pid论文格式

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恒温室房间温度PID控制研究摘要：某恒温实验室的恒温精度为27±0.2℃，但是由于实验室的非凡性，恒温室的内外扰量多且某些随机扰量的大小难于确定，而导致了其恒温精度很难达到预期效果。为了解决这个问题，通过建立恒温室被控对象的数学模型求出其传递函数，然后采用参数寻优方法确定PID控制器的参数，最后采用MATLAB仿真的方法，研究恒温室内外扰量对房间温度的影响。通过研究，可以得出，当设备散热干扰量为17℃以及送风温度干扰量为0.1℃，渗透风干扰量不大于0.3℃时，PID控制才能保证恒温室的恒温精度。关键词：恒温室，PID控制渗透风干扰量参数寻优温度1前言随着科学技术的发展，各类精密产品的生产制造以及特种科学实验都要求具有特定的工作环境，恒温就成为了不可缺少的条件之一。目前我国常见的恒温室的恒温精度为±1℃及±0.5℃，也有±0.1℃。而一些高精度的恒温室如光学仪器厂的刻线室恒温精度已达到了±0.0056℃。但是在某些非凡的科学实验室不仅恒温精度很高，而且干扰量多如渗透风、设备散热、送风温度波动以及电热器供电电压的波动等，且某些干扰量如渗透风其最大值难于确定而没有采用相应的措施控制渗透风扰量，导致了房间温度的波动过大，结果使恒温室的恒温精度很难达到要求。如何使这些非凡的科学实验室恒温精度达到使用要求，也成为了恒温室的空调系统和控制系统设计的一个巨大的难题。由于传统的PID控制算法，其运算简单、调整方便、鲁棒性强，在过程控制中，这种控制算法仍占据相当重要的地位.故目前恒温室的空调系统大部分采用PID控制。但PID控制的效果如何，在很大程度上是取决于控制器参数的正确整定。为此，人们提出了各种不同的参数整定方法，如误差积分最小、固定衰减比、极点配置等方法.这些方法主要是用经典控制理论中的一些设计方法或者依靠现场试验方法来进行PID控制器参数的计算与整定.显然，这就要求操作人员具有较高的理论基础和现场调试经验.而且，被控对象模型参数难以确定以及系统性能稳定性较差，则需频繁地进行参数整定，这必将影响系统的正常运行。对于这些非凡的空调房间温度的控制，由于被控对象具有较大的惯性和迟延，且受各种因素变化的影响，因此对象的传递函数具有非线性和时变特性，采用传统的PID控制难于取得较好的控制效果。本文采用单纯形法寻优PID参数，然后采用MATLAB仿真确定渗透风干扰量的最大值，PID控制才能保证恒温室的恒温精度。2工程概况恒温室建筑面积625m2，层高8m，总送风量27500m3/h，送风温度15℃，房间设计温度27±0.2℃，设备散热量135KW，恒温室建筑墙体、地板采用绝热材料，渗透风来自外部房间其设计温度26±1℃。3恒温室空调过程建模1恒温室空调系统被控对象的数学模型要对一个恒温室空调系统被控对象进行控制，须为其建立一个合适的数学模型。使用数学语言对实际对象进行一些必要的简化和假设：由于该恒温室建筑墙体、地板采用绝热材料，故室内外墙体和地板热量传递忽略不计。恒温室顶棚由盖板组成，存在缝隙，考虑有一定的渗透风，其他地方如门窗的渗透风忽略不计。假如不考虑执行机构的惯性和室温调节对象的传递滞后，根据能量守恒定律，单位时间内进入对象的能量减去单位时间内由对象流出的能量等于对象内能量蓄存量的变化率，表达式和图1如下所示：图1室温自动调节系统数学表达式为：式中：Chrr——恒温室的热容；C——空气的比热；GS——送风量；θ0’——电加热器前的送风温度；θ1——室内空气温度，回风温度；QE——电加热器的热量；Qm——设备散热量；QI——渗透风带入的热量；由式QI＝GIcit式中：GI——渗透风量；θIt——渗透风空气温度；cIt——渗透风空气的比热。把式代入式，整理得式中：T1——调节对象的时间常数，T1=Chrr/；K1——调节对象的放大系数，K1=GSc/；θE——电加热器的调节量，换算成送风温度的变化，θE=QE/GSC；θf——干扰量换算成送风温度的变化，；θf‘——送风温度干扰量，θf‘＝θ0“θIf——渗透风的干扰量，θIf=QI/GSC；θMf——设备散热量的干扰量，θMf=QM/GSC。由式拉普拉斯变换，得假如考虑被控对象传递滞后，则恒温室空调过程的传递函数为：2感温元件和执行调节机构的传递函数感温元件采用热电阻，根据热平衡原理，其热量平衡方程式：式中：C2——热电阻的热容；θ2——热电阻温度；q2——单位时间内空气传给热电阻的热量；α2——室内空气与热电阻表面之间的换热系数；F2——热电阻的表面积；θ1——室内空气温度，回风温度。由式拉普拉斯变换，可得感温元件的传递函数：同样执行调节机构的传递函数：3恒温室特性参数及其他参数的确定恒温室特性即房间的特性，用传递滞后τ、时间常数T1和放大系数K1这三个参数来表示。时间常数T1和放大系数K1由式，η=4，GI=GS×3%，通过式，式计算可以得到，T1=18分，K1=0.971。传递滞后τ由经验公式τ/T1=0.07通过计算则得τ=35分由参考文献的附表6-可以得到感温元件的时间常数和不灵敏区为T3=50秒，2ε=0.05℃。电加热器的比例系数K2=△θ/△N=0.00009，T2=50秒。4单纯形法寻优方法控制系统参数最优化是指对被控对象已知、控制器的结构和形式已确定，需要调整或寻找控制系统的某些参数使整个控制系统在某一性能指标下最佳。单纯形法的思想很简单，若要求一个函数的最大点，则可先计算若干点处的函数值，进行比较，并根据它们的大小关系确定函数的变化趋势作为搜索的参考方向，然后按参考方向搜索直到找到最小值为止。在三维空间内取不同一平面的四个点构成单纯形，如图3所示。图2三维空间的单纯形这四个点X0、X1、XX3对应的函数值为F0、FFF3，比较可看出最大者，则对应点X3作为差点，由此可以推测好点在差点XH的对称点XR处的可能性最大，然后计算XR处的函数值FR，若有FR≥max，说明从XH前进的步长太大，XR并不一定比XH好，因此可以压缩步长在XH与XR之间找一点XS为新点，然后X0，F1，F2中最大者说明情况有所改善，但前进和步长可能还不够，还可以加大步长得XH与XR延长线上的一点XE，若XE对应的函数FE小于FR则以XE作为新点，并以X0、X1、X2构成新的单纯形。最后比较构成新的单纯形的各点处的函数值，若其中最大者和最小者之间的相对差小于预先给定的数E，则说明最小值已经找到，否则继续重复上述步骤直到找到止。5恒温室控制系统仿真整个室温自动调节系统包括调节对象，调节器、感温元件以及PID控制器。根据参数计算结果，最后得到恒温室恒温控制系统如图3所示。图3恒温室恒温控制系统仿真框图？恒温室实验设备散热量相当稳定，由式计算可得，设备散热量干扰量θMf＝17℃是稳定的扰量。而送风温度干扰量主要包括电加热器供电电压的波动和换热器冷冻水温度的波动以及管道温升等引起的送风温度的变化，其值为0.1℃。渗透风干扰量是随机扰量，其随着恒温室外面的房间温度的变化和渗透风风量的变化而变化，它是影响恒温室的房间温度最重要的因数。当渗透风干扰量分别0.1℃、0.2℃、0.3℃、0.4℃时，PID控制的仿真曲线如图4-图7所示。图4θIf为0.1℃时PID控制的仿真曲线图5θIf为0.2℃时PID控制的仿真曲线图6θIf为0.3℃时PID控制的仿真曲线图7θIf为0.4℃时PID控制的仿真曲线分析图4-图可以得出：当渗透风扰量θIf不大于0.3℃时，恒温室房间温度波动小于0.2℃，满足恒温室的恒温精度要求。但是当渗透风扰量θIf为0.4℃时，恒温室房间温度波动大于0.2℃，超出答应的波动范围。6结论通过以上的仿真和分析，可以得出：恒温实验室的恒温精度为27±0.2℃，但是由于实验室的非凡性，恒温室的内外扰量多，只有当设备散热干扰量为17℃以及送风温度干扰量为0.1℃，渗透风干扰量不大于0.3℃时，PID控制才能保证恒温实验室的恒温精度，达到使用的要求。=============Room temperature PID control of room temperatureAbstract: A temperature accuracy of laboratory temperature 27 ± 0.2 ℃, but because of the extraordinary nature of the laboratory, constant temperature room and the volume inside and outside the interference of some random disturbance difficult to determine the size of the volume, which led to the accuracy of its temperature is very difficult to achieve the desired effect. To solve this problem, through the establishment of constant temperature room was charged with the mathematical model of the object to derive its transfer function, and then used to determine optimal parameters of PID controller parameters, and finally the use of MATLAB simulation method to study indoor and outdoor temperature the amount of room disturbance temperature. Through research, can be drawn, when the equipment cooling capacity of 17 ℃ interference and disruption of supply air temperature 0.1 ℃, the volume of infiltration air interference at 0.3 ℃ not more than, PID temperature control room in order to ensure the accuracy of the thermostat. Key words: constant temperature room, PID control parameters of the infiltration volume of wind interference temperature optimization 1 Introduction With the development of science and technology, various types of manufacturing precision products and the special requirements of scientific experiments are with a specific working environment has become a heated one of the conditions indispensable. At present, the temperature of our common room temperature accuracy of ± 1 ℃ and ± 0.5 ℃, also ± 0.1 ℃. And some, such as high-precision optical instrument factory room temperature of the engraved line has reached room temperature accuracy of ± 0.0056 ℃. However, in some extraordinary precision scientific laboratory is not only a high temperature, and interference, such as infiltration of the wind volume, equipment cooling, supply air temperature fluctuations, as well as electric heaters, such as supply voltage fluctuations, and some amount of interference, such as wind penetration of its difficult to determine the maximum value without the use of appropriate measures to control the amount of infiltration of the wind disturbance, leading to fluctuations in room temperature is too large, resulting in the constant temperature room thermostat accuracy requirement is very difficult to achieve. How to make these remarkable scientific accuracy of the use of constant temperature laboratory requirements, but also become a constant temperature room air-conditioning system and control system design of a huge problem. As a result of the traditional PID control algorithm, the computation is simple, convenient adjustment, robustness, and in process control, this control algorithm is still occupied a very important position. Therefore, the current room temperature most of the air-conditioning system using PID control. However, the effects of PID control to a large extent depends on the correct controller parameter tuning. To this end, the people made a variety of parameter tuning methods, such as minimum error integral, fixed attenuation ratio, pole placement and other methods. These methods are mainly used in classical control theory a number of design methods or testing methods rely on the scene to carry out PID control parameters of the calculation and setting. Obviously, this requires the operator has a higher theoretical basis and field testing experience. Moreover, the plant model parameters it is difficult to determine system performance, as well as less stable, it would take frequent tuning parameters, This will affect the normal operation of the system. For these extraordinary control of an air-conditioned room temperature, due to a larger plant with inertia and delay, and by the impact of changes in a variety of factors, so the object of the transfer function with nonlinear and time-varying characteristics, using the traditional PID control difficult to obtain a better control effect. Simplex method using PID parameter optimization, and simulation using MATLAB to determine the amount of infiltration of the maximum wind disturbance, PID control room temperature in order to ensure the accuracy of the thermostat. 2 Project Overview Constant temperature room floor area of 625m2, storey 8m, total air volume of 27500m3 / h, air temperature 15 ℃, room design temperature 27 ± 0.2 ℃, heat dissipation equipment 135KW, constant temperature room building wall, floor insulation materials used, the wind penetration to Since the design of external room temperature 26 ± 1 ℃. 3 air-conditioned room temperature process modeling A constant temperature room air-conditioning systems charged with the mathematical model of the object To a constant temperature room air-conditioning system to control the object to be suitable for the establishment of a mathematical model. The use of mathematical language of the actual number of objects necessary to simplify and assumptions: Room temperature due to the building wall, floor insulation materials used, the indoor and outdoor wall and floor heat transfer is negligible. Room temperature by the flat roof of the existence of the gap, taking a certain degree of infiltration of wind, such as doors and windows in other parts of the wind penetration is negligible. If agencies do not consider the implementation of the inertia and temperature regulation of the transmission lags behind the target, according to the law of conservation of energy, per unit time into the object of energy per unit time minus the outflow of energy from the target object with the same energy with the rate of change of the stock, the expression and Figure 1 as follows: Figure 1 at room temperature automatic adjustment system Mathematical expression is: Where: Chrr - room temperature heat capacity; C - specific heat of air; GS - air traffic; θ0' - electric heater before the air temperature; θ1 - indoor air temperature, return air temperature; QE - the heat electric heater; Qm - equipment heat dissipation; QI - the infiltration of heat into the wind; QI = GIcit by type Type in: GI - the infiltration air flow; θIt - air temperature wind penetration; cIt - the air infiltration heat wind. Skill into the style, finishing a Where: T1 - the object of regulation time constant, T1 = Chrr /; K1 - adjust object magnification factor, K1 = GSc /; θE - regulation of electric heater volume, converted into changes in air temperature, θE = QE / GSC; θf - converted to interfere with the volume of supply air temperature changes ; θf' - interfere with the volume of supply air temperature, θf '= θ0 " θIf - interfere with the volume of wind penetration, θIf = QI / GSC; θMf - equipment heat dissipation capacity of the interference, θMf = QM / GSC. Laplace transform by the style, too If the accused objects to consider transmission lag, then the process of constant temperature air-conditioned room for the transfer function: 2 temperature components and the implementation of the transfer function of regulating agencies Temperature components using thermal resistance, according to the principle of heat balance, the heat balance equation: Where: C2 - thermal resistance of the heat capacity; θ2 - thermal resistance temperature; q2 - units of air time to the thermal resistance of heat; α2 - indoor air and thermal resistance between the surface heat transfer coefficient; F2 - thermal resistance of the surface area; θ1 - indoor air temperature, return air temperature. Laplace transform by type, available thermal transfer function components: Regulating agencies to implement the same transfer function: 3 Room temperature characteristics and other parameters to determine Room temperature properties of the characteristics of that room, with delivery lags τ, time constant T1 and the magnification factor of these three parameters K1 said. Time constant T1 and the magnification factor K1 By type, η = 4, GI = GS × 3%, through the ceremony, can be calculated, T1 = 18 minutes, K1 = 0.971. Transmission lag τ By the empirical formula by calculating the τ/T1 = 0.07 is a τ = 35 minutes References schedule from 6 - temperature components can be time constant and dead-zone for the T3 = 50 seconds, 2ε = 0.05 ℃. The ratio of electric heater coefficient K2 = △ θ / △ N = 0.00009, T2 = 50 seconds. 4 simplex optimization method Control system parameters optimization refers to the object known, controls the structure and form have been identified, need to adjust the control system or to find some of the parameters so that the whole control system of performance indicators in a best. Simplex method of thinking is very simple, if the maximum point of a function may be a number of points to calculate the function values, for comparison, and in accordance with their function to determine the size of the relationship between changes in the trend of the reference as a search direction, and then by reference to the direction of the search until you find the minimum value so far. Three-dimensional space in a plane different from the four points constitute a simplex, as shown in Figure 3. Figure 2 Three-dimensional simplex space These four points X0, X1, XX3 corresponding value is a function of F0, FFF3, comparison can be seen the largest, then the corresponding point as almost X3, it can be a good point that in the nearly symmetric point XH Department is most likely to XR , and then calculating the value function XR Department FR, if FR ≥ max, from the XH big step forward, XR is not necessarily better than the XH, the compression step can be in between the XH and XR point to find the new XS point X0, F1, F2 largest note in the situation has improved, but step forward and might not be enough, you can increase the step-by-step look XH and XR point to extend the online XE, if XE corresponding FE function of small while in the FR as a new point of XE, and X0, X1, X2 constitute a new simplex. Finally, constitute a new comparison of the simplex function value point, if one of the largest and the smallest relative difference between the pre-less than a given number of E, then the minimum has been found, otherwise continue to repeat the above steps until you find the only. 5 Room temperature control system simulation Automatic adjustment of the temperature regulation system includes objects, regulators, and the PID temperature controller components. The calculation results based on parameters, and finally to be room temperature thermostat control system shown in figure 3. Figure 3 Room temperature thermostat control system simulation block diagram? Room temperature a considerable amount of laboratory equipment, heat stability, available from the computing equipment interfere with heat dissipation capacity θMf = 17 ℃ traffic disturbance is stable. Interfere with the volume of supply air temperature and the main electric heater supply voltage fluctuations and the chilled water heat exchanger, as well as fluctuations in temperature, such as pipeline temperature rise caused by changes in air temperature, its value is 0.1 ℃. Interfere with the volume of air infiltration is the amount of random disturbance, the room with the outside temperature changes in room temperature and the amount of infiltration风风changes, it is the impact of room temperature room temperature the most important factor. When the infiltration of wind interference respectively 0.1 ℃, 0.2 ℃, 0.3 ℃, 0.4 ℃, when, PID control of the simulation curve in Figure 4 - as shown in Figure 7. Figure 0.1 ℃ when 4θIf for the simulation curve of PID control Figure 0.2 ℃ when 5θIf for the simulation curve of PID control Figure 0.3 ℃ when 6θIf for the simulation curve of PID control Figure 0.4 ℃ when 7θIf for the simulation curve of PID control Analysis of Figure 4 - plans can be drawn: When the volume of infiltration θIf wind disturbance is not more than 0.3 ℃, the temperature fluctuations in room temperature of less than 0.2 rooms ℃, room temperature thermostat to meet accuracy requirements. However, when the volume of infiltration air disturbance θIf for 0.4 ℃, the temperature fluctuations in room temperature greater than room 0.2 ℃, beyond the agreed scope of the fluctuations. 6 Conclusion Through the above simulation and analysis, can be drawn: Temperature accuracy of laboratory temperature 27 ± 0.2 ℃, but because of the extraordinary nature of the laboratory, both inside and outside the room temperature volume disturbance, interference only when the cooling equipment of air temperature of 17 ℃ and the interference of 0.1 ℃, wind penetration interfere with the volume of not more than 0.3 ℃, when, PID temperature control to ensure the accuracy of laboratory temperature, to achieve the requirements of use.【这是我在网上找的，希望可以帮助你】

只要有论文格式标准和要求就行，调整格式很简单.

题目（黑体不加粗三号居中） 摘要（黑体不加粗四号居中）（摘要正文小4号，写法如下） 首先简要叙述所给问题的意义和要求，并分别分析每个小问题的特点（以下以三个问题为例）。根据这些特点我们对问题1用……的方法解决；对问题2用......的方法解决；对问题3用……的方法解决。对于问题1我们用......数学中的......首先建立了......模型I。在对......模型改进的基础上建立了......。模型II。对模型进行了合理的理论证明和推导，所给出的理论证明结果大约为......。，然后借助于......数学算法和......软件，对附件中所提供的数据进行了筛选，去除异常数据,对残缺数据进行适当补充,并从中随机抽取了3组数据（每组8个采样）对理论结果进行了数据模拟，结果显示，理论结果与数据模拟结果吻合。（方法、软件、结果都必须清晰描述，可以独立成段，不建议使用表格）对于问题2我们用......对于问题3我们用......如果题目单问题，则至少要给出2种模型，分别给出模型的名称、思想、软件、结果、亮点详细说明。并且一定要在摘要对两个或两个以上模型进行比较，优势较大的放后面，这两个（模型）一定要有具体结果。如果在……条件下，模型可以进行适当修改，这种条件的改变可能来自你的一种猜想或建议。要注意合理性。此推广模型可以不深入研究，也可以没有具体结果。关键词：本文使用到的模型名称、方法名称、特别是亮点一定要在关键字里出现，5~7个较合适。注：字数700~1000之间；摘要中必须将具体方法、结果写出来；摘要写满几乎一页，不要超过一页。摘要是重中之重，必须严格执行！。页码：1（底居中）一、问题重述（第二页起黑四号）在保持原题主体思想不变下，可以自己组织词句对问题进行描述，主要数据可以直接复制，对所提出的问题部分基本原样复制。篇幅建议不要超过一页。大部分文字提炼自原题。二、问题分析主要是表达对题目的理解，特别是对附件的数据进行必要分析、描述（一般都有数据附件），这是需要提到分析数据的方法、理由。如果有多个小问题，可以对每个小问题进行分别分析。（假设有3个问题） 问题1的分析对问题1研究的意义的分析。问题1属于......数学问题，对于解决此类问题一般数学方法的分析。对附件中所给数据特点的分析。对问题1所要求的结果进行分析。由于以上原因，我们可以将首先建立一个......的数学模型I,然后将建立一个......的模型II,........对结果分别进行预测，并将结果进行比较.问题2的分析对问题2研究的意义的分析。问题2属于......数学问题，对于解决此类问题一般数学方法的分析。对附件中所给数据特点的分析。对问题2所要求的结果进行分析。由于以上原因，我们可以将首先建立一个......的数学模型I,然后将建立一个......的模型II,......。。对结果分别进行预测，并将结果进行比较. ..............................。。三、模型假设（4号黑体）（以下小4号） 假设题目所给的数据真实可靠；2．3．4．5．6．.................................... 注意：假设对整篇文章具有指导性，有时决定问题的难易。一定要注意假设的某种角度上的合理性，不能乱编，完全偏离事实或与题目要求相抵触。注意罗列要工整。四、定义与符号说明（4号黑体）（对文章中所用到的主要数学符号进行解释小4号）............................ 尽可能借鉴参考书上通常采用的符号，不宜自己乱定义符号，对于改进的一些模型，符号可以适当自己修正（下标、上标、参数等可以变，主符号最好与经典模型符号靠近）。对文章自己创新的名词需要特别解释。其他符号要进行说明，注意罗列要工整。如“～第种疗法的第项指标值”等，注意格式统一，不要出现零乱或前后不一致现象，关键是容易看懂。五、模型的建立与求解（4号黑体）第一部分：准备工作（4号宋体）数据的处理 1、......数据全部缺失，不予考虑。 2、对数据测试的特点，如，周期等进行分析。 3、......数据残缺，根据数据挖掘等理论根据......变化趋势进行补充。 4、对数据特点（后面将会用到的特征）进行提取。（二）聚类分析（进行采样） 用......软件聚类分析和各个不同问题的需要，采得。。。组采样，每组5-8个采样值。将采样所对应的特征值进行列表或图示。预测的准备工作根据数据特点，对总体和个体的特点进行比较，以表格或图示方式显示。第二部分：问题1的...模型（4号宋体）模型I(......的模型)该种模型的一般数学表达式，意义，和式中各种参数的意义。注明参考文献。......模型I的建立和求解说明问题1适用用此模型来解决，并将模型进行改进以适应问题1。借助准备工作中的采样，（用拟合等方法）确定出模型中的参数。给出问题1的数学模型I表达式和图形表示式。给出误差分析的理论估计。3.模型I的数值模拟将模型I进行数值计算，并与附件中的真实采样值（进行列表或图示）比较。对误差进行数据分析。模型II(......的模型)该种模型的一般数学表达式，意义，和式中各种参数的意义。注明参考文献。......模型II的建立和求解说明问题1适用此模型来解决，并将模型进行改进以适应问题1。借助准备工作中的采样，通过确定出模型中的参数。给出问题1的数学模型I表达式和图形表示式。给出误差分析的理论估计。3.模型II的数值模拟将模型II进行数值计算，并与附件中的真实采样值（进行列表或图示）比较。对误差进行数据分析 （三）模型III(......的模型) ........................（四）问题1的三种数学模型的比较。对三种模型的优点和缺点结合原始数据和模拟预测数据进行比较。给出各自得优点和缺点。第三部分：问题2的...个模型（4号宋体）........................。第四部分：问题3的...个模型（4号宋体）........................。六、模型评价与推广对本文中的模型给出比较客观的评价，必须实事求是，有根据，以便评卷人参考。推广和优化，需要挖空心思，想出合理的、甚至可以合理改变题目给出的条件的、不一定可行但是具有一定想象空间的准理想的方法、模型。（大胆、合理、心细。反复推敲，这段500字半页左右的文字，可能决定生死存亡。）七、参考文献（4号黑体）(书写格式如下) [1] 作者名1,作者名2.文章名字.杂志名字，年，卷（期）：起始页码-结束页码[2] 作者名1,作者名2.书名.出版地：出版社，年，起始页码-结束页码[3] 作者名1,作者名2.文章名字. 年，卷（期）：起始页码-结束页码，网页地址。[4] 李传鹏，什么是中国标准书号，，2006-9-18。[5] 徐玖平、胡知能、李军，运筹学（II类），北京：科学出版社，2004。[6] Ishizuka Y, AiyoshiE. Double penalty method for bilevel optimization problems. Annals of Operations Research, 24: 73- 88，1992。注意：5篇以上！八、附件（4号黑体）（正文中不许出现程序，如果要附程序只能以附件形式给出） 数学建模评分参考标准摘要（很重要） 5分数据筛选 35分数学模型 35分数据模拟 15分总体感觉 10分特别注意1.问题的结果要让评卷人好找到；显要位置---独立成段2.摘要中要将方法、结果讲清楚；3.可以有目录也可以不要目录；4.建模的整个过程要清楚，自圆其说，有结果、有创新；5.采样要足够多，每组不少于7个；6.模型要与数据结合，用数据验证过；7.如果数学方法选错，肯定失败；8.规范、整洁；总页数在35~45之间为宜。9.必须有数学模型，同一问题的不同模型要比较；10.数据必须有分析和筛选；11.模型不能太复杂，若用多项式回归分析，次数以3次为好。