﻿ 给排水管道工程中管材造价的确定 Determination of Cost of Pipe Material in Water Supply and Drainage Pipework

Statistics and Application
Vol.07 No.03(2018), Article ID:25489,8 pages
10.12677/SA.2018.73035

Determination of Cost of Pipe Material in Water Supply and Drainage Pipework

Jiakang Du, Yuqing Li, Hongchun Sun, Qingjun Ren

School of Mathematics and Statistics, Linyi University, Linyi Shandong

Received: May 21st, 2018; accepted: Jun. 7th, 2018; published: Jun. 20th, 2018

ABSTRACT

The relationship between cost of pipe material and pipe size in water supply and drainage pipework is studied in this paper. In order to determine the cost formula of pipe material, we established a liner regression and one-dimensional polynomial regression model and fitted the coefficient of model function by least squares. The effectiveness of the algorithm is verified by numerical experiments.

Keywords:Cost of Pipe Material, Regression Model, Least Squares

1. 引言

2. 求解模型建立

$M=\sum _{i=1}^{t}{q}_{i}=\sum _{i=1}^{t}{p}_{i}{y}_{i}$

Table 1. Price list of unit cost and diameter of pipe material

2.1. 一元线性回归模型

$\left\{\begin{array}{l}y=a+bx+\epsilon \\ E\left(\epsilon \right)=0,D\left(\epsilon \right)={\sigma }^{2}\end{array}$

$\left\{\begin{array}{l}{y}_{i}=a+b{x}_{i}+{\epsilon }_{i},i=1,2,\cdots ,n\\ E\left({\epsilon }_{i}\right)=0,D\left({\epsilon }_{i}\right)={\sigma }^{2},{\epsilon }_{i}>0\end{array}$

$Q=Q\left(a,b\right)=\sum _{i=1}^{n}{\epsilon }^{2}=\sum _{i=1}^{n}{\left({y}_{i}-a-b{x}_{i}\right)}^{2}$ (1)

$a,b$ 的估计值 $\stackrel{^}{a},\stackrel{^}{b}$ ，使得

$Q\left(\stackrel{^}{a},\stackrel{^}{b}\right)={Q}_{\mathrm{min}}\left(a,b\right)$ (2)

$\left\{\begin{array}{l}\sum _{i=1}^{n}\left({y}_{i}-\stackrel{^}{a}-\stackrel{^}{b}{x}_{i}\right)=0\\ \sum _{i=1}^{n}{x}_{i}\left({y}_{i}-\stackrel{^}{a}-\stackrel{^}{b}{x}_{i}\right)=0\end{array}$ (3)

$\left\{\begin{array}{l}\stackrel{^}{a}=\overline{y}-\stackrel{^}{b}\overline{x}\\ \stackrel{^}{b}=\frac{\overline{xy}-\overline{x}\text{ }\overline{y}}{\overline{{x}^{2}}-{\overline{x}}^{2}}\end{array}$ (4)

$y=\stackrel{^}{a}+\stackrel{^}{b}x,\overline{y}=\stackrel{^}{b}\left(x-\overline{x}\right)$

2.2. 一元多项式回归

$f\left(x\right)=\sum _{j=0}^{m}{a}_{j}{x}^{j}$ (5)

${P}_{\mathrm{min}}=\sum _{i=1}^{n}{\left[f\left({x}_{i}\right)-{y}_{i}\right]}^{2}=\sum _{i=1}^{n}{\left(\sum _{j=0}^{m}{a}_{j}{x}^{j}-{y}_{i}\right)}^{2}$ (6)

$\frac{\partial P}{\partial {a}_{j}}=2\sum _{i=1}^{n}\left(\sum _{k=0}^{m}{a}_{k}{x}_{i}^{k}-{y}_{i}\right){x}_{i}^{j}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(j=0,1,2,\cdots ,m\right)$ (7)

$\frac{\partial P}{\partial {a}_{j}}=2\sum _{i=1}^{n}\left(\sum _{k=0}^{m}{a}_{k}{x}_{i}^{k}-{y}_{i}\right){x}_{i}^{j}=0$ (8)

$\sum _{k=0}^{m}\left(\sum _{i=i}^{n}{x}_{i}^{j+k}\right){a}_{k}=\sum _{i=i}^{n}{x}_{i}^{j}{y}_{i}\text{}\left(j=0,1,2,\cdots ,m\right)$ (9)

$=\left[\begin{array}{cccc}{c}_{0}& {c}_{1}& \cdots & {c}_{m}\\ {c}_{1}& {c}_{2}& \cdots & {c}_{m+1}\\ ⋮& ⋮& \ddots & ⋮\\ {c}_{m}& {c}_{m+1}& \cdots & {c}_{2m}\end{array}\right]×\left[\begin{array}{c}{a}_{0}\\ {a}_{1}\\ ⋮\\ {a}_{m}\end{array}\right]=\left[\begin{array}{c}{b}_{0}\\ {b}_{1}\\ ⋮\\ {b}_{m}\end{array}\right]$ (10)

$\left\{\begin{array}{l}{c}_{k}=\sum _{i=i}^{n}{x}_{i}^{k},\left(k=0,1,2,\cdots ,m\right)\\ {b}_{k}=\sum _{i=i}^{n}{x}_{i}^{k}{y}_{i},\left(k=0,1,2,\cdots ,m\right)\end{array}$ (11)

$C=\left[\begin{array}{cccc}{c}_{0}& {c}_{1}& \cdots & {c}_{m}\\ {c}_{1}& {c}_{2}& \cdots & {c}_{m+1}\\ ⋮& ⋮& \ddots & ⋮\\ {c}_{m}& {c}_{m+1}& \cdots & {c}_{2m}\end{array}\right]$$A=\left[\begin{array}{c}{a}_{0}\\ {a}_{1}\\ ⋮\\ {a}_{m}\end{array}\right]$$B=\left[\begin{array}{c}{b}_{0}\\ {b}_{1}\\ ⋮\\ {b}_{m}\end{array}\right]$ ，(10)式可以表示为： $CA=B$ 。解出多项式回归的系数 ${a}_{j}\left(j=0,1,2,\cdots ,m\right)$ 就能够得到拟合函数关系：

$f\left(x\right)={a}_{0}+{a}_{1}x+\cdots +{a}_{m}{x}^{m}=\sum _{j=0}^{m}{a}_{j}{x}^{j}$

2.3. 模型比较

$f\left(x\right)={a}_{0}+{a}_{1}x+\cdots +{a}_{m}{x}^{m}$

3. 数值算例

3.1. 具体求解过程

1) 做出散点图

2) 确定数学模型

① 一元线性回归模型。首先设 ${y}_{i}=a+b{x}_{i}+{\epsilon }_{i},i=1,2,\cdots ,n$ 再设

${Q}_{\mathrm{min}}=Q\left(a,b\right)=\sum _{i=1}^{n}{\epsilon }_{i}^{2}=\sum _{i=1}^{n}\left({y}_{i}-a-b{x}_{i}^{2}\right)$

Table 2. Price list of unit cost and diameter of pipe for graphite cast iron pipes

Figure 1. The scatter point diagram of pipe cost and pipe diameter

$\left\{\begin{array}{l}\stackrel{^}{a}=\overline{y}-\stackrel{^}{b}\overline{x}\\ \stackrel{^}{b}=\frac{\overline{xy}-\overline{x}\text{ }\overline{y}}{\overline{{x}^{2}}-{\overline{x}}^{2}}\end{array}$

$\left\{\begin{array}{l}\stackrel{^}{a}=-4.1816\\ \stackrel{^}{b}=53.0580\end{array}$

② 多项式回归模型

$C=\left[\begin{array}{ccc}8& 39& 255\\ 39& 255& 1953\\ 255& 1953& 16371\end{array}\right]$$B=\left[\begin{array}{c}2035.8\\ 13367\\ 104080\end{array}\right]$

${C}^{-1}=\left[\begin{array}{ccc}1.4610& -0.5694& 0.0452\\ -0.5694& 0.2673& -0.0230\\ 0.0452& -0.0230& 0.0021\end{array}\right]$

$A={C}^{-1}B=\left[\begin{array}{c}64.4780\\ 18.0544\\ 3.1994\end{array}\right]$

${a}_{0}=64.4780,{a}_{1}=18.0544,{a}_{2}=3.1994$ 拟合的函数为

$f\left(x\right)=64.4780+18.0544x+3.1994{x}^{2}$

3.2. 误差对比分析

Figure 2. Comparison of two kinds of fitting function values with original data

Figure 3. Numerical error contrast diagram of two models

3.3. 模型代入求解

$M=\sum _{i=1}^{t}{p}_{i}\left(64.4780+18.0544x+3.1994{x}^{2}\right)$

Determination of Cost of Pipe Material in Water Supply and Drainage Pipework[J]. 统计学与应用, 2018, 07(03): 297-304. https://doi.org/10.12677/SA.2018.73035

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