[2023 National Mathematical Modeling Competition] Ideas for solving problem D

Problem-solving ideas for question D in the 2023 National Mathematical Modeling Competition

In order to solve problem 1, problem 2 and problem 3, we can use the dynamic programming method to formulate the production plan, considering the uncertainty factors and the scenario set of many possible situations. First, we need to define some variables and symbols:

• $T$ : Total breeding cycle (number of years).
• $S$ : The number of seasons in each breeding cycle (assuming there are 4 seasons in a year).
• $B$ : Base number of ewes per season.
• $R$ : Number of breeding rams per season.
• $P$ : Number of gestating ewes per season.
• $D$ : Number of ewes in labor per season.
• $L$ : Number of lactating ewes per season.
• $C$ : Number of lambs in the fattening period each season.
• $A$ : The number of ewes in the resting period in each season.

Next, we need to establish a decision variable, that is, the number of breedings in each season, represented by $X$ means.

In order to solve problems 1 and 2, we can formulate the following mathematical model:

Question 1 model:

Objective function:
Maximize the annual number of sheep for slaughter, that is, maximize ${\sum }_{t=1}^T{\sum }_{s=1}^S(2D{\_}_{t,s}+C_{t,s})$。

Restrictions:

1. The basic number of ewes in each season cannot exceed 14: $B_{t,s}\le 14$。
2. The number of breeding rams in each season cannot exceed 4: $R_{t,s}\le 4$。
3. The number of ewes during pregnancy is 85% of the number of ewes during parturition: $P_{t,s}=0.85D_{t,s}$。
4. The number of ewes in the resting period is 15% of the ewes in the farrowing period: $A_{t,s}=0.15D_{t,s}$。
5. The number of lactating ewes is equal to the number of ewes in parturition plus the number of lactating ewes in the previous season: $L_{t,s}=D_{t,s}+L_{t-1,s}$。
6. The number of lambs is equal to the number of lactating ewes in the previous season: $C_{t,s}=L_{t-1,s}$。

Question 2 model:

Objective function:
Maximize the annual number of sheep for slaughter, that is, maximize ${\sum }_{t=1}^T{\sum }_{s=1}^S(2D{\_}_{t,s}+C_{t,s})$。

Constraints:
Same as Problem 1, except we no longer restrict $B_{t,s}$and $R_{t,s}$, but treat them as decision variables that can be freely adjusted each season.

Question 3 model:

Question 3 considers the uncertainty factor and the scenario set of multiple possible situations. To solve problem 3, we can use a dynamic programming approach to make decisions season by season starting from the first season to minimize the expected loss of the overall scenario. Specific steps are as follows:

1. Initialization: Starting from the first season, set $B_{1,1}$and $R_{1,1}$As a decision variable, calculate D 1 , 1 D_{1,1} for all possible cases$D_{1,1}$、$P_{1,1}$、$L_{1,1}$and $C_{1,1}$。

2. Season by season iteration: for each season $s$ , calculated based on the results of the previous season and uncertainty factors$B_{t,s}$and $R_{t,s}$, and then calculate $D_{t,s}$、$P_{t,s}$、$L_{t,s}$and $C_{t,s}$. In each season, choose the decision that minimizes the expected loss.

3. Calculate the expected loss: In each season, calculate the expected loss according to the use of sheep pens in different situations.

4. Termination conditions: Repeat steps 2 and 3 until the breeding cycle ends.

Ultimately, the resulting production plan will take into account the uncertainty factor and make the best decision based on the actual situation in each season to minimize the expected loss. This model can be solved by a computer program to obtain an optimal production plan.