中期質(zhì)量檢查表2011111426陳嫻.doc

韓山師范學(xué)院畢業(yè)論文（設(shè)計(jì)）中期質(zhì)量檢查表（附件五）學(xué) 生自姓名陳嫻專業(yè)數(shù)學(xué)與應(yīng)用數(shù)學(xué)專業(yè)學(xué)號(hào)2011111426課題名稱工作進(jìn)度（對(duì)照開(kāi)題報(bào)告）：工作質(zhì)量（含文獻(xiàn)查閱、外文翻譯、研究方法分析、試驗(yàn)數(shù)據(jù)、設(shè)計(jì)草圖、論文提綱、論文初稿等）：1刁在筠.運(yùn)籌學(xué)（第三版）M.北京：高等教育出版社，2007, 57-62.2張建中，許紹吉.線性規(guī)劃M.北京：科學(xué)出版社，2005.3王艷峰,沈祖培.輸氣管道優(yōu)化設(shè)計(jì)新模型J. 油氣儲(chǔ)運(yùn). 2004.4姜啟源.數(shù)學(xué)模型（第四版）M.北京：高等教育出版社，2011, 85-95.The model of the transportation of natural gas pipeline Chen Xian Abstract: Through the analysis of gas suppliers and residential areas, we arrange the proper scheme of pipeline transport, make the pipeline transportation costs minimum, prompting the biggest profit. According to the specific circumstances, linear programming model is established, and the related constraints and objective function will be the constraint optimization problem.In the MATLAB and LINGO software we can find the optimal solutions to prove that the method has its feasibility,and the optimization design of urban natural gas pipeline has a practical guiding significance.Key words: The transportation of natural gas pipeline; Linear programming; The optimization design研究方法分析：建立線性規(guī)劃模型，利用約束條件和目標(biāo)函數(shù)求解約束優(yōu)化問(wèn)題，并找出最佳的解決方案，在MATLAB和LINGO軟件中證明該方法是可行的，以及管道運(yùn)輸?shù)膬?yōu)化對(duì)城市燃?xì)庠O(shè)計(jì)具有一定的指導(dǎo)意義.實(shí)驗(yàn)數(shù)據(jù)附錄一：求解問(wèn)題1的LINGO程序附錄二：附錄一的運(yùn)行結(jié)果附錄三：求解問(wèn)題2的LINGO程序附錄四：附錄三的運(yùn)行結(jié)果附錄正文附錄一：求解問(wèn)題1的LINGO程序max =2.8*x11+3.1*x12+3*x13+3.2*x14+3*x15+3*x21+2.9*x23+3*x24 +2.8*x25+2.9*x31+3.1*x32+2.8*x33+3.1*x35; x11+x12+x13+x14+x15=3.2;x21+x23+x24+x25=3.9;x31+x32+x33+x34=1.5;x11+x21+x31=1.8;x12+x32=1.3;x13+x23+x33=1.3;x14+x24=1.6;x15+x25+x35=1.5;x11+x21+x31=1.8;x12+x32=1.3;x13+x23+x33=1.3;x14+x24=1.6;x15+x25+x35=1.3;x14+x24=1.6;x15+x25+x35=1.5;x11+x21+x31=1.8;x12+x32=1.3;x13+x23+x33=1.3;x14+x24=1.6;x15+x25+x35=2.4;end附錄四：附錄三的運(yùn)行結(jié)果 Global optimal solution found. Objective value: 2.340000 Infeasibilities: 0.000000 Total solver iterations: 8 Variable Value Reduced Cost X11 0.000000 0.3000000 X12 1.000000 0.000000 X13 0.000000 0.000000 X14 1.800000 0.000000 X15 0.000000 0.1000000 X21 2.200000 0.000000 X23 1.300000 0.000000 X24 0.000000 0.1000000 X25 0.000000 0.2000000 X31 0.000000 0.2000000 X32 1.500000 0.000000 X33 0.000000 0.2000000 X35 2.400000 0.000000 Row Slack or Surplus Dual Price 1 2.340000 -1.000000 2 0.000000 -0.2000000 3 0.000000 -0.3000000 4 0.000000 -0.2000000 5 0.7000000 0.000000 6 0.000000 0.000000 7 0.7000000 0.000000 8 0.2000000 0.000000 9 0.000000 -0.1000000 10 0.4000000 0.000000 11 0.5000000 0.000000 12 0.000000 0.1000000 13 0.8000000 0.000000