實驗一一、實驗目的熟悉掌握VISIO繪圖工具二、實驗要求繪制2-3個復雜的圖形三、實驗內容用VISIO軟件繪制《基礎工業(yè)工程》（易樹平主編）教材P15頁圖1-9圖形和圖p99頁圖4-39圖四、實驗結果實驗結果如下圖：圖1-1

圖1-2

實驗二一、實驗目的學習和掌握運用軟件解決規(guī)劃問題。二、實驗要求要求掌握該軟件的編程方法，用該軟件解決一類復雜題目求出其解。三、實驗內容用LINDO或GLPS軟件解決《運籌學》教材中的多目標規(guī)劃問題（或其它規(guī)劃問題）的解。四、實驗過程及結果【案例】工程建設與財政平衡決策問題1、問題描述某市政府為改善其基礎設施，在近3年內要著手如下5項工程的建設，按重要性排序的工程建設項目名稱及造價如表2－1所示。表2－1工程建設項目名稱及造價表該市政府的財政收入主要來自國家財政撥款、地方稅收和公共事業(yè)收費。3年內該三項總收入分別估計為e1，e2和e3。除此之外就靠向銀行貸款和發(fā)行債券，3年中可貸款的上限為U11、U12和U13，，年利率為g；可發(fā)行債券的上限為U21、U22和U23，年利率為f。銀行還貸款期限為1年(假定貸款在年初付出)，債券則由下年起每年按一定比例(r)歸還部分債主的本金。市政府應如何作出3年的投資決策。要求：（1）給定具體數(shù)據(jù)：b1＝700，b2＝500，b3＝800，b4＝400，b5＝680；e1＝700，e2＝900，e3＝1200，U11＝300，U12＝400，U13＝450，U21＝300，U22＝350，U23＝350，f＝0.055，g＝0.05，r=0.2。用軟件求滿意解；（2）對結果進行分析，列出3年詳細的項目投資計劃、資金分配表和平衡表，資金是否有缺口，寫出分析報告。2、建模分析設x1t((t=1,2,3)為第t年向銀行貸款數(shù)，x2t(t=1,2,3)為第t年發(fā)行債券數(shù)，(i=1,2,…,5;t=1,2,3)為項目i在第t年的完工率（投資比例），見表2－2。表2－2各年貸款、發(fā)行債券及各工程每年完工率表除上述變量外，為了寫出平衡式，引進第1年的起始財政平衡變量z0和每年末的財政平衡變量z1、z2和z3。（1）決策變量：為了列出目標規(guī)劃決策模型，決策變量如表a－2所示。（2）約束和目標：注意問題中有的目標(例如歷年財政平衡)實際上是硬約束，其中不含偏差變量，因此引入松弛變量si(i=1,2,…,7)作等式的平衡。（3）財政平衡約束條件：①變量的上限限制和財政平衡目標：變量包括決策變量、財政平衡變量和保證財政平衡的人工變量。表a－2所列變量都有上界限制的，把這些有上界約束的變量寫成目標形式，對平衡變量應使z0為零，使zl，z2，z3為正值，故除z0外其它平衡變量都引進了正偏差變量，而且把使z0為零和使其它平衡變量為正作“硬約束”的規(guī)定。因此有式中：為正偏差變量，s4+k是松弛變量(等價于負偏差變量)，z0是第1年年初的可用資金，假設z0＝0，則約束z0－s4=0可以去掉。zk是第k年年末剩余（k＋1年年初可用）資金，所有變量非負。②根據(jù)財政平衡的意義，可列出3年中每年的財政平衡約束條件，即(該年銀行貸款)+(該年發(fā)行債券)+(該年財政收入)—(該年各項工程撥款)—(該年銀行還款)—(該年債券還款)—(該年銀行貸款付息)—(該年債券付息)+(起始平衡)—(最終平衡)＝0。則有第一年：第二年：第三年：（4）目標函數(shù)：對問題目標函數(shù)的要求有如下幾點：①硬約束為1級目標，以首先保證各年財政平衡，這可使這些約束條件的相應松弛變量的和為最小；②力圖盡量獲得銀行貸款和發(fā)行債券，以解決工程建設的資金問題；③保證頭兩項工程的優(yōu)先完成(按重點順序加權)；④按重點順序加權，抓緊后三項工程的建設；⑤爭取每個項目在3年內都完工；⑥使各年最終財政平衡變量為最小。因此，目標函數(shù)可列出：

3、程序設計整理得到目標規(guī)劃數(shù)學模型：注：以上所有變量都為非負4、軟件求解將給定具體數(shù)據(jù)：b1＝700，b2＝500，b3＝800，b4＝400，b5＝680；e1＝700，e2＝900，e3＝1200，U11＝300，U12＝400，U13＝450，U21＝300，U22＝350，U23＝350，f＝0.055，g＝0.05，r=0.2，帶入模型中，進行求解。

在Lindo下按照目標規(guī)劃的層次算法求解目標規(guī)劃，P1層次模型為：STx11<=300,x12<=400,x13<=450x21<=300,x22<=350,x23<=350y11+d11_=1,y12+d12_=1,y13+d13_=1y21+d21_=1,y22+d22_=1,y23+d23_=1y31+d31_=1,y32+d32_=1,y33+d33_=1y41+d41_=1,y42+d42_=1,y43+d43_=1y51+d51_=1,y52+d52_=1,y53+d53_=1y11+y12+y13+d1_=1y21+y22+y23+d2_=1y31+y32+y33+d3_=1y41+y42+y43+d4_=1y51+y52+y53+d5_=1z1+s5-d6=0,z2+s6-d7=0z3+s7-d8=0,z0-s4=0700y11+500y21+800y31+400y41+680y51-x11-0.945x21+z1+s1=700700y12+500y22+800y32+400y42+680y52+1.05x11-x12+0.244x21-0.945x22-z1+z2+s2=900700y13+500y23+800y33+400y43+680y53+1.05x12-x13+0.233x21+0.244x22-0.945x23-z2+z3+s3=1200END

輸入lindo求解LPOPTIMUMFOUNDATSTEP7OBJECTIVEFUNCTIONVALUE1)0.0000000E+00VARIABLEVALUEREDUCEDCOSTS10.0000001.000000S20.0000001.000000S30.0000001.000000S40.0000001.000000S50.0000001.000000S60.0000001.000000S70.0000001.000000X110.0000000.000000X120.0000000.000000X130.0000000.000000X210.0000000.000000X220.0000000.000000X230.0000000.000000Y110.0000000.000000D11_1.0000000.000000Y120.1714290.000000D12_0.8285710.000000Y130.8285710.000000D13_0.1714290.000000Y210.0000000.000000D21_1.0000000.000000Y220.0000000.000000D22_1.0000000.000000Y230.0000000.000000D23_1.0000000.000000Y310.8750000.000000D31_0.1250000.000000Y320.1250000.000000D32_0.8750000.000000Y330.0000000.000000D33_1.0000000.000000Y410.0000000.000000D41_1.0000000.000000Y420.0000000.000000D42_1.0000000.000000Y430.0000000.000000D43_1.0000000.000000Y510.0000000.000000D51_1.0000000.000000Y521.0000000.000000D52_0.0000000.000000Y530.0000000.000000D53_1.0000000.000000D1_0.0000000.000000D2_1.0000000.000000D3_0.0000000.000000D4_1.0000000.000000D5_0.0000000.000000Z10.0000000.000000D60.0000000.000000Z20.0000000.000000D70.0000000.000000Z3620.0000000.000000D8620.0000000.000000Z00.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)300.0000000.0000003)400.0000000.0000004)450.0000000.0000005)300.0000000.0000006)350.0000000.0000007)350.0000000.0000008)0.0000000.0000009)0.0000000.00000010)0.0000000.00000011)0.0000000.00000012)0.0000000.00000013)0.0000000.00000014)0.0000000.00000015)0.0000000.00000016)0.0000000.00000017)0.0000000.00000018)0.0000000.00000019)0.0000000.00000020)0.0000000.00000021)0.0000000.00000022)0.0000000.00000023)0.0000000.00000024)0.0000000.00000025)0.0000000.00000026)0.0000000.00000027)0.0000000.00000028)0.0000000.00000029)0.0000000.00000030)0.0000000.00000031)0.0000000.00000032)0.0000000.00000033)0.0000000.00000034)0.0000000.000000NO.ITERATIONS=7RANGESINWHICHTHEBASISISUNCHANGED:OBJCOEFFICIENTRANGESVARIABLECURRENTALLOWABLEALLOWABLECOEFINCREASEDECREASES11.000000INFINITY1.000000S21.000000INFINITY1.000000S31.000000INFINITY1.000000S41.000000INFINITY1.000000S51.000000INFINITY1.000000S61.000000INFINITY1.000000S71.000000INFINITY1.000000X110.000000INFINITY0.000000X120.000000INFINITY0.000000X130.000000INFINITY0.000000X210.000000INFINITY0.000000X220.000000INFINITY0.000000X230.000000INFINITY0.000000Y110.000000INFINITY0.000000D11_0.0000000.000000INFINITYY120.0000000.0000000.000000D12_0.0000000.0000000.000000Y130.0000000.0000000.000000D13_0.0000000.0000000.000000Y210.000000INFINITY0.000000D21_0.0000000.000000INFINITYY220.000000INFINITY0.000000D22_0.0000000.000000INFINITYY230.000000INFINITY0.000000D23_0.0000000.000000INFINITYY310.0000000.0000000.000000D31_0.0000000.0000000.000000Y320.0000000.0000000.000000D32_0.0000000.0000000.000000Y330.000000INFINITY0.000000D33_0.0000000.000000INFINITYY410.000000INFINITY0.000000D41_0.0000000.000000INFINITYY420.000000INFINITY0.000000D42_0.0000000.000000INFINITYY430.000000INFINITY0.000000D43_0.0000000.000000INFINITYY510.000000INFINITY0.000000D51_0.0000000.000000INFINITYY520.0000000.000000INFINITYD52_0.000000INFINITY0.000000Y530.000000INFINITY0.000000D53_0.0000000.000000INFINITYD1_0.000000INFINITY0.000000D2_0.0000000.000000INFINITYD3_0.000000INFINITY0.000000D4_0.0000000.000000INFINITYD5_0.000000INFINITY0.000000Z10.000000INFINITY0.000000D60.000000INFINITY0.000000Z20.000000INFINITY0.000000D70.000000INFINITY0.000000Z30.0000000.0000000.000000D80.0000000.0000000.000000Z00.000000INFINITY1.000000RIGHTHANDSIDERANGESROWCURRENTALLOWABLEALLOWABLERHSINCREASEDECREASE2300.000000INFINITY300.0000003400.000000INFINITY400.0000004450.000000INFINITY450.0000005300.000000INFINITY300.0000006350.000000INFINITY350.0000007350.000000INFINITY350.00000081.000000INFINITY1.00000091.000000INFINITY0.828571101.000000INFINITY0.171429111.000000INFINITY1.000000121.000000INFINITY1.000000131.000000INFINITY1.000000141.000000INFINITY0.125000151.000000INFINITY0.875000161.000000INFINITY1.000000171.000000INFINITY1.000000181.000000INFINITY1.000000191.000000INFINITY1.000000201.000000INFINITY1.000000211.000000INFINITY0.000000221.000000INFINITY1.000000231.0000000.1714290.828571241.000000INFINITY1.000000251.0000000.1500000.125000261.000000INFINITY1.000000271.0000000.0000000.852941280.0000000.000000INFINITY290.0000000.000000INFINITY300.000000620.000000INFINITY310.000000INFINITY0.00000032700.000000100.000000120.00000833900.000000580.000000120.000008341200.000000INFINITY620.000000因為s1+s2+s3+s4+s5+s6+s7=0,在P2層次模型中加入s1+s2+s3+s4+s5+s6+s7=0,得注：由于求解過程太多不便列出，后面只給出最后一步的過程及結果因為d1_+d2_+d3_+d4_+d5_=0，故在p5的層次優(yōu)化模型上加上d1_+d2_+d3_+d4_+d5_=0,得STx11<=300,x12<=400,x13<=450x21<=300,x22<=350,x23<=350y11+d11_=1,y12+d12_=1,y13+d13_=1y21+d21_=1,y22+d22_=1,y23+d23_=1y31+d31_=1,y32+d32_=1,y33+d33_=1y41+d41_=1,y42+d42_=1,y43+d43_=1y51+d51_=1,y52+d52_=1,y53+d53_=1y11+y12+y13+d1_=1y21+y22+y23+d2_=1y31+y32+y33+d3_=1y41+y42+y43+d4_=1y51+y52+y53+d5_=1z1+s5-d6=0,z2+s6-d7=0z3+s7-d8=0,z0-s4=0700y11+500y21+800y31+400y41+680y51-x11-0.945x21+z1+s1=700700y12+500y22+800y32+400y42+680y52+1.05x11-x12+0.244x21-0.945x22-z1+z2+s2=900700y13+500y23+800y33+400y43+680y53+1.05x12-x13+0.233x21+0.244x22-0.945x23-z2+z3+s3=1200s1+s2+s3+s4+s5+s6+s7=02d11_+2d12_+2d13_+d21_+d22_+d23_<63d31_+3d32_+3d33_+2d41_+2d42_+2d43_+d51_+d52_+d53_<12d1_+d2_+d3_+d4_+d5_=0END輸入lindo求解LPOPTIMUMFOUNDATSTEP14OBJECTIVEFUNCTIONVALUE1)0.0000000E+00VARIABLEVALUEREDUCEDCOSTD60.0000000.000000D70.0000000.000000D80.0000000.000000X110.0000000.000000X120.0000000.000000X13280.0000000.000000X210.0000000.000000X220.0000000.000000X230.0000000.000000Y110.0000000.000000D11_1.0000000.000000Y120.0000000.000000D12_1.0000000.000000Y131.0000000.000000D13_0.0000000.000000Y210.5600000.000000D21_0.4400000.000000Y220.4400000.000000D22_0.5600000.000000Y230.0000000.000000D23_1.0000000.000000Y310.5250000.000000D31_0.4750000.000000Y320.0000000.000000D32_1.0000000.000000Y330.4750000.000000D33_0.5250000.000000Y410.0000000.000000D41_1.0000000.000000Y420.0000000.000000D42_1.0000000.000000Y431.0000000.000000D43_0.0000000.000000Y510.0000000.000000D51_1.0000000.000000Y521.0000000.000000D52_0.0000000.000000Y530.0000000.000000D53_1.0000000.000000D1_0.0000000.000000D2_0.0000000.000000D3_0.0000000.000000D4_0.0000000.000000D5_0.0000000.000000Z10.0000001.000000S50.0000001.000000Z20.0000001.000000S60.0000001.000000Z30.0000001.000000S70.0000001.000000Z00.0000000.000000S40.0000000.000000S10.0000000.000000S20.0000000.000000S30.0000000.000000ROWSLACKORSURPLUSDUALPRICES2)300.0000000.0000003)400.0000000.0000004)170.0000000.0000005)300.0000000.0000006)350.0000000.0000007)350.0000000.0000008)0.0000000.0000009)0.0000000.00000010)0.0000000.00000011)0.0000000.00000012)0.0000000.00000013)0.0000000.00000014)0.0000000.00000015)0.0000000.00000016)0.0000000.00000017)0.0000000.00000018)0.0000000.00000019)0.0000000.00000020)0.0000000.00000021)0.0000000.00000022)0.0000000.00000023)0.0000000.00000024)0.0000000.00000025)0.0000000.00000026)0.0000000.00000027)0.0000000.00000028)0.0000001.00000029)0.0000001.00000030)0.0000001.00000031)0.0000000.00000032)0.0000000.00000033)0.0000000.00000034)0.0000000.00000035)0.0000000.00000036)0.0000000.00000037)0.0000000.00000038)0.0000000.000000NO.ITERATIONS=14

RANGESINWHICHTHEBASISISUNCHANGED:OBJCOEFFICIENTRANGESVARIABLECURRENTALLOWABLEALLOWABLECOEFINCREASEDECREASED61.000000INFINITY1.000000D71.000000INFINITY1.000000D81.000000INFINITY1.000000X110.000000INFINITY0.000000X120.000000INFINITY0.000000X130.0000000.0000000.000000X210.000000INFINITY0.000000X220.000000INFINITY0.000000X230.000000INFINITY0.000000Y110.000000INFINITY0.000000D11_0.0000000.000000INFINITYY120.0000000.0000000.000000D12_0.0000000.0000000.000000Y130.0000000.000000INFINITYD13_0.000000INFINITY0.000000Y210.0000000.0000000.000000D21_0.0000000.0000000.000000Y220.0000000.0000000.000000D22_0.0000000.0000000.000000Y230.000000INFINITY0.000000D23_0.0000000.000000INFINITYY310.0000000.0000000.000000D31_0.0000000.0000000.000000Y320.000000INFINITY0.000000D32_0.0000000.000000INFINITYY330.0000000.0000000.000000D33_0.0000000.0000000.000000Y410.000000INFINITY0.000000D41_0.0000000.000000INFINITYY420.000000INFINITY0.000000D42_0.0000000.000000INFINITYY430.0000000.000000INFINITYD43_0.000000INFINITY0.000000Y510.0000000.0000000.000000D51_0.0000000.0000000.000000Y520.0000000.000000INFINITYD52_0.000000INFINITY0.000000Y530.000000INFINITY0.000000D53_0.0000000.000000INFINITYD1_0.000000INFINITY0.000000D2_0.000000INFINITY0.000000D3_0.000000INFINITY0.000000D4_0.000000INFINITY0.000000D5_0.0000000.000000INFINITYZ10.000000INFINITY1.000000S50.000000INFINITY1.000000Z20.000000INFINITY1.000000S60.000000INFINITY1.000000Z30.000000INFINITY1.000000S70.000000INFINITY1.000000Z00.0000000.000000INFINITYS40.0000000.000000INFINITYS10.000000INFINITY0.000000S20.000000INFINITY0.000000S30.000000INFINITY0.000000RIGHTHANDSIDERANGESROWCURRENTALLOWABLEALLOWABLERHSINCREASEDECREASE2300.000000INFINITY300.0000003400.000000INFINITY400.0000004450.000000INFINITY170.0000005300.000000INFINITY300.0000006350.000000INFINITY350.0000007350.000000INFINITY350.00000081.0000000.0000000.00000091.0000000.0000000.000000101.0000000.0000000.000000111.0000000.0000000.000000121.0000000.0000000.000000131.0000000.0000000.000000141.0000000.0000000.475000151.0000000.0000001.000000161.0000000.0000000.525000171.0000000.0000001.000000181.0000000.0000001.000000191.0000000.0000000.000000201.0000000.0000001.000000211.0000000.0000000.411765221.0000000.0000001.000000231.0000000.0000000.000000241.0000000.0000000.000000251.0000000.2125000.000000261.0000000.0000000.000000271.0000000.2500000.000000280.0000000.000000INFINITY290.0000000.000000INFINITY300.0000000.000000INFINITY310.000000INFINITY0.00000032700.000000280.000000170.00000033900.000000280.000000170.000000341200.000000280.000000170.000000350.000000INFINITY0.000000366.0000000.0000000.0000003712.000000INFINITY0.000000380.0000000.0000000.000000至此求解完成

五、解的分析由軟件求出如下表的解：銀行第三年貸款280萬元。各項目每年完工率如下表2-3。表2-3各項目每年完工率項目投資計劃表如表2-4表a-4項目投資計劃表

資金平衡表如表2-5表a-5資金平衡表分析報告：工程需要資金總額3080萬元，三年可用資金2800萬元，銀行貸款280萬元，資金缺口3080－(2800+280)+280×0.05＝14(萬元)。按照上述分配，第一年和第二年不貸款和發(fā)行債券，財政收入完全用于工程建設，第三年貸款，貸款金額加上財政收全部用于工程建設，第三年完成全部工程。

實驗三一、實驗目的學習掌握計算機仿真的基本原理，對所給出的問題建模，寫出其算法，并得出解二、實驗要求要求對所給出的問題建模，寫出其算法，并得出解。三、實驗內容用系統(tǒng)建模與仿真的方法編寫中國期刊網(wǎng)中的論文“基于故障樹分析法的供應鏈可靠性診斷與仿真研究”一文中的算法，并求其解。四、實驗過程及結果本實驗中，給出的案例包含三個供應商、一個制造商以及兩個分銷商。1、首先，我們按照實驗要求分析案例求解的算法，在分析算法前，理應理清三者的關系，做出他們之間的供應鏈結構如下圖3-1：圖3-12、做出該供應鏈的失效診斷模型如下圖3-2：

圖3-23、由以上邏輯關系分析算法:假設N表示系統(tǒng)的仿真次數(shù)，Tgij(i=1,2,3,j=1,2,3,4,5,6,7,8)表示由第i個供應商發(fā)生第j類失效原因的次數(shù)，Tzk(k=6,7)表示制造商發(fā)生的第k類失效原因的次數(shù)，Tfst(s=1,2,t=1,2,3,4,5,6)表示由第s個分銷商發(fā)生第t類失效原因的次數(shù)，T表示系統(tǒng)總失效次數(shù)。算法步驟如下：（1）運用蒙特卡羅方法，產(chǎn)生系列隨機數(shù)Rij(i=1,2,3,j=1,2,3,4,5,6,7,8),以判斷供應商中各底事件的發(fā)生狀態(tài)，而判斷供應商是否發(fā)生失效。若失效，則記下是由哪一類或幾類失效因素引起，并記Tgij=Tgij+1(i=1,2,3,j=1,2,3,4,5,6,7,8)，T=T+1，進行下一次仿真；若不失效，則轉下一步。（2）再產(chǎn)生系列隨機數(shù)Rk(k=6,7)，以判斷制造商中各底事件的發(fā)生狀態(tài),從而判斷制造商是否發(fā)生失效。若失效，則記下是由哪一類或幾類失效因素引起，并且Tzk=Tzk+1(k=6,7)，T=T+1，進行下一次仿真；若不失效則轉下一步。（3）再產(chǎn)生系列隨機數(shù)Rst(s=1,2,t=1,2,3,4,5,6)，以判斷分銷商中各底事件的發(fā)生狀態(tài)，從而判斷分銷商是否發(fā)生失效。若失效，則記下是由哪一類或幾類失效因素引起，并且Tfst=Tfst+1，T=T+1(s=1,2,t=1,2,3,4,5,6)，進行下一次仿真；若不失效，則表明該次供應鏈運行正常，仿真轉入下一次供應鏈運行。（4）重復以上過程，直至仿真結束，然后計算相關可靠性指標和各底事件重要度。整個仿真過程如下圖3-3所示：

圖3-3下面給出案例中個事件的概率度：

由以上數(shù)據(jù)及其分析進行編程，并在C++中運行：#include"stdio.h"#include"time.h"#include"math.h"#defineNmax50000floatran(){longi;floatr;i=rand();r=i/32767.0;returnr;main(){longTc1=0,Tz1=0,Tz2=0,Tg11=0,Tg13=0,Tg14=0,Tg16=0,Tg24=0,Tg26=0,Tg28=0,Tg31=0,Tg33=0,Tg34=0,Tf11=0,Tf12=0,Tf15=0,Tf16=0,Tf21=0,Tf25=0,Tf26=0,T=0,c=0,z=0,g=0,g1=0,g2=0,g3=0,f=0,f1=0,f2=0,N=0,TN=0;floatc1=0,z1=0,z2=0,g11=0,g13=0,g14=0,g16=0,g24=0,g26=0,g28=0,g31=0,g33=0,g34=0,f11=0,f12=0,f15=0,f16=0,f21=0,f25=0,f26=0,GZD;srand((unsigned)time(NULL));for(N=1;N<=Nmax;N++){c1=ran();if(c1>=0.03&&c1<=1)c1=0;elsec1=1;if(c1==0){g11=ran();g13=ran();g14=ran();g16=ran();if(g11>=0.39&&g11<=1)g11=0;elseg11=1;if(g13>=0.08&&g13<=1)g13=0;elseg13=1;if(g14>=0.12&&g14<=1)g14=0;elseg14=1;if(g16>=0.15&&g16<=1)g16=0;elseg16=1;if((g11+g13+g14+g16)>=1)g1=1;elseg1=0;g24=ran();g26=ran();g28=ran();if(g24>=0.04&&g24<=1)g24=0;elseg24=1;if(g26>=0.18&&g26<=1)g26=0;elseg26=1;if(g28>=0.10&&g28<=1)g28=0;elseg28=1;if((g24+g26+g28)>=1)g2=1;elseg2=0;g31=ran();g33=ran();g34=ran();if(g31>=0.08&&g31<=1)g31=0;elseg31=1;if(g33>=0.17&&g31<=1)g33=0;elseg33=1;if(g34>=0.060&&g34<=1)g34=0;elseg34=1;if((g31+g33+g34)>=1)g3=1;elseg3=0;if((g1+g2+g3)>=3){g=1;T=T+1;Tg11=Tg11+g11;Tg13=Tg13+g13;Tg14=Tg14+g14;Tg16=Tg16+g16;Tg24=Tg24+g24;Tg26=Tg26+g26;Tg28=Tg28+g28;Tg31=Tg31+g31;Tg33=Tg33+g33;Tg34=Tg34+g34;continue;elseg=0;if(c1==1){g31=ran();g33=ran();g34=ran();if(g31>=0.08&&g31<=1)g31=0;elseg31=1;if(g33>=0.17&&g31<=1)g33=0;elseg33=1;if(g34>=0.060&&g34<=1)g34=0;elseg34=1;if((g31+g33+g34)>=1){g=1;T=T+1;Tg31=Tg31+g31;Tg33=Tg33+g33;Tg34=Tg34+g34;Tc1=Tc1+1;continue;elseg=0;z1=ran();z2=ran();if(z1<=0.020){z1=1;T=T+1;Tz1=Tz1+z1;continue;}elsez1=0;if(z2<=0.0380){z2=1;T=T+1;Tz2=Tz2+z2;continue;}elsez2=0;f11=ran();f12=ran();f15=ran();f16=ran();if(f11>=0.034&&f11<=1)f11=0;elsef11=1;if(f12>=0.14&&f12<=1)f12=0;elsef12=1;if(f15>=0.012&&f15<=1)f15=0;elsef15=1;if(f16>=0.010&&f16<=1)f16=0;elsef16=1;if((f11+f