CS计算机代考程序代写 Mathematics 340 Homework 9 Due April 9th 􏰁 Only part of the problems may be graded. But, you have to submit all the problems.

Mathematics 340 Homework 9 Due April 9th 􏰁 Only part of the problems may be graded. But, you have to submit all the problems.
􏰁 Submit only pdf files.
1. [Vanderbei 13.2] On Planet Claire, markets are highly volatile. Here’s some recent historical data:
(a) Calculater1,r2,r3,andexpressthelinearprogramtomaximizeμtimestheexpected return minus the mean absolute deviation.
(b) Find the value μM where if μ > μM the optimal solution is to put all the investment into the investment with highest expected return. Hint: The optimal dictionary is worked out in Vanderbei section 13.1.2. You only need to consider the objective function.
(c) Optional: Find every portfolio on Planet Claire’s efficient frontier.
Year-Month
Hair Products (R1)
Cosmetics (R2)
Cash (R3)
2007-04 2007-03 2007-02 2007-01
1.0 2.0 2.0 0.5
2.0 2.0 0.5 2.0
1.0 1.0 1.0 1.0
Solution:
(a) Our variables are x1,x2,x3,y1,y2,y3,y4. We calculate r1 = 11, r2 = 13, r3 = 1. 88
The deviations are given by
t
R1 − r1
R2 − r2
R3 − r3
4 3 2 1
−3 8
5 8
5 8
−7 8
3 8
3 8
−9 8
3 8
0 0 0 0.
The linear program is to Maximize:
􏰀3 μ
j=1
1 􏰀4 xj rj − 4 yt
t=1
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Mathematics 340 Homework 9 Due April 9th
Subject to
3
􏰀xj􏰂Rj(t)−rj􏰃−yt ≤0 t∈{1,2,3,4} j=1
3
−􏰀xj􏰂Rj(t)−rj􏰃−yt ≤0 t∈{1,2,3,4} j=1
3
􏰀xj =1 j=1
along with xj ≥ 0 for j ∈ {1,2,3} and yt ≥ 0 for t ∈ {1,2,3,4} (nonnegativity for yt is not really necessary as it is implied by the constraints).
(b) The investment with highest expected returns is Cosmetics (R2). There is a basic feasible solution with x2 = 1, x1 = x3 = 0, and yt = | 􏰆3j=1 xj 􏰂R2(t) − r2􏰃|.
We let wt+ and wt− be the slack variables for the inequality constraints for each t and u+ and u− the slack variables for the final equality constraint. The dictionary looks like (some shortcuts are possible here)
Z = 0 +μr1 x1 +μr2 x2 +μr3 x3 −1y1 −1y2 −1y3 −1y4 4444
w+= 0 +7x −3x +y 181821
w−= 0 −7x +3x +y 181821
w+= 0 −5x +9x +y 281822
w−= 0 +5x −9x +y 281822
w+= 0 −5x −3x +y 381823
w−= 0 +5x +3x +y 381823
w+= 0 +3x −3x +y 481824
w−= 0 −3x +3x +y 481824
u+ = 1 −x1 −x2 −x3 u− = −1 +x1 +x2 +x3
First let us pivot with x2 entering and u+ leaving. We can then get to the basic
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Mathematics 340 Homework 9 Due April 9th
feasible solution above by pivoting in all the y variables.
μr2 +μ(r1 −r2)x1 −μr2 u+ +μ(r3 −r2)x3
3 −10×1 −3u+ −3×3 +w1+
Z =
w+= +9 −14x −9u+ −9x +y 28818832
−1y1 −1y2 −1y3 −1y4 4444
y1 = w− =
8888
3 −10x −3u+ −3x +y 18818831
+w2−
+ w 3+
w−= +3 +2x −3u+ −3x 38818833
y 4 = 3 − 6 x 1 − 3 u + − 3 x 3 8888
w−= +3 −6x −3u+ −3x 4881883 4
x2 = 1 −x1 −u+ −x3 u− = 0 −u+
We update the objective function
Z = μr2 − 1􏰇3 + 9 + 3 + 3􏰈 48888
+μ(r1 −r2)x1 − 1􏰇− 10 − 14 + 2 − 6􏰈x1 48888
+μr2u− −1􏰇3+9+3+3􏰈u− 48888
+μ(r3 −r2)x3 − 1􏰇− 3 − 9 − 3 − 3􏰈x3 48888
= μ 13 − 9 8 16
μ􏰂− 2)x1 + 7×1 88
− μ 13 u− − 9 u− 8 16
+μ􏰂−5􏰃x3+ 9×3. 8 16
y2 = 9 −14×1 −9u+ −9×3 8888
y 3 = 3 + 2 x 1 − 3 u + − 3 x 3 8888
+y
+ w 4+ +y
Forthistobeoptimalweneedμ2 ≥7 soμ≥7 andμ5 ≥ 9 soμ≥ 9.Clearly
the larger is μM = 7. 2
8 8 2 8 16 10
2. Consider the following investments
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Mathematics 340 Homework 9 Due April 9th
􏰁 Canada bonds: Vanguard Canadian Aggregate Bond Index ETF (VAB)
􏰁 Canada stocks: Vanguard FTSE Canada All Cap Index ETF (VCN)
􏰁 Non-Canada stocks: Vanguard FTSE Global All Cap ex Canada Index ETF (VXC) 􏰁 Nixed 60% Stocks, 30% Canada: Vanguard Balanced ETF Portfolio (VBAL)
and their historical prices for the last year (in accompanying .csv file).
(a) Use PulP to compute the optimal portfolio with risk/return parameter μ = 0, 0.5, 1 given the historical data. (Hint: see accompanying .ipynb file for a start, you will also need to upload the .csv file to syzygy)
(b) Repeat part (a) after adding the constrants appearing on the midterm: no more than 90% Stocks and no more than 80% Canada. How have the solutions changed?
Solution:
See HW9P2Sol for the Jupyter file. (a) Solutions are
(b) The new solutions are
μ
VAB
VBAL
VCN
VXC
0
1
0
0
0
0.5
0.2
0.4
0
0.4
1
0
0
0
1
μ
VAB
VBAL
VCN
VXC
0
0.71
0.29
0
0
0.5
0.2
0.4
0
0.4
1
0.1
0
0
0.9
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Mathematics 340 Homework 9 Due April 9th 3. The following investment opportunities in Apple stock (AAPL) are available:
􏰁 A call option to purchase the stock at a price of 130$ in one month (h1(S1) = max{0, S1 − 130}), at a cost of 3 dollars.
􏰁 A call option to purchase the stock at a price of 140$ in one month (h2(S1) = max{0, S1 − 140}), at a cost of 1 dollar.
You want to price the call order to purchase the stock in one month at a price of 135: h3(S1) = max{0, S1 − 135}.
(a) Find all convex combinations of h1 and h2 that dominate h3. Considering only those two alternatives, what is the maximum price for the new call option to satisfy the no-arbitrage condition.
(b) Find all convex combinations of h1 and h2 that are dominated by h3. What does this say about the minimum price for the new call option.
(c) Given that the current market price of AAPL is 123$ and the monthly returns of AAPL for the last 5 months were 0.93, 1.11, 1.10, 1.00, and 0.93, what is the expected return and mean absolute deviation (MAD) of the new call option (h3(S1) = max{0, S1 − 135}) in one month.
Solution:
(a) For S1 ≥ 140, We require that
(S1 − 130)x2 + (S1 − 140)x3 ≥ (S1 − 135),
which reduces to or equivalently
−130×2 − 140×3 ≥ −135. −10×3 ≥ −5,×3 ≤ 1.
2 (S1 − 130)x2 ≥ S1 − 135,
We also require
whichishardesttosatisfyatS1 =140so10x2 ≥5,orx2 ≥ 1,whichisthe
same.
The LP for the maximum price is to minimize
3×2 + x3
subject to x2 ≥ 1. Anyways, the answer is 2. So the maximum price we could 2
offer is 2.
2
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Mathematics 340 Homework 9 Due April 9th
(b) The only strategy that is dominated by the new call option is when x3 = 1. This says that the minimum price is 1.
(c) The expected return is
0 + 0 + 1􏰄1.11 ∗ 123 − 135􏰅 + 1􏰄1.10 ∗ 123 − 135􏰅 + 0 = 0.37. 55
Even at a price of 1 dollar it does not seem to be a good buy. The MAD is
3 ∗ 1|0.37| + 1|1.11 ∗ 123 − 135 − 0.37| + 1|1.10 ∗ 123 − 135 − 0.37| = 0.47. 555
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