Lecture 21
November 18, 2024
Text: VSRIKRISH to 22333
Certainty (e.g. all parameters and relationships between decision variables and outcomes are known) is a key assumption of linear programming.
But most systems problems aren’t actually deterministic.
A Common Approach:
Alternative:
We can represent uncertainty through scenarios and (conditional) sequential events, then rewrite the decision problem.
\[ \min_x \mathbb{E}_\xi \left[f(x; \xi)\right] \]
Two Stage:
\[\min_x f(x) + \mathbb{E}_\xi\left[Q(x; \xi)\right]\]
where \(Q(x; \xi)\) is the recourse decision made after realizing the uncertainty \(\xi\).
This framework can work for:
A farmer can grow wheat, corn, and sugar beets on 500 ha of land. How much land should they allocate to each crop?
From prior experience, the farmer knows that in an average year, yields are:
What are the decision variables?
Variable | Definition |
---|---|
\(x_i\) | ha of crop \(i\) planted |
\(y_i\) | T of crop \(i\) purchased |
\(z_i\) | T of crop \(i\) sold |
One twist: sold beets will be represented by two different \(z\)-variables (\(z_3\) and \(z_4\)), based on whether we exceed 6000 T.
\[ \begin{align} \min_{x, y, z} & \quad 150x_1 + 230 x_2 + 260 x_3 + 238 y_1 + 210 y_2 - 170 z_1 \\ & \qquad -150z_2 - 36z_3 - 10z_4 \\[1ex] \text{subject to:} \quad & x_1 + x_2 + x_3 \leq 500 \\[0.5ex] & 2.5 x_1 + y_1 - z_1 \geq 200 \\[0.5ex] & 3 x_2 + y_2 - z_2 \geq 240 \\[0.5ex] & z_3 + z_4 \leq 20 x_3 \\[0.5ex] & z_3 \leq 6000 \\[0.5ex] & x_i, y_i, z_i \geq 0 \end{align} \]
Variable | Wheat | Corn | Beets |
---|---|---|---|
Area (ha) | 120 | 80 | 300 |
Yield (T) | 300 | 240 | 6000 |
Sales (T) | 100 | – | 6000 |
Purchased (T) | – | – | – |
This solution yields a profit of $118,600.
But yields tend to vary from year to year. From prior experience, the historical variability is:
To simplify things, let’s assume these yields vary consistently across crops and that each scenario (average, good, bad) has an equal probability of occurrence.
What should change about our decision variables?
Variable | Definition |
---|---|
\(x_i\) | ha of crop \(i\) planted |
\(y_{ij}\) | T of crop \(i\) purchased in scenario \(j\) |
\(z_{ij}\) | T of crop \(i\) sold in scenario \(j\) |
This is typical of a two-stage decision problem:
How can we formulate an objective?
First choice: what statistic are we trying to optimize?
Let’s choose the expected value \(\mathbb{E}\left[\text{Profit}\right]\).
Other possible choices: quantiles (robust optimization) to hedge against worst-case outcomes, variance (to minimize year-on-year fluctuations).
Since we have a discrete (and small) set of possible outcomes, we can use a scenario tree to write out the possible outcomes.
Our new objective becomes:
\[\begin{alignedat}{2} &\min_{x, y, z} & & \quad 150x_1 + 230x_2 + 260x_3 \\[0.5ex] & &&\qquad -\frac{1}{3} \left(170z_{11} + 150z_{21} + 36z_{31} + 10z_{41} - 238y_{11} - 210 y_{21}\right) \\[0.5ex] & &&\qquad -\frac{1}{3} \left(170z_{12} + 150z_{22} + 36z_{32} + 10z_{42} - 238y_{12} - 210 y_{22}\right) \\[0.5ex] & &&\qquad -\frac{1}{3} \left(170z_{13} + 150z_{23} + 36z_{33} + 10z_{43} - 238y_{13} - 210 y_{23}\right) \end{alignedat}\]
\[\begin{alignedat}{2} &x_1 + x_2 + x_3 \leq 500 \\[0.5ex] &\color{blue}3x_1 + y_{11} - z_{11} \geq 200, && \qquad \color{blue}3.6x_2 + y_{21} - z_{21} \geq 240 \\[0.5ex] &\color{blue}z_{31} + z_{41} \leq 24x_3, && \qquad \color{blue}z_{31} \leq 6000 \\[0.5ex] &\color{purple}2.5x_1 + y_{12} - z_{12} \geq 200, && \qquad \color{purple}3x_2 + y_{22} - z_{22} \geq 240 \\[0.5ex] &\color{purple}z_{32} + z_{42} \leq 20x_3, && \qquad \color{purple}z_{32} \leq 6000 \\[0.5ex] &\color{red}2x_1 + y_{13} - z_{13} \geq 200, && \qquad \color{red}2.4x_2 + y_{23} - z_{23} \geq 240 \\[0.5ex] &\color{red}z_{33} + z_{43} \leq 16x_3, && \qquad \color{red}z_{33} \leq 6000 \\[0.5ex] &x_i, y_i, z_i \geq 0 \end{alignedat}\]
Variable | Wheat | Corn | Beets |
---|---|---|---|
Deterministic Area (ha) | 120 | 80 | 300 |
Stochastic Area (ha) | 170 | 80 | 250 |
How do we interpret the differences?
Can we directly compare the deterministic and stochastic solution values that we have obtained so far?
Year | Deterministic Profit | Stochastic Profit |
---|---|---|
Good | $148,000 | $167,000 |
Average | $118,600 | $109,350 |
Bad | $56,800 | $48,820 |
Expected Profit from Deterministic Solution under uncertainty: $107,240
We can now compare the difference in expected profits under uncertainty, which is the fair comparison.
This is called the value of the stochastic solution (VSS), or sometimes the expected value of including uncertainty (EVIU).
\[\begin{aligned} VSS &= \mathbb{E}[\text{Stochastic Profit}] - \mathbb{E}[\text{Deterministic Profit}] \\[0.5ex] &= \$108,390 - \$107,240 \\[0.5ex] &= \$1,150. \end{aligned}\]
Another relevant quantity, the expected value of perfect information (EVPI), concerns the value associated with better forecasts.
If the farmer had perfect foresight and could allocate acreage \(x\) accordingly and sell/purchase \(z\) and \(y\) optimally (versus taking the 1/3-probability stochastic solution),
\[\mathbb{E}[\text{Profits} | \text{perfect information}] = \sum_j p_j \times \text{Profits}_j = \$115,405.\]
\[\text{EVPI} = \$115,405 - \$108,390 = \$7,015\]
The EVPI can be interpreted as the amount the farmer might be willing to pay for an improved forecast.
Wednesday: Limits of Mathematical Programming
Monday: Prelim 2 Review, Wrap-Up
HW5: Due 12/5.