% 20% 25% 30% 35% 40%
Risk (Standard Deviation)
Ex
pe
cte
d R
etu
rn
Lilly
FordHP
Kellogg
Merck
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Repeat Part 2 with shorting allowed.
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75767778798081828384858687888990919293949596979899100
A B C D E F G HRequired Return Risk Return Ford Lilly Kellogg Merck HP
0.000 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.010 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.020 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.030 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.040 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.050 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.060 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.070 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.080 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.090 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.100 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.110 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.120 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.130 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.140 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.150 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.160 0.114 0.166 0.311 0.432 0.009 -0.074 0.3220.170 0.114 0.170 0.300 0.432 0.013 -0.059 0.3150.180 0.115 0.180 0.274 0.431 0.021 -0.026 0.3000.190 0.116 0.190 0.249 0.430 0.029 0.007 0.2860.200 0.119 0.200 0.224 0.429 0.038 0.039 0.2710.210 0.123 0.210 0.198 0.428 0.046 0.072 0.2560.220 0.128 0.220 0.173 0.426 0.054 0.105 0.2410.230 0.133 0.230 0.148 0.425 0.063 0.138 0.2260.240 0.139 0.240 0.122 0.424 0.071 0.171 0.211
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Efficient Frontier
0%
5%
10%
15%
20%
25%
30%
35%
40%
45%
50%
0% 5% 10% 15% 20% 25% 30% 35% 40%
Risk (Standard Deviation)
Ex
pe
cte
d R
etu
rn
Lilly
FordHP
Kellogg
Merck
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72
• Invest in Vanguard mutual funds under university retirement plan
• No shorting
• Max 8 mutual funds
• Rebalance once per year
• Tools used: • Excel Solver
• Basic Stats (mean, stdev, correl, beta, crude version of CAPM)
Juran’s Lazy Portfolio
Decision Models -- Prof. Juran
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73Decision Models -- Prof. Juran
1999 DJ S&P2000 7.8% -10.1%2001 3.9% -13.0%2002 -14.4% -23.4%2003 31.5% 26.4%2004 15.1% 9.0%2005 10.4% 3.0%2006 15.3% 13.6%2007 8.6% 3.5%2008 -41.5% -38.5%2009 45.0% 23.5%2010 17.8% 12.8%2011 -8.5% 0.0%2012 5.6% 13.4%2013 6.0% 29.6%2014 6.6% 11.4%
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$-
$0.50
$1.00
$1.50
$2.00
$2.50
1999 2001 2003 2005 2007 2009 2011 2013
$1 Invested 12/31/1999
DJ
S&P
DJ S&PMean 7.3% 4.1%StDev 19.6% 18.8%
Correl 0.817Beta 0.853
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Summary
• Basic Optimization: Linear programming– Graphical method– Spreadsheet Method
• Extension: Nonlinear programming– Portfolio optimization