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摘要: 不等式约束部分变量含误差(partial errors-in-variables, PEIV)模型目前主要采用线性化方法和非线性规划类算法, 前者计算效率较低, 后者基于最优化理论, 计算复杂, 未能与经典平差理论建立联系, 难以在测量实际中推广。在整体最小二乘准则下, 根据最优解的Kuhn-Tucker条件, 将不等式约束整体最小二乘解的计算转化为二次规划问题, 并提出改进的Jacobian迭代法求解二次规划。所提方法不需要对观测方程线性化, 与经典最小二乘法具有相同的形式, 易于编程实现。数值实例表明, 所提方法形式简洁, 具有良好的计算效率, 是经典最小二乘平差理论的有益拓展。
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关键词:
- 不等式约束 /
- 部分变量含误差模型 /
- 整体最小二乘 /
- Jacobian迭代法 /
- 经典最小二乘
Abstract:Objectives The inequality constrained partial errors-in-variables(ICPEIV) model is mainly solved by linear approximation method and nonlinear programming algorithms. The linear approximation method is computationally inefficient and the nonlinear programming algorithms are complicated because they are based on optimization theory. The nonlinear programming algorithms are impracticable to apply in surveying fields because the connections between nonlinear programming methods and classical adjustment have not been established.Methods Under the total least squares(TLS) criterion, the inequality constrained TLS problem is transformed into the quadratic programming according to the Kuhn-Tucker condition. Then an improved Jacobian iteration approach is proposed to solve the quadratic programming.Results The proposed method does not require the linearization process and has the same form with the classical least squares which is easy to code.Conclusions The numerical examples show that the proposed method is efficient in computation and concise in form and it is a beneficial extension of classical least squares theory. -
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表 1 算例1的模拟实验结果
Table 1 Simulation Results of Example 1
ICWTLS解 $ \hat {\boldsymbol{\beta}}_1 $ $ \hat {\boldsymbol{\beta}}_2 $ $ \hat {\boldsymbol{\beta}}_3 $ $ \hat {\boldsymbol{\beta}}_4 $ 外迭代次数 平均内迭代次数 计算时间/s $ \hat {\boldsymbol{\beta}}_{{\rm{LA}}} $ −0.100 000 −0.100 000 0.168 581 0.399 765 23 2 0.038 $ \hat {\boldsymbol{\beta}}_{{\rm{SQP}}} $ −0.100 000 −0.100 000 0.168 579 0.399 766 5 2 0.020 $ \hat {\boldsymbol{\beta}}_{{\rm{CLS}}} $ −0.100 000 −0.100 000 0.168 547 0.399 777 22 132 0.025 表 2 算例2的模拟实验结果
Table 2 Simulation Results of Example 2
ICWTLS解 $ \hat {\boldsymbol{\beta}}_1 $ $ \hat {\boldsymbol{\beta}}_2 $ $ \hat {\boldsymbol{\beta}}_3 $ $ \hat {\boldsymbol{\beta}}_4 $ 外迭代次数 平均内迭代次数 计算时间/s $ \hat {\boldsymbol{\beta}}_{{\rm{LA}}} $ 0.127 524 −0.576 759 0.426 986 0.243 459 4 2 0.018 $ \hat {\boldsymbol{\beta}}_{{\rm{SQP}}} $ −0.100 000 −0.100 000 0.168 579 0.399 766 5 2 0.020 $ \hat {\boldsymbol{\beta}}_{{\rm{CLS}}} $ 0.127 524 −0.576 759 0.426 986 0.243 459 14 27 0.009 -
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