[docs] add multiobjective portfolio example (#3227)

jump-dev · Feb 22, 2023 · 6a422d3 · 6a422d3
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diff --git a/docs/Project.toml b/docs/Project.toml
@@ -35,7 +35,7 @@ JSON = "0.21"
 JSONSchema = "1"
 Literate = "2.8"
 MathOptInterface = "=1.12.0"
-MultiObjectiveAlgorithms = "=0.1.1"
+MultiObjectiveAlgorithms = "=0.1.3"
 Plots = "1"
 SCS = "=1.1.3"
 SQLite = "1"

diff --git a/docs/src/tutorials/nonlinear/introduction.md b/docs/src/tutorials/nonlinear/introduction.md
@@ -39,7 +39,7 @@ will help you know where to look for certain things.
    new to JuMP.
    * [Rocket Control](@ref)
    * [Optimal control for a Space Shuttle reentry trajectory](@ref)
-   * [Quadratic portfolio optimization](@ref)
+   * [Portfolio optimization](@ref)
  * The [Tips and tricks](@ref nonlinear_tips_and_tricks) tutorial contains a
    number of helpful reformulations and tricks you can use when modeling
    nonlinear programs. Look here if you are stuck trying to formulate a problem

diff --git a/docs/src/tutorials/nonlinear/portfolio.jl b/docs/src/tutorials/nonlinear/portfolio.jl
@@ -18,7 +18,7 @@
 # OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE  #src
 # SOFTWARE.                                                                      #src
 
-# # Quadratic portfolio optimization
+# # Portfolio optimization
 
 # **Originally Contributed by**: Arpit Bhatia
 
@@ -29,10 +29,19 @@
 # This tutorial solves the famous Markowitz Portfolio Optimization problem with
 # data from [lecture notes from a course taught at Georgia Tech by Shabbir Ahmed](https://www2.isye.gatech.edu/~sahmed/isye6669/).
 
+# ## Required packages
+
 # This tutorial uses the following packages
+
 using JuMP
+import DataFrames
 import Ipopt
+import MultiObjectiveAlgorithms as MOA
+import Plots
 import Statistics
+import StatsPlots
+
+# ## Formulation
 
 # Suppose we are considering investing 1000 dollars in three non-dividend paying
 # stocks, IBM (IBM), Walmart (WMT), and Southern Electric (SEHI), for a
@@ -88,7 +97,7 @@ import Statistics
 # this form. We can also write this equation as:
 
 # ```math
-# \operatorname{Var}\left[\sum_{i=1}^{3} \tilde{r}_{i} x_{i}\right] =x^{T} Q x
+# \operatorname{Var}\left[\sum_{i=1}^{3} \tilde{r}_{i} x_{i}\right] =x^\top Q x
 # ```
 
 # Where $Q$ is the covariance matrix for the random vector $\tilde{r}$.
@@ -97,77 +106,150 @@ import Statistics
 
 # ```math
 # \begin{aligned}
-# \min x^{T} Q x \\
-# \text { s.t. } \sum_{i=1}^{3} x_{i} \leq 1000.00 \\
-# \overline{r}^{T} x \geq 50.00 \\
+# \min x^\top Q x \\
+# \text { s.t. } \sum_{i=1}^{3} x_{i} \leq 1000 \\
+# \overline{r}^\top x \geq 50 \\
 # x \geq 0
 # \end{aligned}
 # ```
 
-# After that long discussion, let's now use JuMP to solve the portfolio
-# optimization problem for the data given below.
-
-# | Month        |  IBM     |  WMT    |  SEHI  |
-# |--------------|----------|---------|--------|
-# | November-00  |  93.043  |  51.826 |  1.063 |
-# | December-00  |  84.585  |  52.823 |  0.938 |
-# | January-01   |  111.453 |  56.477 |  1.000 |
-# | February-01  |  99.525  |  49.805 |  0.938 |
-# | March-01     |  95.819  |  50.287 |  1.438 |
-# | April-01     |  114.708 |  51.521 |  1.700 |
-# | May-01       |  111.515 |  51.531 |  2.540 |
-# | June-01      |  113.211 |  48.664 |  2.390 |
-# | July-01      |  104.942 |  55.744 |  3.120 |
-# | August-01    |  99.827  |  47.916 |  2.980 |
-# | September-01 |  91.607  |  49.438 |  1.900 |
-# | October-01   |  107.937 |  51.336 |  1.750 |
-# | November-01  |  115.590 |  55.081 |  1.800 |
-
-stock_data = [
-    93.043 51.826 1.063
-    84.585 52.823 0.938
-    111.453 56.477 1.000
-    99.525 49.805 0.938
-    95.819 50.287 1.438
-    114.708 51.521 1.700
-    111.515 51.531 2.540
-    113.211 48.664 2.390
-    104.942 55.744 3.120
-    99.827 47.916 2.980
-    91.607 49.438 1.900
-    107.937 51.336 1.750
-    115.590 55.081 1.800
-]
-
-# Calculating stock returns
-
-stock_returns = Array{Float64}(undef, 12, 3)
-for i in 1:12
-    stock_returns[i, :] =
-        (stock_data[i+1, :] .- stock_data[i, :]) ./ stock_data[i, :]
-end
-stock_returns
+# ## Data
+
+# For the data in our problem, we use the stock prices given below, in monthly
+# values from November 2000, through November 2001.
 
-# Calculating the expected value of monthly return:
+df = DataFrames.DataFrame(
+    [
+        93.043 51.826 1.063
+        84.585 52.823 0.938
+        111.453 56.477 1.000
+        99.525 49.805 0.938
+        95.819 50.287 1.438
+        114.708 51.521 1.700
+        111.515 51.531 2.540
+        113.211 48.664 2.390
+        104.942 55.744 3.120
+        99.827 47.916 2.980
+        91.607 49.438 1.900
+        107.937 51.336 1.750
+        115.590 55.081 1.800
+    ],
+    [:IBM, :WMT, :SEHI],
+)
 
-r = Statistics.mean(stock_returns; dims = 1)
+# Next, we compute the percentage return for the stock in each month:
 
-# Calculating the covariance matrix Q
+returns = diff(Matrix(df); dims = 1) ./ Matrix(df[1:end-1, :])
 
-Q = Statistics.cov(stock_returns)
+# The expected monthly return is:
 
-# JuMP Model
+r = vec(Statistics.mean(returns; dims = 1))
 
-portfolio = Model(Ipopt.Optimizer)
-set_silent(portfolio)
-@variable(portfolio, x[1:3] >= 0)
-@objective(portfolio, Min, x' * Q * x)
-@constraint(portfolio, sum(x) <= 1000)
-@constraint(portfolio, sum(r[i] * x[i] for i in 1:3) >= 50)
-optimize!(portfolio)
+# and the covariance matrix is:
 
-objective_value(portfolio)
+Q = Statistics.cov(returns)
 
-#-
+# ## JuMP formulation
+
+model = Model(Ipopt.Optimizer)
+set_silent(model)
+@variable(model, x[1:3] >= 0)
+@objective(model, Min, x' * Q * x)
+@constraint(model, sum(x) <= 1000)
+@constraint(model, r' * x >= 50)
+optimize!(model)
+solution_summary(model)
+
+# The optimal allocation of our assets is:
 
 value.(x)
+
+# So we spend \$497 on IBM, and \$503 on SEHI. This results in a variance of:
+
+scalar_variance = value(x' * Q * x)
+
+# and an expected return of:
+
+scalar_return = value(r' * x)
+
+# ## Multi-objective portfolio optimization
+
+# The previous model returned a single solution that minimized the variance,
+# ensuring that our expected return was at least \$50. In practice, we might
+# be willing to accept a slightly higher variance if it meant a much increased
+# expected return. To explore this problem space, we can instead formulate our
+# portfolio optimization problem with two objectives:
+#
+#  1. to minimize the variance
+#  2. to maximize the expected return
+#
+# The solution to this biobjective problem is the
+# [efficient frontier](https://en.wikipedia.org/wiki/Efficient_frontier) of
+# modern portfolio theory, and each point in the solution is a point with the
+# best return for a fixed level of risk.
+
+model = Model(() -> MOA.Optimizer(Ipopt.Optimizer))
+set_silent(model)
+
+# We also need to choose a solution algorithm for `MOA`. For our problem, the
+# efficient frontier will have an infinite number of solutions. Since we cannot
+# find all of the solutions, we choose an approximation algorithm and limit the
+# number of solution points that are returned:
+
+set_optimizer_attribute(model, MOA.Algorithm(), MOA.EpsilonConstraint())
+set_optimizer_attribute(model, MOA.SolutionLimit(), 50)
+
+# Now we can define the rest of the model:
+
+@variable(model, x[1:3] >= 0)
+@constraint(model, sum(x) <= 1000)
+@expression(model, variance, x' * Q * x)
+@expression(model, expected_return, r' * x)
+## We want to minimize variance and maximize expected return, but we must pick
+## a single objective sense `Min`, and negate any `Max` objectives:
+@objective(model, Min, [variance, -expected_return])
+optimize!(model)
+solution_summary(model)
+
+# The algorithm found 50 different solutions. Let's plot them to see how they
+# differ:
+
+objective_space = Plots.hline(
+    [scalar_return];
+    label = "Single-objective solution",
+    linecolor = :red,
+)
+Plots.vline!(objective_space, [scalar_variance]; label = "", linecolor = :red)
+Plots.scatter!(
+    objective_space,
+    [value(variance; result = i) for i in 1:result_count(model)],
+    [value(expected_return; result = i) for i in 1:result_count(model)];
+    xlabel = "Variance",
+    ylabel = "Expected Return",
+    label = "",
+    title = "Objective space",
+    markercolor = "white",
+    markersize = 5,
+    legend = :bottomright,
+)
+for i in 1:result_count(model)
+    y = objective_value(model; result = i)
+    Plots.annotate!(objective_space, y[1], -y[2], (i, 3))
+end
+
+decision_space = StatsPlots.groupedbar(
+    vcat([value.(x; result = i)' for i in 1:result_count(model)]...);
+    bar_position = :stack,
+    label = ["IBM" "WMT" "SEHI"],
+    xlabel = "Solution #",
+    ylabel = "Investment (\$)",
+    title = "Decision space",
+)
+Plots.plot(objective_space, decision_space; layout = (2, 1), size = (600, 600))
+
+# Perhaps our trade-off wasn't so bad after all! Our original solution
+# corresponded to picking a solution #17. If we buy more SEHI, we can increase
+# the return, but the variance also increases. If we buy less SEHI, such as a
+# solution like #5 or #6, then we can achieve the corresponding return without
+# deploying all of our capital. We should also note that at no point should we
+# buy WMT.