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Maximizing profit through mathematical modeling
Train meal price problem

1 Abstract: Service providers on trains are in a monopoly position and have unique advantages, which makes food easy to sell at higher market prices. The higher the price, the greater the profit, but there is an upper limit for passengers. If the price continues to increase, more and more passengers will give up buying, and the total profit of service providers will change accordingly. In this model, we choose the supply and demand sides as a system, and study how to multiply the sales price by the demand from the supplier's point of view to reduce the cost, that is, maximize the sales benefit.

2 restatement of the problem:

Long-distance trains need to provide some on-board services because of their long time. Providing three meals a day is the main service. Due to the high cost of all aspects on the train, the price of food on the train is also slightly higher. Take K452 train from Chengdu to Urumqi as an example. Breakfast every day is a bowl of porridge, an egg and some pickles, and the price is 10 yuan. Lunch at noon and in the evening, the price is 15 yuan. Due to the high price, passengers usually bring instant noodles, bread and other foods. Instant noodles and bread are also sold on the train, but the price is also expensive. For example, instant noodles sold in 3 yuan are generally sold in 5 yuan. Of course, due to the limited passenger capacity of the train and the limited number of meals provided, it is normal to raise the price appropriately. But there must be a limit to the high price, not too high. If there are 1000 passengers on the bus, 500 of them have the need to buy food on the bus, but the box lunch on the bus can only supply 200 people per meal; In addition, the car can also provide instant noodles for 100 people every meal. Please design a price plan according to the actual situation, so that the train can get the maximum benefit in catering sales.

3 problem analysis:

The price and profit rise at the same time, resulting in a decrease in the number of people asking to buy;

Lower prices and lower profits have also led to an increase in the number of people asking to buy;

Everything sold on the train has an optimal price and gets the maximum benefit at the same time;

This is an optimization problem, and the key is to find the optimal price.

4 problem hypothesis:

1. Every time the price of box lunch increases by 1 yuan, 20 people will choose to give up buying, that is, b1= 20;

2. Every time the price of instant noodles increases by 1 yuan, 36 people will choose to give up buying, that is, b2 = 36.

3. Because 500 people need to buy food in the car, it is assumed that 500 breakfasts can be provided;

4. Every time the price of breakfast increases by 1 yuan, 30 people will choose to give up buying, that is b3 = 30.

5. Here, the prices of various catering markets are taken as the cost price, that is, q 1 = 10 yuan (box lunch), q2 = 3 yuan (instant noodles) and q3 =5 yuan (breakfast);

6. the sales volume x depends on the price p, and x (p) is a decreasing function.

7. Further suppose: x (p) = a–BP, a, b > 0;;

5 symbol description:

Q: Take the prices of various catering markets as the cost price here, that is, the cost price of food;

P: the selling price of food;

A: Absolute demand (the demand when P is very small), that is, the number of buyers at the lowest price;

B: When the price rises by 1 yuan, the buyers decrease (the sensitivity of demand to price);

I: income; U: profit; C: expenditure;

X: the number of people who need to buy a certain food;

The corresponding subscripts 1, 2 and 3 represent breakfast, box lunch and instant noodles respectively; For example: x 1, x2, X3, x3 respectively represent the number of people who buy box lunch, instant noodles and breakfast;

6 model establishment and solution:

Adopt the algorithm of first unification and then separation;

Income I (p) = px; Expenditure c (p) = qx; Profit u (p) = I (p)-C (p); Maximize p u (p);

Meet the optimal price p* that maximizes the profit U(p).

u(p)= I(p)–C(p)

=(p–q)(a–BP)

= -bpp + ( a + bq)p - aq

because

Q/2 ~ half of the cost;

B ~ When the price rises 1 unit, the sales volume decreases (the sensitivity of demand to price) b p*

A ~ absolute demand (demand when P is very small) a p*

7 For bento:

According to the hypothesis, Q 1 = 10, b1= 20; Because 500 people have the need to buy food in the car, the box lunch in the car can only supply 200 people per meal; So: a1= 500; Number of buyers: x1= 500–20p1;

From p* = q/2+a/2 * b, we can get: p * = q1/2+a1=12+500/2 * 20 =/kloc.

From 500–20 *17.5 =150

P 1 > is obtained from 500–20p/kloc-0: = 200.

Therefore, when the maximum profit is obtained, p1=15;

Instant noodles can also be obtained in the same way:

Q2 = 3, b2 = 36 Because 500 people need to buy food in the car, but the box lunch in the car can only serve 200 people per meal, and there are still 300 people who need instant noodles at this time; So a2 = 300, and the number of buyers X = 300-36p2;

From p* = q/2+a/2 * b, we can get: P * = Q2/2+A2/2 * B2 = 3/2+300/2 * 36 = 6.3;

From 300–36 * 6.3 = 73

From 300–36 * p2 > =100; Get p2

P2 = 5.5; when U(p) takes the maximum value;

Breakfast is served in the same way:

Q3 = 5, b3 = 30 Because 500 people have the requirement to buy food in the car, it is: a1= 500; Number of purchasers: x1= 500–30p3;

Because the supply is not constrained, the formula can be used directly:

From p* = q/2+a/2 * b, we can get: p * = Q3/2+A3/2 * B3 = 5/2+500/2 * 30 =11.5;

So when the maximum profit is obtained, p1=11.5;

Analysis and test of 10 results;

1) The prices of box lunch, instant noodles and breakfast are 15 yuan, 5.5 yuan, 1 1.5 yuan respectively;

Through calculation, all the prices are not more than 3 times of the market price, and the provided (except breakfast) can be sold out to get the maximum benefit;

These calculation results should be acceptable to consumers;

The price calculated by this model is close to the actual train price, which shows that it is reasonable.

The modeling and solving process of the model has been tested, and the in-depth test needs to be completed by practice.

1 1 model evaluation;

Advantages: A quadratic function is used to solve the price optimization problem, so as to obtain the maximum benefit; Simplify a complicated and changeable problem;

Disadvantages: the sensitivity of demand to price is not accurate only by its own experience, and the sensitivity problem only uses a function once, which is too simplistic and may not show its real model well;

Improvement direction: make the sensitivity problem accurate and closer to reality; Functionally, you can try to use a higher level variety;

Popularize new ideas: the establishment of mathematical model must be considered from many aspects, and its ultimate goal is to simplify complex problems and mathematize practical problems; Mathematical problems are life-oriented; But all this has a premise: it must be based on strict mathematical rules and mathematical theories.