Optimization of engineering quantity.
In view of the first question, in order to get the earthwork volume of the canal excavation, this paper divides it into three sections for comparison.
Hermite interpolation and cubic spline interpolation, and finally piecewise cubic Hermite interpolation fitting is carried out on the known data points to obtain the curve equation y f x about the channel. , the integral of the canal curve equation is
The channel length is 14550 7650 2 1 dx yL, and the channel length is m 5.7522 through MATLAB.
Therefore, the final solution is that the total earthwork volume of canal excavation is 3 m 135405LSV? .
Aiming at the second question, on the basis of the first question, this paper establishes the upper bound function model of integral: let 7650 a? ,
1x satisfies 1 2 1.
six
x
a
V
S y dt？ 1i x after I x is found? Satisfy 1 2 1
six
I
I
x
x
V
This will
Divide the total earthwork into six parts and get 7 .8736x 1? ，2 .9862x 2？ ， 10956 x3？ ， 12 1 16 x4？ ,
13353x 5？ Then determine the coordinates y and x of the sextant.
In view of the third question, there are three variables on the highway along the canal, namely k ji x, X, X, in order to make
The transportation workload is the smallest, so this paper establishes an unconstrained programming model and solves it with MATLAB to get the minimum transportation volume.
The quantity is 4 8 1027.7 m? . The location coordinates of the canal during the construction of the two highways are 7.519296B and 4.3 1 1683c respectively.
Keywords: Hermite interpolation MATLAB integral upper bound function unconstrained programming
I. Restatement of the problem
Dig a canal in a certain area, and you will know several points that the canal passes through.
The first problem is to solve the total stone quantity of channel construction;
Question 2: If the channel is divided into six sections, and each section has the same earthwork volume, where should the section point be taken?
Settings;
Question 3: Build a road parallel to the canal. Earthwork for river excavation will be transported to A (9500,4000).
In order to facilitate transportation, it is planned to choose two points on the road along the canal to build a temporary road to A, so as to
The total workload of earthwork transportation is the smallest.
Second, the analysis of the problem
Aiming at the first question, this question needs to dig the total earthwork of the canal, and the main purpose is to know the cross-sectional area of the canal.
The purpose is to find out the length of the canal. Knowing the positions of several points through which the channel passes, in order to get the length of the channel, this paper
It is considered that the channel curve can be obtained by interpolation fitting and the channel length can be obtained by curve integration. Interpolation and fitting
There are many methods, spline interpolation will be smoother, but it may not keep the original shape, considering that it should be better preserved.
The shape of the channel, so this paper chooses Hermite method for interpolation fitting.
For the second question, the canal should be divided into six equal parts, and the earthwork volume of each section is the same. This problem is the inverse problem of a function.
Therefore, in order to solve this problem, this paper can consider using the upper integral function when the channel curve function is known.
Solve, so as to determine the x point, and then get the y point.
In order to solve the third problem, it is necessary to build roads to transport earthwork and minimize the workload of transportation. This problem
In order to plan the problem, in the second problem, this paper knows that X is related to the earthwork volume V, and because of the transportation work,
Quantity is equal to the product of earthwork quantity and distance. Therefore, this paper adopts unconstrained programming model to minimize the workload.
Just value.
Third, the model hypothesis.
1. The two temporary roads built are straight lines.
2. The road function curve along the canal is almost the same as the curve function of the canal.
Four. Symbolic description of xf channel curve equation
Five, earthwork volume
S channel cross-sectional area l channel length
The abscissa of the point on the ninth river.
The ordinate of this point on the Erie Canal.
IW earthwork transportation workload
1L temporary highway 2L temporary highway
The establishment and solution of verb (verb's abbreviation) model
5. 1 question 1
5. 1. 1 interpolation and fitting
Draw a scatter diagram of the known points passing through the canal (figure 1).
0.8 0.9 1. 1. 1.2 1.3 1.4 x 10 4
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
X/m
Y/m
Canal scatter diagram
Figure 1. Canal scatter diagram
Method 1. Known data points are interpolated by Hermite method. 3 Set the number of known functions xfy? At 1 n? The function values on different nodes are ii xfy n 10 x, l, x, x? n，L， 1，0i？ And derivative value i' i' x fy? , the polynomial xH is required to be at most 2 n+1 time, so i i yxH? I'm xH? N, 1, 0i Hermite interpolation polynomial is:? ? 2 ' i i i i i i H x h x x a y y y？
Among them,
2
n
ij 0j j i
j i x x xx h，？ n ij 0j j i i x x 1 a .
Use MATLAB to get the interpolation curve (Figure 2).
0.8 0.9 1. 1. 1.2 1.3 1.4 x 10 4
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5500
6000
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7000
Hermite interpolation curve and original data points
X/m
Y/m
Original data points of Hermite interpolation curve
Figure 2. Hermite interpolation curve and original data points
Method two. Spline difference interpolation of known data points. three
Define spline function:
Mathematically, piecewise polynomials with certain smoothness are called spline functions. Specifically, given the division of interval a and b.
0 1 1nn :a x x x x b
If the function () sx satisfies: 1. Between each cell, 1, (0, 1, 1) symplectic? The upper () sx is a polynomial of degree k;
2. () Does SX have 1 k on A and B? Order continuous derivative.
So () sx is about division? K- spline function, whose graph is called k-spline curve. 0 1,,, n x x x represents
Is a spline node, 1 2 1,,, n x x x? It is called the inner node, and 0, n xx is called the boundary point, so that the whole spline function
Write (,) p Sk? , called k-spline function space.
Obviously, the polyline is a linear spline curve.
If () (,) p sx S k, then () sx is about partition? K-degree polynomial spline function of. Polynomial of degree k
The general form of spline function is
1
0 1 ( ) ( ) ! ! I know what you are talking about.
Where (0, 1,,) i ik and (1, 2, 1) jjn? Is an arbitrary constant and
(), (), (1, 2, 1) 0, kjjjjkjxxxxxxxxxx This article uses 3 k? Case: It is a cubic spline function. Cubic spline function: for partition 0 1 1nn :a x x x x b on a and b, then
1 2 3 3 32
3 0 1
1 ( ) ( ) ( ,3) 2! 3! 3! n j jp j aa s x x x x x x S？
In ...
3 3( )，()，( 1，2，， 1) 0，jj j j x x x x x x j n xx
Cubic spline function difference:
Because 3 () (,3) ps x S contains 3 n? A coefficient to be determined, so it should need 3 n? Interpolation condition, already
Given the interpolation node i x and the corresponding function value () (0, 1, 2,) IIF X Y I N, where is 1 n? One condition,
Two more boundary conditions are needed.
Commonly used cubic spline functions have three boundary conditions:
(1) 3 0 3 (), () n Sayy Sby. The spline interpolation function established by this intermediate boundary condition is called () fx.
Complete cubic spline interpolation function.
Special, 0'0 n yy? , the spline curve is in a horizontal state at the endpoint.
If () fx？ I don't know, can I find 3() sx? And () fx? Approximately equal at the endpoints. At this moment
0 1 23,, x x x x as the node to make a cubic Newton interpolation polynomial () a Nx, with 1 23,,, n N N N N N X X X? Zuo Yi
A cubic Newton interpolation polynomial () b Nx, which requires
( ) ( )，())ab s a N a s b N b
The cubic spline established by this boundary condition is called Lagrange cubic spline interpolation function of () fx.
(2) 3 0 3 3 (), () Sayy Sby. The special 0nyy is called the natural boundary condition.
(3)3333(0)(0)(0)(0)SASB, (33 (0) (0) s a s b needed here? )
This situation is called periodic condition.
Interpolation curve is obtained by cubic spline interpolation with MATLAB (Figure 3).
0.8 0.9 1. 1. 1.2 1.3 1.4 x 10 4
2000
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X/m
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Cubic spline interpolation curve and original data points
Cubic spline interpolation curve original data points
Figure 3. Cubic spline interpolation curve and original data points
Comparison between Hermite interpolation and cubic spline interpolation
five
：
The construction method of function s(x) provided by SPLINE is exactly the same as that of function p(x) in PCHIP, except that
O x
y
0: 00 a. m.
1M
2M
1nM？
n BM？
Figure 4
However, the selection method of the slope at X(j) is different.
The second derivative D2s(x) of the spline function S (x) in X(j) is also continuous, and the following results are obtained.
Results:
(1) spline is smoother, that is, D 2s (x) is continuous.
(2) If the data is the value of a smoothing function, the spline is more accurate.
(3) If the data is not smooth, PCHIP has no overshoot and little fluctuation.
(4) It is less difficult to establish PC HIP.
(5) The estimation difficulty of these two functions is the same.
Cubic spline is smoother than Hermite interpolation, the second derivative of spline is continuous, and Hermite interpolation is first order.
Derivative continuity. Discontinuous second derivative means discontinuous curvature. People's eyes can perceive graphics.
Discontinuity of curvature. On the other hand, Hermite interpolation is conformal, while spline interpolation is not necessarily conformal.
By comparing Hermite interpolation with cubic spline interpolation, there is no obvious difference on this issue. Get better.
In order to ensure the shape of the graph and reduce the error, Hermite interpolation is adopted in this paper.
5. 1.2 Solving the Channel Length
When the number of sides increases infinitely, the circumference of a circle can be determined by the limit of the circumference of the inscribed regular polygon of the circle.
A similar method can be used to establish the arc length of plane continuous curve and calculate the arc length by definite integral.
Let AB be the two endpoints of the curve arc. Take a point on the arc AB:
0 1 2 1 1,,,,,, I n a m m m m m m m b, and use this to connect adjacent polylines (figure
4)。
When the number of points increases infinitely and each segment is 1ii MM? When they all shrink to a point, if the length of this broken line
1
1
n
two
I
Hmm? Limit exists, then this limit is called the arc length of curve arc AB, and this curve arc AB is called
It can be long.
Since the arc length of a smooth curve can be calculated, the arc length can be calculated by definite integral.
Let the curve arc pass through the parametric equation:
()，()xt t yt？ ? ?
Here, where (), () tt is,
In the continuous derivative, and () () tt are not zero at the same time, now.
Calculate the length of the curve arc. Take the parameter t as an integer variable, and its change interval is,
. Corresponding to,
Any small interval
，t t dt？ The length s of the small arc segment? Is approximately equal to the length () () xy of the corresponding chord 22? , because
()()()x t dt t dx t dt？ ?
()()()y t dt t dy t dt？ ?
So, s? Approximation of (arc length difference), that is, the arc length element is
2 2 2 2 2 2 2 2 2()()()()()()()()ds dx dy t dt t dt t dt
So the arc length is
22 ( ) ( ) s t t dt？
When the curve arc is composed of rectangular coordinate equations
()()y f x a x b？
Given that () fx has a first-order continuous derivative with respect to ab, the curve arc is defined by a parametric equation.
()
()xx
a x b
y f x？
So the arc length is
2 1b a s y dx？ ?
By integrating the curve function of the canal after interpolation, the length of the canal is obtained. 14550 7650 2 1 dx yL solved by MATLAB to get m 5.7522? L
5. 1.3 Solving Earthwork Quantity
The length and cross-sectional area of the canal are known.
Then:
2m 18228 10？
3m 135405LSV？
5.2 Question 2
Let the function () fx be continuous on the interval ab, and let x be a point on ab. Observe that the definite integral of () fx on ax is between divisions.
()
x
a f x dx？
First of all, because () fx is still continuous on ax, definite integral exists. Here, x means
The upper limit of definite integral is the integral variable. Because the definite integral has nothing to do with the sign of the integral variable, it is
For the sake of clarity, the integral variable can be changed to other symbols, such as t, so the above definite integral can be
write in
()
x
A f t dt?
If the upper limit X changes arbitrarily in the interval ab, the definite integral has a corresponding value for each given value of X, so a function is defined on ab, which is called () x? ：
()()()x a x f t dt a x b？
()x？ It is an upper integral function.
In this paper, an integral upper bound function model is established for the second problem:
2 1
six
x
a
V
S y dt？ Taking the starting point a as the lower integral limit, the first upper integral limit, that is, the first bisector, is obtained.
The bisector is the lower integral limit, and the second upper integral limit is obtained, that is, the second bisector, and so on. Change products
Divide the upper and lower limits into five equal parts, divide the channel into six equal parts, and each section has the same earthwork volume.
Solving with MATLAB (see Appendix 8.2), the coordinates of bisector are: 4.52 14, 7.8736, 2.4797, 2.9862, 6.4 197 10956, 2.3730,1.
The earthwork volume of each section is 3 m8. 1253.
5.3 Question 3 According to Question 2, the earthwork volume V is related to the channel curve function. Establish xFV first? Model.
There are three variables k ji x, X, X in the road along the canal, and the temporary road to be built needs to ensure the transport workers.
The amount of work is the smallest, so the earthwork excavated on the left side of point D is transported to point B, and the channel is excavated on the right side of point D..
All the earth and stone are transported to C, and finally the earth and stone are transported from B and C to A (see Figure 4 for schematic diagram).
y = f(x)
(xk)
(Xu Ji) d
(xi)
L2
L 1
A
C
B
Figure 4. Schematic diagram of temporary highway construction in aqueduct
The transportation workload is equal to the earthwork quantity multiplied by the transportation distance, so for the transportation workload on the canal curve, this paper
The established model is:
Take i 0 x ~x section as an example, assuming that the length of the canal in this section is
No, no, no, no.
, the earthwork quantity of this section is
SL
n V V，V i 0i0 i0？
,
Then:
)LnL(V)L2L(V)LL(VW i0 i0 i0
)n2 1(LVVLn i0？ ?
n2 1n
luncheon voucher
What time? i 0 i 0 x x
2
x
x
2 i0 dx y 1 dxy 1S 2 1 LV 2 1 W，n
In the same way, you can get1kjjkijw, w, w.
Then:
iji0 1kjk 1 WWWWW
14550 14550 2 2 2 2 1 1 1 1 1 122 kk J k k xx x x x x x S y dx y y dx y y dx J ii x 2x x 7650 2x 7650 2d xy 1 dxy 65438
2 0k 2 0k 14550 x 22 0i 2 0i
x
7650
2 2y yxxdxy 1 syyxxdxy 1 SW
j
j
In order to minimize the transportation workload, that is, the sum of 1 W and 2 W, this paper establishes an unconstrained clearance.
Lottery model: kj2kji 1x, x, xwx, x, xwmin?
Namely:
? ? ? ? 2 0i 2 0i x 7650 2 2x 2 2x 7650 2y yxxdxy 1 dxy 1 2 1 dxy 1 2 1Min j j j I I
Zazie Hoko
? 2 0k 2 0k 14550 x 2 2 14550 x 2 2 x 2y yxxdxy 1 dxy 1 2 1 dxy 1 2 1 jkk j
Solving with MATLAB: (See Appendix 8.3 for program code)
The minimum transportation workload is: 4 8 1027.7 m? The coordinates of b and c are: 7.519296B and 4.3888 1 1683C respectively.
The advantages and disadvantages of intransitive verbs.
Advantages:
1. By comparing Hermite interpolation with cubic spline interpolation, the score of canal length is obtained.
Don't be 7522.5m and 7524.438+0 m, there is no obvious difference in this problem.
2. For the transportation planning problem, accurately reflect the optimal solution.
Disadvantages:
1, for the curve integral of interpolation function, the derivative of the curve is approximate and there is some error.
2. The calculation of planning problems is very large. The advantage of using MATLAB algorithm is not obvious.