A paper on mathematical modeling

Mathematical modeling paper

Mathematical modeling in topic life

college

Specialized courses

Student name

achievement

date month year

How to arrange the transportation of steel, coal and water from several supply points to some demand points?

Can this plan maximize profits? Due to the limitation of volume and weight, how to pack all kinds of goods together to get the highest profit? A number of tasks are assigned to some candidates, because everyone's expertise is different and the benefits of completing the tasks are different. How to distribute to maximize the total income? This paper will discuss the methods to solve these problems through mathematical modeling through the following examples.

Keywords: most profitable, 0- 1 variable

First, the problem of tap water transmission

Problem A city has four residential areas, A, B, C and D. The tap water is supplied by three reservoirs, A, B and C. The daily basic domestic water consumption of the four areas must be 8050 102000 tons respectively. However, due to the shortage of water, the three reservoirs can only supply 60704000 tons of tap water respectively. Due to regional differences, the water transfer management fees paid by water supply companies to deliver water from reservoirs to various districts are different (see the table below), and other management fees are all 400 yuan per thousand tons. According to the company's regulations, users in all districts charge per thousand tons of 950 yuan according to the unified standard. In addition, all four districts have applied to the company for additional water consumption, namely10,20,30,50 kilotons. How should the company allocate water supply to make more profits?

Water transfer management fee (per thousand tons RMB) a, b, c and d.

a 160 130 220 170

b 140 130 190 150

c 190 200 230-

problem analysis

Water supply distribution is a plan to distribute the water supply from three reservoirs to four regions, with the goal of maximizing profits. According to the data given in the title, the water supply of A, B and C districts is170kt, which is less than the sum of the basic domestic water consumption of 270kt and the extra water consumption of the four districts, so it can always be sold for profit. Therefore, the daily total income of the water supply company is 950 * (60+70+40). Similarly, the company's other daily management fee of 400 * (60+70+40) = 68,000 yuan has nothing to do with the water supply scheme. Therefore, if the profit is the largest, only the water transfer management fee is the smallest. In addition, the water supply scheme is naturally limited by the three-time water supply capacity and the four-time water supply demand.

Model structure

The decision variables are A, B, C and the water supply of three reservoirs (i= 1, 2,3) to four communities (j= 1, 2,3,4) respectively. Let the daily water supply of the reservoir from I to J be Xiji. Because there is no water pipeline between the destination of fish and the destination of reservoir C, that is, X34=0, the decision variable is only 1 1.

From the above analysis, we can see that the goal of the problem can be changed from profit maximization to water transfer management fee minimization, so there are

min = 160 * x 1 1+ 13+220 * x 170 * x 14+ 140 * x 2 1+ 130 * x22+ 190 * x23+65438

There are two kinds of constraints: one is the supply constraint of the reservoir, and the other is the demand constraint of each district. Because water supply can always be sold and profitable, the water supply limit of a reservoir can be expressed as

x 1 1+x 12+x 13+x 14 = 60；

x 2 1+x22+x23+x24 = 70；

x 3 1+x32+x33 = 40；

Considering the basic water consumption of songs, the demand limit can be expressed as

80 & lt= x 2 1+x 1 1+x 3 1；

50 & lt= x 12+x22+x32；

10 & lt； = x 13+x23+x33；

20 & lt= x 14+x24；

x 2 1+x 1 1+x 3 1 & lt； =90;

x 12+x22+x32 & lt； =70;

x 13+x23+x33 & lt； =40;

x 14+x24 & lt； =70;

Model solving

Input the above formula into LINGO to solve, and get the following output:

Find the optimal solution at step 10.

Target value: 25800.00

Variable value reduces cost

x 1 1 0.000000 20.00000

X 12 60.00000 0.000000

x 13 0.000000 40.00000

X 14 0.0000000 20.00000

X2 1 50.00000 0.000000

X22 0.0000000 0.0000000

X23 0.0000000 10.00000

X24 20.00000 0.000000

X3 1 30.00000 0.000000

X32 0.0000000 20.00000

X33 10.00000 0.000000

The water supply scheme is: Reservoir A supplies 60,000 tons of water to Area B, reservoirs A and D supply 5,020,000 tons respectively, and reservoir C supplies 30 10000 tons to A and C respectively. The water transfer management fee is 25,800 yuan, and the profit161500-68,000-25,800 yuan = 67,700 yuan.

Two. Cargo loading

Problem A Jet engine oil has three cargo holds: the front cargo hold, the middle cargo hold and the rear cargo hold. The maximum cargo volume that the three cargo holds can carry is limited, as shown in the following table, and in order to keep the balance of the aircraft, the weight of the cargo held in the three cargo holds in the Middle Ages must be in direct proportion to its maximum allowable weight.

Front, middle and rear cabins.

Weight limit (ton) 15 26 12

Volume limit (cubic meter) 8000 9000 6000

There are four kinds of goods shipped on this flight. See the table below for relevant information. The last column is the profit after shipment. How to arrange shipment to maximize the profit of cargo plane's flight?

Weight (ton) Space profit (yuan per thousand tons)

Commodities 1 20 480 3500

Commodity 2 18 650 4000

Goods 3 35 600 3500

Commodity 4 15 390 3000

There are no other requirements for the shipment of goods in the model assumption, and we can make the following assumptions:

(1) Every commodity can be divided into any small size;

(2) Each cargo can be randomly distributed in one or more cargo holds;

(3) A variety of goods can be mixed and matched to ensure seamless.

Model structure

Decision variables: Xij represents the weight (ton) of type I cargo loaded in the j-th cargo hold, and cargo hold j= 1, 2, 3 represent the front hold, middle hold and rear hold respectively.

The goal of decision-making is to maximize profits, that is,

max = 3500 *(x 1 1+x 12+x 13)+4000 *(x 2 1+x22+x23)+3500 *(x 3 1+x32+x33)+3000 *(x 4 1+x42+x43)；

Constraints include the following four aspects:

(1) The total weight constraint of loading four kinds of goods, namely

x 1 1+x 12+x 13 & lt； =20;

x 2 1+x22+x23 & lt； = 18;

x 3 1+x32+x33 & lt； =35;

x 4 1+x42+x43 & lt； = 15;

(2) Weight limitation of three cargo spaces, namely

x 1 1+x 2 1+x 3 1+x 4 1 & lt； = 15;

x 12+x22+x32+x42 & lt； =26;

x 13+x23+x33+x43 & lt； = 12;

(3) Space limitation of three cargo spaces, namely

480 * x 1 1+650 * x 2 1+600 * x 3 1+390 * x 4 1 & lt； =8000;

480 * x 12+650 * x22+600 * x32+390 * x42 & lt； =9000;

480 * x 13+650 * x23+600 * x33+390 * x43 & lt； =6000;

(4) Balance constraint of loading weight of three cargo holds, namely

(x 1 1+x 2 1+x 3 1)/ 15 =(x 12+x22+x32+x42)/26；

(x 12+x22+x32+x42)/26 =(x 13+x23+x33+x43)/ 12；

Model solving

Input the above model into LINGO to solve, and you can get:

Find the optimal solution at step 10.

Objective value: 155340. 1

Variable value reduces cost

x 1 1 0.5055 147 0.0000000

X 12 6.562500 0.000000

X 13 2.286953 0.000000

x 2 1 1.93439 0.0000000

X22 0.0000000 2526.843

x23 6.0656 1 1 0.0000000

x 3 1 0.000000 0.4547474 e- 12

X32 0.0000000 1783.654

X33 1.599359 0.000000

x 4 1 0.000000 1337.740

X42 15.00000 0.000000

X43 0.0000000 1337.740

In fact, we might as well round off the optimal solution, and the result is that the cargo 1 loaded in the front cabin 1 ton, 7 tons in the middle cabin and 2 tons in the rear cabin; Cargo 2 is loaded in the front cabin 12 tons, and the rear cabin is 6 tons; The goods are packed in 3 tons and put into the back cabin; No.4 cargo is loaded in the middle cabin, 15 tons. Maximum profit 155340 yuan.

Three. Selection of medley relay team

Question A Class is going to select four swimmers from five to form a relay team to participate in the school's 4 * 100 meter medley relay competition. The average 100-meter scores of four Chinese characters of five athletes are shown in the following table. How should the selected athletes form a relay team?

Methyl ethyl propyl butyl pentyl

Butterfly1` 0657` 21`181`1kloc-0/` 07.

Backstroke1`151` 061` 071`141`1.

Breaststroke1` 271` 061` 241` 091` 23.

Freestyle 58'' 6 53'' 59'' 4 57'' 2 1' 02

Problem analysis: Four of the five players were selected to form the relay team. None of them had swimming strokes, and the four people used different words, so the relay team performed best. An easy-to-think method is exhaustive method, and there are five schemes to form a relay pair! = 120, and the optimal scheme can be found by calculating and comparing one by one. Obviously, this is not a good way to solve this kind of problem. As the scale of the problem becomes larger, the calculation amount of the exhaustive method will be unacceptable.

The 0- 1 variable can be used to indicate that relay team members are not selected, thus establishing a 0- 1 programming model of this problem and solving it with the help of the county's mathematical software.

Establishment and solution of the model

Let the player i= 1, 2, 3, 4, 5; That is, butterfly, backstroke, breaststroke and freestyle are swimming strokes j= 1, 2, 3 and 4 respectively. The word 100 meter used by player I's J has the best score of Cij(s), which is both.

cij I = 1 I = 2 I = 3 I = 4 I = 5

j = 1 66 57.2 78 70 67

j = 2 75 66 67 74 7 1

J = 3 87 66 84 69 83

J = 4 58 53 59 57.2 62

The variable Xij of 0- 1 is introduced. If player I is selected to participate in swimming stroke J, Xij-= 1, otherwise Xij=0. According to the requirements of forming a relay team, Xij should meet two constraints:

First, no one can be selected as one of the four Chinese characters at most. Remember that i= 1, 2,3,4,5 should have ∑ xij "=1;

Second, each stroke must have one person, and only 1 person can be selected. Remember that for a, 2, 3, 4, there should be ∑ xij =1;

When player I is selected for swimming stroke J, CijXij indicates his achievement, otherwise CijXij=0. So the relay team's performance can be expressed as ∑∑CijXij, which is the objective function of the problem.

Bring the data given by the topic into this model and enter LINGO:

min = 66 * x 1 1+75 * x 12+87 * x 13+58.6 * x 14+57.2 * x 2 1+66 * x22+66 * x23+53 * x24+78 * x 3 1+67 * x32+84 * x33+59.4 * x34+70 * x4

Suffer

x 1 1+x 12+x 13+x 14 & lt； = 1;

x 2 1+x22+x23+x24 & lt； = 1;

x 3 1+x32+x33+x34 & lt； = 1;

x 4 1+x42+x43+x44 & lt； = 1;

x 1 1+x 2 1+x 3 1+x 4 1+x 5 1 = 1；

x 12+x22+x32+x42+x52 = 1；

x 13+x23+x33+x43+X53 = 1；

x 14+x24+x34+x44+X54 = 1；

@ bin(x 1 1)； @ bin(x 12)； @ bin(x 13)； @ bin(x 14)； @ bin(x 2 1)； @ bin(X22)； @ bin(X23)； @ bin(X24)； @ bin(x 3 1)； @ bin(X32)； @ bin(X33)； @ bin(X34)； @ bin(x 4 1)； @ bin(X42)； @ bin(X43)； @ bin(X44)； @ bin(x 5 1)； @ bin(X52)； @ bin(X53)； @ bin(X54)；

The following results were obtained

Find the optimal solution at step 12.

Target value: 25 1.8000

Branch count: 0

Variable value reduces cost

x 1 1 0.000000 66.00000

X 12 0.0000000 75.00000

X 13 0.0000000 87.00000

x 14 1.000000 58.60000

x 2 1 1.000000 57.20000

X22 0.0000000 66.00000

X23 0.0000000 66.00000