According to the conditions and requirements of the problem, this paper establishes a model, which is a multivariable linear programming model. By solving this model, the problem can be completely solved.
The profit objective function is additive, and the supply of agricultural products is in short supply, that is, all agricultural products can be sold, so the total sales volume remains unchanged. Therefore, the minimum transportation cost algorithm can be directly applied to solve the maximum profit problem of the base.
The modeling process is divided into eight parts: 1. Restatement of the problem; 2. Model assumptions; 3. Symbol and text description; 4. Problem analysis; 5. Establishment of the model; 6. Solve the model; 7. Analysis of model results; 8. Evaluation and further discussion of the model.
In the modeling, the preservation of agricultural products and the cost other than transportation cost are not considered.
Making a transportation plan for agricultural products is a plan to arrange eight kinds of vegetables to be sold to six markets.
In the linear programming model, profit = total sales-total transportation fee. The total sales are unchanged, so only the total transportation cost is considered. Because the objective function of the model is multivariable and linear, the profit obtained is = total sales-total transportation fee = 1505885 yuan.
In this paper, some results of the model are analyzed, and the model has wide applicability.
Finally, the problem is discussed in depth.
Keywords: transportation scheme model, hypothesis, problem analysis model, result analysis model, evaluation and discussion
1. Restatement of the problem
Direct topic
2. Model assumptions
(1) When making the transportation plan of agricultural products, the data shall be accurate to 1 unit, that is, to the ton. This assumption theoretically guarantees more accurate results.
(2) Damage to vegetables during transportation is not considered.
(3) There is no need to consider other expenses except freight during the transportation of vegetables.
3. Symbol and text description
Y stands for the profit from selling vegetables at the base;
I= 1, 2…, 8, where 1 stands for Chinese cabbage, 2 for potatoes, 3 for tomatoes, 4 for beans, 5 for cucumbers, 6 for pumpkins, 7 for eggplants and 8 for zucchini;
J= 1, 2, …, 6 where 1 stands for market A, 2 stands for market B, 3 stands for market C, 4 stands for market D, 5 stands for market E and 6 stands for market F;
It represents the total amount of vegetables I (I = 1, 2,8) transported to the market of j (j = 1, 2,6).
4. Problem analysis
Making a transportation plan for agricultural products is a plan to arrange the vegetables in No.8 Middle School to be sold in six markets. The goal is to get the maximum profit. Judging from the data given in the title, it is obvious that the demand for vegetables is in short supply, so all the vegetables in the base can be sold at a profit, so the total income of the base is the income after all the vegetables are sold, which has nothing to do with the transportation plan. Therefore, in order to maximize profits, we must adjust the transportation plan of agricultural products. In addition, as can be seen from the table, the supply of agricultural products is in short supply, so the transportation scheme is limited by the base supply and market demand.
Use a linear programming model.
5. Establishment of the model
basic model
Decision variable: let the total amount of class I agricultural products transported to the j th market be.
Objective function: let the profit be y yuan. Get from the topic
MaxY = 400 x 1 1+400 x 12+400 x 13+400 x 14+400 x 15+400 x 16-80 x 1 1- 130 x 12- 150 X6
Constraints:
The total supply of raw materials for various agricultural products must not exceed the supply, and the supply exceeds the demand, so the supplied agricultural products can be sold, that is to say,
x 1 1+x 12+x 13+x 14+x 15+x 16 = 826
x 2 1+x22+x23+x24+x25+x26 = 594
x 3 1+x32+x33+x34+x35+x36 = 600
x 4 1+x42+x43+x44+x45+x46 = 356
x 5 1+x52+x53+x54+x55+x56 = 423
x 6 1+x62+x63+x64+x65+x66 = 890
x 7 1+x72+x73+x74+x75+x76 = 600
x 8 1+x82+x83+x84+x85+x86 = 500
Market demand The quantity of various agricultural products transported to each market shall not exceed the demand of the corresponding market for the corresponding agricultural products, that is,
x 1 1 & lt； = 160; x 12 & lt； = 130; x 13 & lt； =200; x 14 & lt； = 150; x 15 & lt； = 140; x 16 & lt； = 180;
x 2 1 & lt； =60; x22 & lt= 180; x23 & lt= 160; x24 & lt= 100; x25 & lt=20; x26 & lt= 130;
x 3 1 & lt； = 100; x32 & lt= 140; x33 & lt=200; x34 & lt=60; x35 & lt=80; x36 & lt=90;
x 4 1 & lt； =70; x42 & lt=90; x43 & lt= 140; x44 & lt= 100; x45 & lt=40; x46 & lt=80;
x 5 1 & lt； =50; x52 & lt= 100; x53 & lt= 130; x54 & lt=90; x55 & lt=90; x56 & lt=70;
x 6 1 & lt； =200; x62 & lt=2 10; x63 & lt= 130; x64 & lt= 100; x65 & lt=240; x66 & lt= 150;
x 7 1 & lt； = 120; x72 & lt= 150; x73 & lt=90; x74 & lt= 150; x75 & lt= 100; x76 & lt=90;
x 8 1 & lt； =60; x82 & lt=90; x83 & lt= 150; x84 & lt= 140; x85 & lt= 100; x86 & lt=80。
Non-negative constraints cannot be negative, that is, > =0.
6. Solve the model
6. 1 algorithm idea
This problem is realized by linear programming, and the algorithm is relatively simple and clear. By finding out that the total sales revenue of agricultural products MINUS the total transportation cost equals the profit, the optimal solution is obtained by linear programming.
6.2 the solution of the model
According to Schedule 4, the transportation scheme is as follows:
Total quantity transported to each market (unit; Tons)
Organize yourself and fill in the contents of the internship.