﻿ 基于DEA模型贵州省高校研究生培养的有效性分析

# 基于DEA模型贵州省高校研究生培养的有效性分析The Effective Analysis of Graduate Student Education in Universities for Guizhou Province Based on DEA Model

Abstract: The scale and effectiveness of postgraduate education play an important guiding role in talents cultivation. We study the postgraduate education for the six universities with academic postgrad-uate education in Guizhou province. Selecting the structure of source to students, the process of education, structure of teaching staff, the average funding investment and the hardware facilities of each student for input index of the model, and the postgraduate’s published papers (core journals and above), the ratio of the outstanding papers on the provincial level and the first employment rate for the output index of model, DEA-C2R is established and solved by excel. The conclusion shows that the investments of the six universities have reached the largest level. But all of them are non-scale effective. It indicates that we can appropriately change some inputs, so that the scale of postgraduate education is the effectiveness of DEA.

1. 引言

《国家中长期教育改革和发展规划纲要(2010-2020)》第(十九)条指出提高人才培养质量，牢固确立人才培养在高校工作中的中心地位，着力培养信念执著、品德优良、知识丰富、本领过硬的高素质专门人才和拔尖创新人才。作为人才培养的一个重要环节——硕士研究生教育承担两大主要任务：一是为各行各业培养高层次应用型人才，另外是为博士阶段输送优秀生源。随着我国经济社会的快速发展，各行各业实际部门对高层次应用人才的需求持续增长，对研究生质量都有着更大的需求，有更迫切的愿望。

2. 数量指标的确定

2.1. 指标的选取

2.2. 指标值的确定

Table 1. Values of input index

Table 2. Values of output index

3. DEA-C2R模型的建立与求解

3.1. C2R模型的建立

${h}_{j}=\frac{\underset{r=1}{\overset{3}{\sum }}{\mu }_{r}{y}_{rj}}{\underset{i=1}{\overset{5}{\sum }}{\lambda }_{i}{x}_{ij}}$ .

$\left({\text{C}}^{\text{2}}\text{R}\right)\left\{\begin{array}{l}\mathrm{max}Z={h}_{{j}_{0}},\\ \text{s}\text{.t}\text{.}\text{ }{h}_{j}\underset{_}{\underset{_}{<}}1,\\ \text{ }\text{ }\lambda \ge 0,\mu \ge 0,\end{array}$ (1)

$m=\frac{1}{{\lambda }^{\text{T}}{x}_{{j}_{0}}},\text{ }\alpha =m\lambda ,\text{ }\beta =m\mu ,$

$\left(\text{LP}\right)\left\{\begin{array}{l}\mathrm{max}W={\beta }^{\text{T}}{y}_{{j}_{0}},\\ \text{s}\text{.t}\text{.}\text{ }{\alpha }^{\text{T}}{x}_{j}-{\beta }^{\text{T}}{y}_{j}\ge 0,\\ \text{ }\text{ }{\alpha }^{\text{T}}{x}_{{j}_{0}}=1,\\ \text{ }\text{ }\alpha \ge 0,\beta \ge 0,\end{array}$ (2)

1) 若 ${\alpha }^{0},{\beta }^{0}$ 为线性规划LP(2)的最优解，那么 ${\alpha }^{0},{\beta }^{0}$ 也是DEA-C2R模型(1)的最优解，并且最优值相等；

2) 若 ${\lambda }^{0},{\mu }^{0}$ 为DEA-C2R模型(1)的最优解，那么

${\alpha }^{0}={m}^{0}{\lambda }^{0},\text{ }{\beta }^{0}={m}^{0}{\mu }^{0}$

${m}^{0}=\frac{1}{{\lambda }^{0}{}^{\text{T}}{x}_{{j}_{0}}}.$

$W={\beta }^{0\text{T}}{y}_{{j}_{0}}=1$ ,

$W={\beta }^{0\text{T}}{y}_{{j}_{0}}=1$ ,

3.2. C2R模型的求解

$\left\{\begin{array}{l}\mathrm{max}\text{\hspace{0.17em}}W=0.09{\beta }_{1}+0.005{\beta }_{2}+0.73{\beta }_{3},\\ \text{s}\text{.t}\text{.}\text{ }0.53{\alpha }_{1}+0.02{\alpha }_{2}+0.54{\alpha }_{3}+0.07{\alpha }_{4}+92.8{\alpha }_{5}-0.09{\beta }_{1}-0.005{\beta }_{2}-0.73{\beta }_{3}\ge 0,\\ \text{ }\text{ }0.51{\alpha }_{1}+0.08{\alpha }_{2}+0.29{\alpha }_{3}+0.09{\alpha }_{4}+173.7{\alpha }_{5}-0.051{\beta }_{1}-0.003{\beta }_{2}-0.81{\beta }_{3}\ge 0,\\ \text{ }\text{ }0.60{\alpha }_{1}+0.06{\alpha }_{2}+0.35{\alpha }_{3}+0.13{\alpha }_{4}+174.4{\alpha }_{5}-0.14{\beta }_{1}-0.014{\beta }_{2}-0.60{\beta }_{3}\ge 0,\\ \text{ }\text{ }0.61{\alpha }_{1}+0.05{\alpha }_{2}+0.65{\alpha }_{3}+0.13{\alpha }_{4}+119.8{\alpha }_{5}-0.19{\beta }_{1}-0.01{\beta }_{2}-0.76{\beta }_{3}\ge 0,\\ \text{ }\text{ }0.64{\alpha }_{1}+0.06{\alpha }_{2}+0.49{\alpha }_{3}+0.10{\alpha }_{4}+162{\alpha }_{5}-0.13{\beta }_{1}-0.04{\beta }_{2}-0.74{\beta }_{3}\ge 0,\\ \text{ }\text{ }0.12{\alpha }_{1}+0.11{\alpha }_{2}+0.61{\alpha }_{3}+0.12{\alpha }_{4}+128.2{\alpha }_{5}-0.25{\beta }_{1}-0.024{\beta }_{2}-0.76{\beta }_{3}\ge 0,\\ \text{ }\text{ }0.53{\alpha }_{1}+0.02{\alpha }_{2}+0.54{\alpha }_{3}+0.07{\alpha }_{4}+92.8{\alpha }_{5}=1\\ \text{ }\text{ }{\alpha }_{1},{\alpha }_{2},{\alpha }_{3},{\alpha }_{4},{\alpha }_{5},{\beta }_{1},{\beta }_{2},{\beta }_{3}\ge 0\end{array}$ (3)

${\alpha }_{1}^{0}={\left(0,0,0,0,0.10\right)}^{\text{T}},{\beta }_{1}^{0}={\left(0,0,1.37\right)}^{\text{T}}$ (4)

${\alpha }_{2}^{0}={\left(0,0,0,0.95,0.004\right)}^{\text{T}},{\beta }_{2}^{0}={\left(0,0,1.23\right)}^{\text{T}}$ (5)

${\alpha }_{3}^{0}={\left(0,0,0,2.23,0.001\right)}^{\text{T}},{\beta }_{3}^{0}={\left(3.51,0,0.85\right)}^{\text{T}}$ (6)

${\alpha }_{4}^{0}={\left(0,1.1,0,0,0.007\right)}^{\text{T}},{\beta }_{4}^{0}={\left(2.24,0,0.76\right)}^{\text{T}}$ (7)

${\alpha }_{5}^{0}={\left(0,2.8,1.2,0,0.001\right)}^{\text{T}},{\beta }_{5}^{0}={\left(2.07,1.8,0.89\right)}^{\text{T}}$ (8)

${\alpha }_{6}^{0}={\left(0.53,0,0,0,0.007\right)}^{\text{T}},{\beta }_{6}^{0}={\left(0,0,1.32\right)}^{\text{T}}$ (9)

4. 结论

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