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<h2 class="hd hd-2 unit-title">Problem 1</h2>
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PART A
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<p>Consider the following nonlinear constrained minimization problem:</p>
<span>
<p class="equation">\[\left.\begin{array}{rrcl} \min &amp; \sqrt{x_2} \\ \text {s.t.:} &amp; &amp; &amp; \\ &amp; x_2 \geq (\frac{5}{2} x_1)^3 \\ &amp; x_2 \geq (-\frac{3}{2}x_1 + 1)^3 \\ &amp; x_2 \geq 0 \end{array}\right\}\]</p>
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<p>A plot can be found below</p>
<img width="500" src="/assets/courseware/v1/aad3a0beffcba220c6528ef299dd36b2/asset-v1:MITx+15.053x+3T2016+type@asset+block/R6P1plot.png"/>
<p>
<i>Hint: Several algorithms will work, but we suggest the method of moving asymptotes, the details of which are not necessary to understand for the time being.</i>
</p>
<p>To two decimals, what is the objective value?</p>
<p>You can solve using spreadsheet optimization or using Julia. If you use Julia, you will need the following additional syntax <pre><code>
Pkg.add("NLopt")
yourModelVariable=Model(solver=NLoptSolver(algorithm=:LD_MMA))
@NLobjective(yourModelVariable, Max/Min, Function)
@NLconstraint(yourModelVariable, Inequality))
setvalue(variable, initialValue)
</code></pre></p>
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<h2 class="hd hd-2 unit-title">Problem 2</h2>
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PART A
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<p>An investor has $5000 and two potential investments. Let \(x_j \text{ for } j = 1 \text{ and } j = 2 \) denote his allocation to investment \(j \) in thousands of dollars. From historical data, investments 1 and 2 have an expected annual return of 20 and 16 percent, respectively. Also, the total risk involved with investments 1 and 2, as measured by the variance of total return, is given by \(2x_1^2 +x_2^2 +(x_1+x_2)^2\), so that risk increases with total investment and with the amount of each individual investment. The investor would like to maximize his expected return and at the same time minimize his risk. Clearly, both of these objectives cannot, in general, be satisfied simultaneously. There are several possible approaches. For example, he can minimize risk subject to a constraint imposing a lower bound on expected return. Alternatively, expected return and risk can be combined in an objective function, to give the model:</p>
<span>
<p class="equation">\[\left.\begin{array}{rrcl} \max &amp; f(x) = 20x_1 + 16x_2 &#8722; \theta[2x_1^2 + x_2^2 + (x_1 + x_2)^2] \\ \text {s.t.:} &amp; &amp; &amp; \\ &amp; x_1 + x_2 \leq 5 \\ &amp; x_1, x_2 \geq 0 \\ &amp; 0 \leq x_i \leq 5, i \in \{1,2\} \end{array}\right\}\]</p>
</span>
<p>Suppose \(\theta=1\). To two decimals, what is the objective value?</p>
<p>You can solve using spreadsheet optimization or using Julia. If you use Julia, you will need the following additional syntax: <pre><code>
Pkg.add("NLopt")
yourModelVariable=Model(solver=NLoptSolver(algorithm=:LD_MMA))
@NLobjective(yourModelVariable, Max/Min, Function)
@NLconstraint(yourModelVariable, Inequality))
</code></pre></p>
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PART B
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<p>Next, we would like to study the objective for different values of \(\theta\). Specifically, for \(\theta \in \{0, 1, \ldots, 5\} \), we would like to determine where the objective is maximized. </p>
<p>For which of these six values value of \(\theta\) is the optimum objective value maximized?</p>
<p>If you use Julia and JuMP, then create a vector containing the values of \(\theta\), and then use a loop. If you use a spreadsheet, solve the problem six times and record your answers.</p>
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PART C
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<p>Next, we would like to plot the expected return as a function of \(\theta\) for \(\theta \in \{0, \ldots, 5\} \) .</p>
<p>
<i>If you use Julia and JuMP, you will need the following additional syntax</i>
<pre><code>
Pkg.add("PyPlot")
using PyPlot
plot(xVector, yVector, label="your label", xlabel="...", ylabel="...")
</code></pre>
</p>
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PART A
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<p>Consider the following second-order cone programming problem:</p>
<span>
<p class="equation">\[\left.\begin{array}{rrcl} \min &amp; t \\ \text {s.t.:} &amp; &amp; &amp; \\ &amp; ||x||_2 \leq t \\ &amp; x_1+x_2 \geq 1 \\ &amp; 0 \leq x_i \leq 1, i \in \{1,2\} \end{array}\right\}\]</p>
</span>
<p>You can solve using spreadsheet optimization or using Julia. If you use Julia, you will need the following additional syntax: <pre><code>
Pkg.add("ECOS")
using ECOS
@constraint(yourModelVariable, soc,norm2{x[i], i=start:end} [insert inequality] t)
yourModelVariable=Model(solver=ECOSSolver())
@show getvalue(variable of interest)
</code></pre></p>
<p>To two decimals, what is the optimal objective value?</p>
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\(\)
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<p> For each part, determine whether the function is convex or not over its domain.</p>
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</a>
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<p> \( f(x) = |x-1| \) for \(-3 \le x \le 3 \). </p>
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<p> \( g(x) = x^2 -3 \) for \(-3 \le x \le 3 \). </p>
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Convex
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PART C
</h3>
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</a>
</p>
<p> \( h(x) = \max\{|x-1| , x^2-3\} \) for \( -3 \le x \le 3. \) </p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_2925dae3ab7d496ca438f506cae8baef_2_1">
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Not Convex
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Convex
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PART D
</h3>
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<img width="600" height="439" alt="Plot1" src="/assets/courseware/v1/6fb2f6bd7ed06236840b3b406ad2d78c/asset-v1:MITx+15.053x+3T2016+type@asset+block/R6P4dd.png"/>
</a>
</p>
<p> \(h(x) = \min\{|x-1| , x^2-3\} \) for \( -3 \le x \le 3. \) </p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_ffd6a5f5b999492f9fe35266cd27239b_2_1">
<fieldset aria-describedby="status_ffd6a5f5b999492f9fe35266cd27239b_2_1">
<div class="field">
<input type="radio" name="input_ffd6a5f5b999492f9fe35266cd27239b_2_1" id="input_ffd6a5f5b999492f9fe35266cd27239b_2_1_choice_0" class="field-input input-radio" value="choice_0"/><label id="ffd6a5f5b999492f9fe35266cd27239b_2_1-choice_0-label" for="input_ffd6a5f5b999492f9fe35266cd27239b_2_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_ffd6a5f5b999492f9fe35266cd27239b_2_1">
Convex
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<div class="field">
<input type="radio" name="input_ffd6a5f5b999492f9fe35266cd27239b_2_1" id="input_ffd6a5f5b999492f9fe35266cd27239b_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="ffd6a5f5b999492f9fe35266cd27239b_2_1-choice_1-label" for="input_ffd6a5f5b999492f9fe35266cd27239b_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_ffd6a5f5b999492f9fe35266cd27239b_2_1">
Not Convex
</label>
</div>
<span id="answer_ffd6a5f5b999492f9fe35266cd27239b_2_1"/>
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PART E
</h3>
<div class="problem-progress" id="block-v1:MITx+15.053x+3T2016+type@problem+block@2163a96241a14a63a5b4eebc0b224868-problem-progress"></div>
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<p>
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<img width="600" height="439" alt="Plot1" src="/assets/courseware/v1/35933cdc22755a229e89bca9ef283591/asset-v1:MITx+15.053x+3T2016+type@asset+block/R6P4e2.png"/>
</a>
</p>
<p> \( h(x) = |x-1| - (x^2-3) \) for \( -3 \le x \le 3. \) </p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_2163a96241a14a63a5b4eebc0b224868_2_1">
<fieldset aria-describedby="status_2163a96241a14a63a5b4eebc0b224868_2_1">
<div class="field">
<input type="radio" name="input_2163a96241a14a63a5b4eebc0b224868_2_1" id="input_2163a96241a14a63a5b4eebc0b224868_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="2163a96241a14a63a5b4eebc0b224868_2_1-choice_1-label" for="input_2163a96241a14a63a5b4eebc0b224868_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_2163a96241a14a63a5b4eebc0b224868_2_1">
Not Convex
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</div>
<div class="field">
<input type="radio" name="input_2163a96241a14a63a5b4eebc0b224868_2_1" id="input_2163a96241a14a63a5b4eebc0b224868_2_1_choice_0" class="field-input input-radio" value="choice_0"/><label id="2163a96241a14a63a5b4eebc0b224868_2_1-choice_0-label" for="input_2163a96241a14a63a5b4eebc0b224868_2_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_2163a96241a14a63a5b4eebc0b224868_2_1">
Convex
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</div>
<span id="answer_2163a96241a14a63a5b4eebc0b224868_2_1"/>
</fieldset>
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<span class="status unanswered" id="status_2163a96241a14a63a5b4eebc0b224868_2_1" data-tooltip="Not yet answered.">
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</span>
</div>
</div></div>
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<span id="solution_2163a96241a14a63a5b4eebc0b224868_solution_1"/>
</div></div>
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