Finding Mathematical Modeling Answers, Mathematical Modeling Finding Answers

Guo Dun: The average interval between 100 ships arriving at the port (hours) = 1182 100=, the number of ships arriving at the port within 1 year (ships) = 365 24, the average loading and unloading time of 100 ships (hours) = 1120 100=, the ** unloading time (hours) = 741 of ships arriving at the port within 1 year, the maximum number of ships that can be unloaded in 1 year (ships) = 741, the total detention time of ships (hours) = 741 in 1 year, The total ship detention fee payable by the port within 1 year (yuan) = 5600 100 = 560 000, and the condition given is about 600,000 yuan.

The average interval between the arrival of 100 ships at the port is hours, and the average loading and unloading time of 100 ships is hours, and in terms of their average time, they are balanced or basically balanced. Why does the average detention time of the ship pay about 600,000 yuan per year? This is due to the imbalance between the arrival of ships at the port and the loading and unloading times of the vessels.

Therefore, it is necessary to analyze the degree of imbalance, and the most important thing is to analyze which ships arrive at the port at intervals and the number of ships, which leads to the detention of ships, and their detention time.

Obviously, the short interval between the arrival of ships at the port greatly exceeded the loading and unloading capacity of the port. There are 38 ships in the arrival interval from 5 to 10 hours, the total interval between ships is 302 hours, and the average interval time is 302 38=

hours), which represents the average vessel dwell time due to the short interval between 38 vessels arriving at the port;

2013-5-22 - Snapshot: Calculation of unloading time and demurrage loss during subchartering.

Demurrage is a form of compensation for the time that a vessel stays in port for more than the agreed operating time.

Demurrage is paid because the vessel has been detained for longer than the agreed operating time.

We assume that the loading and unloading operation time agreed in the contract is 24 hours, and the agreed time for port loading and unloading operation is not given in this question, and from the information provided by the snapshot, we assume that the agreed loading and unloading operation time is an average loading and unloading time of hours, which is severe for the port, and the detention period should be more than the time spent in the port after exceeding the loading and unloading time hours.

Therefore, loading and unloading in the case of 12 to 16 hours will also increase the detention time of the ship, in this case a total of 41 ships, ** unloading time is 511 hours, the average loading and unloading time is 511 41 = hours), which is due to the average detention time of 41 ships caused by the long loading and unloading time.

The total dwell time of the two = hours).

The average detention time of 100 ships = hours), the correct result of the total ship detention fee payable by the port within 1 year should be (yuan) = (130,000 yuan).

The condition given is that "the average detention time of the ship is an hour, and the detention fee of about 600,000 yuan is paid every year." "I don't know what the basis is.

Is it to increase the number of berths for loading and unloading? The above doubts need to be clarified before submission.

There are many problems in the design of some topics.

2] Jin Lie Gongxue, Zixiao Gong Leaned.

The equation relation of this problem is that time is equal and distance is equal. The details are as follows:

1.Divide the students other than the class leader into four groups of 11 students each.

2.At 7 a.m., the class president and the first group of 11 students got on the school bus, and the remaining three groups of students walked to their destination. After 1 hour of driving, the school bus dropped off the first batch of students and drove back and forth, and the first batch of students walked to their destination.

The school bus drove back for x2 hours to meet the three groups of students who were walking, and drove the second group of students to their destination, leaving the third and fourth groups of students to continue walking. The school bus traveled for x3 hours to drop off the second group of students and drive back and forth, and the second group of students walked to their destination. The school bus drove back for x4 hours, met the third and fourth groups of students, drove the third group of students to the destination, and the fourth group of students continued walking.

The school bus traveled for x5 hours to drop off the third group of students and drive back and forth, and the third group of students walked to their destination. The school bus traveled x6 hours to meet the fourth group of students, and carried the fourth group of students to their destination after x7 hours. At this time, four batches of students arrived at their destination at the same time.

The squad leader was in the car the whole time.

3.Start the column equation (if you're not interested, you can just look at section 5).

For the first batch of students, school bus time * school bus speed + walking time * walking speed = distance, get 70 * x1 + 5 * (x2 + x3 + x4 + x5 + x6 + x7) =;

The same goes for the first. The second, third and fourth batches are: 70*x3+5*(x1+x2+x4+x5+x6+x7)=;

70*x5+5*(x1+x2+x3+x4+x6+x7)=；

70*x7+5*(x1+x2+x3+x4+x5+x6)=；

4.On the way back, the school bus met the students three times. 70*x1-70*x2=5*(x1+x2)；

70*x3-70*x4=5*(x3+x4);

70*x5-70*x6=5*(x5+x6);

5.These 7 equations are independent of each other, 7 unknowns, and can be solved (it looks troublesome, but it is actually very simple).

6.In fact, there is a simpler way of thinking, that is, each group of students departs and arrives at the same time, except for the bus ride and walking (ignoring the boarding and boarding time), therefore, each group of students takes the bus and walks for the same time. That is to say, each batch of students takes the same amount of time on the school bus, that is, x1 = x3 = x5 = x7; At the same time, the time for the school bus to return to meet the next batch of students is also equal, i.e., x2=x4=x6=13 15*x1.

This greatly simplifies the calculation, i.e. 70x1+5*(3+13 15*3)x1=,x1=11 140, total time x=363 700

7.To sum up, the fastest is 31 minutes and 7 seconds (363 700 hours) for everyone to arrive at the same time.

I hope you are satisfied, please point out any mistakes!

Linear programming models.

Full-time attendant:

9 12 + 13 17: x1 person.

9~13 + 14~17: x2

Half-time waiter:

9~13: x3

10~14: x4

11~15: x5

12~16: x6

13~17: x7

Objective function: min

Constraints: 9 10 time period not less than 4:

x1 + x2 + x3 >=4;

10 11 period is not less than 3:

x1 + x2 + x3 + x4 >=3;

The same can be written all the time:

x1+x2+x3+x4+x5>=4;

x2+x3+x4+x5+x6>=6;

x1+x4+x5+x6+x7>=5;

x1+x2+x5+x6+x7>=6;

x1+x2+x6+x7>=8;

x1+x2+x7>=8;

There is also a limit on the total number of waiters in half time:

x3+x4+x5+x6+x7<=3.

Notice again that this is integer programming, and run the following statement in Mathematica:

linearprogramming[, automatic, integers]

The result is that the values from x1 to x7 correspond to each other.

Don't talk nonsense if you don't understand the first floor, read for yourself on page 66.

Question A: Digital camera positioning.

Digital camera positioning has a wide range of applications in traffic supervision (e-police) and other aspects. The so-called digital camera positioning refers to the use of digital cameras to take photos of objects and determine the position of certain landmark points on the surface of the object. The most commonly used positioning method is binocular positioning, which uses two cameras to locate the position.

For a feature point on an object, the image of the object is captured by two cameras fixed in different positions, and the coordinates of the point on the image plane of the two cameras are obtained respectively. As long as the exact relative position of the two cameras is known, the coordinates of the feature point in the coordinate system of the fixed camera can be obtained by geometric methods, that is, the position of the feature point is determined. Therefore, for binocular positioning, it is crucial to accurately determine the relative position of the two cameras, a process called system calibration.

One of the methods of calibration is: draw a number of points on a flat plate, take pictures with these two cameras at the same time, obtain the image points of these points on their image planes respectively, and use the geometric relationship of these two groups of image points to obtain the relative positions of these two cameras. However, we cannot directly obtain a "point" without geometric dimensions on either the object plane or the image plane.

The actual practice is to draw a number of circles (called targets) on the plane of objects, and their centers are geometric points. Their images are generally distorted, as shown in Figure 1, so the image of the center of the circle must be accurately found from the images of these circles on the target, and calibration can be achieved.

Figure 1 Image of a circle on a target.

Someone designed the target as follows, take a square with a side length of 100mm, take four vertices (corresponding to a, c, d, e) as the center of the circle, and 12mm as the radius as the circle. Take B at a distance of 30 mm from point A on the edge of the AC as the center of the circle, and 12 mm as the radius to make a circle, as shown in Figure 2.

Figure 2 Schematic diagram of the target.

Figure 3 Image of the target.

1) Establish mathematical models and algorithms to determine the image coordinates of the center of the circle on the target in the image plane of the camera, where the origin of the coordinate system is taken as the focal point of the camera, and the x-y plane is parallel to the image plane;

2) for the target and its image given by Fig. 2 and Fig. 3 respectively, calculate the image coordinates of the center of the circle on the target on the image plane, the image distance of the camera (i.e., the distance from the focal point to the image plane) is 1577 pixel units (1 mm is about one pixel unit), and the camera resolution is 1024 786;

3) Design a method to test your model and discuss the accuracy and stability of the method;

4) Establish a mathematical model and method for giving the relative positions of two fixed cameras with this target.

A Publisher: 設 yā indicates the remuneration received by the author in A Publisher;

Assuming that the price of each book in Publishing House A is a yuan, it is expected that the new work can sell X (A) books in Publishing House A; So.

When x(a)<=3000, ya = ;

When x(a)>3000, ya = 3000* + x(a)-3000) (

B Publisher: Note yb indicates the remuneration received by the author in A Publisher;

Assuming that the price of each book in Publishing House B is B yuan, it is expected that the new work can sell X (B) books in Publishing House B; So.

When x(b)<=4000, yb = 0;

When x(b)>4000, yb = (x(b)-4000) (

After sorting it out, comparing the two publishing houses, there are.

A Publisher: When x(a)<=3000, ya = ;

When x(a)>3000, ya = (60a+6000).

B Press: When x(b)<=4000, yb = 0;

When x(b) > 4000, yb = (400b+12000).

It can be seen that when the number of publications is less than 4000, a publishing house must be chosen; When the number of publications exceeds 4000, the first thing to do is to compare the number of x(a) and x(b) that can be sold by the two publishers; Then there is the price of each book of the two publishers, A and B. Of course, the variables x(a) and x(b) and a and b depend on the competitiveness of the two publishers; When a more accurate estimate of these variables is made, compare the sizes of ya and yb to further determine which publisher to choose.

Let the temperature of the object be t, and the rate of change of temperature is dt dt, where t is the time, and the temperature of the water is t1, then the temperature difference between the egg and the water is t-t1

From the meaning of the question: t-t1=kdt dt (where k is the proportionality constant) (1) equation (1) is reduced to : dt=kdt (t-t1) (2) to (2) both sides of the same integration and sorting out to get:

t=k*ln（t-t1）+c

Then the known data can be substituted to determine the coefficients k and c, here there is a hidden condition of the problem is that the temperature of the water has not changed t1 is always 18, and finally after determining k and c, you can find the time taken by the egg to 20, and then subtract 5 minutes to get how long it will take.

If you still don't understand, just ask privately.

There is a table missing, but it can be answered according to the requirements in the question.

So get <= (ad time).

16 (publicity time).

The optimal number of solutions is x=4 and y=1

But there is no schedule, so this answer is incorrect. Thank you.

The title of your contest.