HW4-2023-Solutions

1 Problem 1 (60 points): Two-stage Stochastic Optimization Part a) Set ? 0 , ? 0 as the land given to plant rice and corn respectively, and ? 𝑖 , ? 𝑖 as the land for rice and crop given up under scenario i (i = 1,2,3,4,5 and stands for vary dry, dry, normal, wet and very wet conditions respectively). The problem could be formulated as follows. There are several different formulations of the objective function. Two possible formulations are given as follows. Note, it is also correct if you choose the decision variables at the second stage as the amount of the crops you irrigate at the second stage instead of the crops you give up at the second stage. One formulation for the objective is given as follows. In this formulation, in the first stage, you only have planting cost, and then in the second stage, you pay the water cost (only for the potion you decide to keep), and you get benefit (only out of the potion that you decide you keep). This is actually the most natural way to think of this problem. max ??? ????𝑖? = −[?𝑙???𝑖?? ????] ?𝑖??? ??𝑎?? + [????𝑖? − ????? ???? ] ?????? ??𝑎?? = (−0.25? 0 − 0.1? 0 ) × 1000 + ∑ ? 𝑖 (2.0(? 0 − ? 𝑖 ) + 0.5(? 0 − ? 𝑖 ) − 0.05 × 3(? 0 − ? 𝑖 ) − 0.05 × 1(? 0 − ? 𝑖 )) 5 𝑖=1 × 1000 = 1600? 0 + 350? 0 − ∑ ? 𝑖 (1850 ? 𝑖 + 450 ? 𝑖 ) 5 𝑖=1 ($) An alternative formulation is given as follows, which calculates the net profit by subtracting the net profit you give up at the second stage from the net profit you will gain if you keep planting all the crops you planted at the first stage. max ??? ????𝑖? = [????𝑖? − ?𝑙???𝑖?? ???? − ????? ????] ?𝑖??? ??𝑎?? − [????𝑖? − ????? ???? ] ?𝑖??? ?? 𝑖? ?ℎ? ?????? ??𝑎?? = (2.0 ? 0 + 0.5 ? 0 − 0.25? 0 − 0.1? 0 − 0.05 × 3? 0 − 0.05 × 1? 0 ) × 1000 − ∑ ? 𝑖 (2.0 ? 𝑖 + 0.5 ? 𝑖 − 0.05 × 3? 𝑖 − 0.05 × 1? 𝑖 ) 5 𝑖=1 × 1000 = 1600? 0 + 350? 0 − ∑ ? 𝑖 (1850 ? 𝑖 + 450 ? 𝑖 ) 5 𝑖=1 ($) Here, ? 𝑖 is the probability for each possible scenario, and their value could be found in the following table. ? 1 ? 2 ? 3 ? 4 ? 5 0.2 0.25 0.40 0.1 0.05

2 And the problem should be solved subject to the following constraints. 1. Land constraints. ? 0 + ? 0 ≤ 100 (𝑖 = 1,2,3,4,5) 2. Water constraints. 3(? 0 − ? 𝑖 ) + (? 0 − ? 𝑖 ) ≤ 𝐴??𝑖𝑙??𝑙? ????? 𝑖 Here, 𝐴??𝑖𝑙??𝑙? ????? 𝑖 (unit ha.foot) is the amount of water available under each scenario, and their values could be found in the following table. 𝐴??𝑖𝑙??𝑙? ????? 1 𝐴??𝑖𝑙??𝑙? ????? 2 𝐴??𝑖𝑙??𝑙? ????? 3 𝐴??𝑖𝑙??𝑙? ????? 4 𝐴??𝑖𝑙??𝑙? ????? 5 20 45 60 75 90 3. Total given up limitations. ? 𝑖 ≤ ? 0 ? 𝑖 ≤ ? 0 4. Non-negative constraints. ? 𝑖 , ? 𝑖 ≥ 0 (𝑖 = 0,1,2,3,4,5. ) Solving the problem by excel solver, and we can get the following solution. First stage Very dry Dry Normal Wet Very Wet x (ha) 25.00 18.33 10.00 5.00 0.00 0.00 y (ha) 0.00 0.00 0.00 0.00 0.00 0.00 And under this condition, the total profit will be $ 24891.67 . Part b) The shadow price of water under different conditions are shown as follows. Very dry Dry Normal Wet Very Wet Shadow Price ($/ha.foot) 123.3333 154.1667 246.6667 9.166667 0 As we can see from the table above, the shadow price for water first goes up with water availability from very dry to normal and then goes down with water availability from normal to very wet. There are three different factors that could affect the shadow price of water. The first factor is that whether the water constraint is a binding constraint under the scenario. If water constraint is not binding under one scenario, the shadow price of water will be zero. For example, under very wet condition, the water constraint is not binding, namely, there is still abundant water. Therefore, the total profit will not go up with unit increment in the water availability under such scenario. Second, the probability of a scenario also affects the shadow price of water. Scenarios with higher possibility have higher shadow price, as it has more possibility to add the profit with one unit increase in water availability under scenarios with higher possibility. Actually, the shadow price of water under scenarios of very dry, dry and normal is proportional to the possibility of the scenarios. Third, whether increasing water availability under one scenario will affect the first stage decision.

4 average net profit is higher in the deterministic case ($27,333.33) than the stochastic case ($24,891.67). By having complete knowledge of future water availability, the farmer is able to plant less (saving planting costs) to maximize the net profit. Without complete knowledge of future water availability (probabilities of each scenario are known), the farmer must plant more (incurring higher planting costs) and later give up those crops to maximize the expected net profit. The table below compares the total irrigated land for each crop for the stochastic and deterministic models: Stochastic Deterministic Scenario Rice Irrigated (ha) Corn Irrigated (ha) Net Profit ($) Rice Irrigated (ha) Corn Irrigated (ha) Net Profit ($) Very Dry 6.67 0 $24,891.67 6.67 0 $10,666.67 Dry 15 0 15 0 $24,000.00 Normal 20 0 20 0 $32,000.00 Wet 25 0 25 0 $40,000.00 Very Wet 25 0 30 0 $48,000.00 The amount of rice irrigated is very similar between the models. They differ for the Very Wet scenario, which is highly unlikely. The stochastic model recommends planting less than the optimal amount of rice for the Very Wet scenario because the scenario is unlikely (probability = 0.05) and would incur more planting costs for the other, more likely scenarios.

5 Problem 2 (40 points) Decision variables: ? 1 : ???? ?? ?𝑖??; ? 2 :???? ?? ???? Objective function: max ? = 2? 1 + 0.5? 2 Constraints: { ? 1 + ? 2 ≤ 100 𝑃???(????? ??? ≤ ????? ???𝑖𝑙??𝑖𝑙𝑖??) ≥ 𝛼 ⇒ 3? 1 + ? 2 ≤ 𝜇 − 𝜎Φ −1 (𝛼) For question (a) Plug in 𝛼 = 0.95 , Φ −1 (0.95) = 1.645 , 𝜇 = 60 , 𝜎 = 10 to the 2 nd constraint, we can obtain the solution: ? 1 ∗ = 14.52 ? 2 ∗ = 0, ? ∗ = 29.03 For question (b) Replace the 2 nd constraint with 3? 1 + ? 2 ≤ 60 , solve the model and we obtain: ? 1 ∗ = 20, ? 2 ∗ = 0, ? ∗ = 40 The uncertainty of water availability will cause less profit. For question (c) Repeating the procedure of question (a) with 𝜎 ∈ {8,10,12} , we can obtain the following table: We can observe that with larger uncertainty of water availability, i.e., greater 𝜎 , the optimal planting strategy will end up with more conservative strategy and consequently the profit will be less. 𝜎 ? 1 ∗ ? 2 ∗ ? ∗ 8 15.61 0 31.23 10 14.52 0 29.03 12 13.42 0 26.84