運籌學(xué)實驗報告-lingo軟件的使用-習(xí)題代碼

1、運籌學(xué)實驗報告姓名：學(xué)號：相關(guān)問題說明：實驗性質(zhì)和教學(xué)目的本實驗是運籌學(xué)課內(nèi)安排的上機操作實驗。目的在于了解、熟悉計算機Lingo 軟件在運籌學(xué)模型求解中的作用，激發(fā)學(xué)習(xí)興趣，提高學(xué)習(xí)效果，增強自身的動手能力，提高實際應(yīng)用能力。實驗基本要求要求學(xué)生：1. 實驗前認真做好理論準備，仔細閱讀實驗指導(dǎo)書；2. 遵從教師指導(dǎo)，認真完成實驗任務(wù)，按時按質(zhì)提交實驗報告。主要參考資料1 LINGO 軟件2 LINGO8.0 及其在環(huán)境系統(tǒng)優(yōu)化中的應(yīng)用，天津大學(xué)出版社，20053.優(yōu)化建模與LINDO/LINGO軟件，清華大學(xué)出版社，20054運籌學(xué)編寫組主編，運籌學(xué)（修訂版），清華大學(xué)出版社，19905藍伯

2、雄主編，管理數(shù)學(xué)（下）運籌學(xué)，清華大學(xué)出版社，19976胡運權(quán)主編，運籌學(xué)習(xí)題集（修訂版），清華大學(xué)出版社，19957胡運權(quán)主編，運籌學(xué)教程（第二版），清華大學(xué)出版社，2003實驗內(nèi)容1、線性規(guī)劃問題：maxz 8x1 6x29x1 8x2127x1 11x2 24s.t. 129x1 11x2 13x1, x20（1）給出原始代碼；（2）計算結(jié)果（包括靈敏度分析，求解結(jié)果粘貼）；（3）回答下列問題（手寫）：a）最優(yōu)解及最優(yōu)目標函數(shù)值是多少；b）資源的對偶價格各為多少，并說明對偶價格的含義；c）為了使目標函數(shù)值增加最多，讓你選擇一個約束條件，將它的常數(shù)項增加一個單 位，你將選擇哪一個約束條件？

3、這時目標函數(shù)值將是多少？d）對x2的目標函數(shù)系數(shù)進行靈敏度分析；e）對第2個約束的約束右端項進行靈敏度分析；f ）結(jié)合本題的結(jié)果解釋“ Reduced Cost”的含義。對偶價格就是說約束方程右端變量增加 1對目標函數(shù)值的影響答案：（1）代碼max =8*x1+6*x2;9*x1+8*x2<=12;7*x1+11*x2<=24;9*x1+11*x2<=13;x1>=0;x2>=0;(2)計算結(jié)果Global optimal solution found.Objective value:Total solver iterations:10.666672Variabl

4、eValueReduced CostX11.3333330.000000X20.0000001.111111RowSlack or Surplus Dual Price110.666671.00000020.0000000.8888889314.666670.00000041.0000000.00000051.3333330.0000000.0000000.000000Ranges in which the basis is unchanged:Objective Coefficient RangesCurrentAllowableAllowableVariableCoefficientInc

5、reaseDecreaseX18.000000INFINITY1.250000X26.0000001.111111INFINITYRighthand Side RangesRowCurrentAllowableAllowableRHSIncreaseDecrease212.000001.00000012.00000324.00000INFINITY14.66667413.00000INFINITY1.00000050.01.333333INFINITY60.00.0INFINITY(3)a)b)c)d)e)f)2、運輸問題:已知6個發(fā)點8個收點的最小費用運輸問題。產(chǎn)銷量及單位運價如下表。銷地“

6、-a、 產(chǎn)地BBB3B4B5RBB8產(chǎn) 量A629742r 51955 A24553853247A5213748342A767992P 71152 A52365726541A5922814332銷量6055514341524338給出原始代碼；（2）計算結(jié)果（決策變量求解結(jié)果粘貼）68Cij XijMin Z =6i 1Xij <=bj(j=1.8 ) 銷量約束8Xij = ai ( i=1.6 ) 產(chǎn)量約束j1Xij > 0 (i=1.6 ; j=18)代碼：model:!6 發(fā)點8 model :!6 發(fā)點8收點運輸問題;sets :warehouses/wh1.wh6/: c

7、apacity;vendors/v1.v8/: demand;links(warehouses,vendors): cost, volume;endsetsmin=sum(links: cost*volume); ! 目標函數(shù);for(vendors(J):sum(warehouses(I): volume(I,J)<=demand(J);! 需求約束;for(warehouses(I):sum(vendors(J): volume(I,J)=capacity(I);! 產(chǎn)量約束;! 這里是數(shù)據(jù);data :capacity=55 47 42 52 41 32;demand=60 55

8、51 43 41 52 43 38;cost=6 2 9 7 4 2 5 94 5 5 3 8 5 3 25 2 1 3 7 4 8 37 6 7 9 9 2 7 12 3 6 5 7 2 6 55 9 2 2 8 1 4 3;enddataend答案Global optimal solution found.Objective value:473.0000Infeasibilities:0.000000Total solver iterations:9Model Class:LPTotal variables:48Nonlinear variables:Integer variables:0

9、015Total constraints:Nonlinear constraints:Total nonzeros:144Nonlinear nonzeros:0VariableValue Reduced CostCAPACITY( WH1)55.000000.000000CAPACITY( WH2)47.000000.000000CAPACITY( WH3)42.000000.000000CAPACITY( WH4)52.000000.000000CAPACITY( WH5)41.000000.000000CAPACITY( WH6)32.000000.000000DEMAND( V1)60

10、.000000.000000DEMAND( V2)55.000000.000000DEMAND( V3)51.000000.000000DEMAND( V4)43.000000.000000DEMAND( V5)41.000000.000000DEMAND( V6)52.000000.000000DEMAND( V7)43.000000.000000DEMAND( V8)38.000000.000000COST( WH1, V1)6.0000000.000000COST( WH1, V2)2.0000000.000000COST( WH1, V3)9.0000000.000000COST( W

11、H1, V4)7.0000000.000000COST( WH1, V5)4.0000000.000000COST( WH1, V6)2.0000000.000000COST( WH1, V7)5.0000000.000000COST( WH1, V8)9.0000000.000000COST( WH2, V1)4.0000000.000000COST( WH2, V2)5.0000000.000000COST( WH2, V3)5.0000000.000000COST( WH2, V4)3.0000000.000000COST( WH2, V5)8.0000000.000000COST( W

12、H2, V6)5.0000000.000000COST( WH2, V7)3.0000000.000000COST( WH2, V8)2.0000000.000000COST( WH3, V1)5.0000000.000000COST( WH3, V2)2.0000000.000000COST( WH3, V3)1.0000000.000000COST( WH3, V4)3.0000000.000000COST( WH3, V5)7.0000000.000000COST( WH3, V6)4.0000000.000000COST( WH3, V7)8.0000000.000000COST( W

13、H3, V8)3.0000000.000000COST( WH4, V1)7.0000000.000000COST( WH4, V2)6.0000000.000000COST( WH4, V3)7.0000000.000000COST( WH4, V4)9.0000000.000000COST( WH4, V5)9.0000000.000000COST( WH4, V6)2.0000000.000000COST( WH4, V7)7.0000000.000000COST( WH4, V8)1.0000000.000000COST( WH5, V1)2.0000000.000000COST( W

14、H5, V2)3.0000000.000000COST( WH5, V3)6.0000000.000000COST( WH5, V4)5.0000000.000000COST( WH5, V5)7.0000000.000000COST( WH5, V6)2.0000000.000000COST( WH5, V7)6.0000000.000000COST( WH5, V8)5.0000000.000000COST( WH6, V1)5.0000000.000000COST( WH6, V2)9.0000000.000000COST( WH6, V3)2.0000000.000000COST( W

15、H6, V4)2.0000000.000000COST( WH6, V5)8.0000000.000000COST( WH6, V6)1.0000000.000000COST( WH6, V7)4.0000000.000000COST( WH6, V8)3.0000000.000000VOLUME( WH1, V1)0.0000004.000000VOLUME( WH1, V2)55.000000.000000VOLUME( WH1, V3)0.0000007.000000VOLUME( WH1, V4)0.0000005.000000VOLUME( WH1, V5)0.0000002.000

16、000VOLUME( WH1, V6)0.0000000.000000VOLUME( WH1, V7)0.0000003.000000VOLUME( WH1, V8)0.0000008.000000VOLUME( WH2, V1)0.0000001.000000VOLUME( WH2, V2)0.0000002.000000VOLUME( WH2, V3)0.0000002.000000VOLUME( WH2, V4)43.000000.000000VOLUME( WH2, V5)0.0000005.000000VOLUME( WH2, V6)0.0000002.000000VOLUME( W

17、H2, V7)4.0000000.000000VOLUME( WH2, V8)0.0000000.000000VOLUME( WH3, V1)0.0000004.000000VOLUME( WH3, V2)0.0000001.000000VOLUME( WH3, V3)42.000000.000000VOLUME( WH3, V4)0.0000002.000000VOLUME( WH3, V5)0.0000006.000000VOLUME( WH3, V6)0.0000003.000000VOLUME( WH3, V7)0.0000007.000000VOLUME( WH3, V8)0.000

18、0003.000000VOLUME( WH4, V1)0.0000005.000000VOLUME( WH4, V2)0.0000004.000000VOLUME( WH4, V3)0.0000005.000000VOLUME( WH4, V4)0.0000007.000000VOLUME( WH4, V5)0.0000007.000000VOLUME( WH4, V6)14.000000.000000VOLUME( WH4, V7)0.0000005.000000VOLUME( WH4, V8)38.000000.000000VOLUME( WH5, V1)41.000000.000000V

19、OLUME( WH5, V2)0.0000001.000000VOLUME( WH5, V3)0.0000004.000000VOLUME( WH5, V4)0.0000003.000000VOLUME( WH5, V5)0.0000005.000000VOLUME( WH5, V6)0.0000000.000000VOLUME( WH5, V7)0.0000004.000000VOLUME( WH5, V8)0.0000004.000000VOLUME( WH6, V1)0.0000004.000000VOLUME( WH6, V2)0.0000008.000000VOLUME( WH6,

20、V3)0.0000001.000000VOLUME( WH6, V4)0.0000001.000000VOLUME( WH6, V5)0.0000007.000000VOLUME( WH6, V6)32.000000.000000VOLUME( WH6, V7)0.0000003.000000VOLUME( WH6, V8)0.0000003.000000RowSlack or SurplusDual Price1473.0000-1.000000219.000000.00000030.0000000.00000049.0000000.00000050.0000000.000000641.00

21、0000.00000076.0000000.000000839.000000.00000090.0000001.000000100.000000-2.000000110.000000-3.000000120.000000-1.000000130.000000-2.000000140.000000-2.000000150.000000-1.0000003、一般整數(shù)規(guī)劃問題：某服務(wù)部門各時段（每 2h為一時段）需要的服務(wù)員人數(shù)見下表。按規(guī)定，服務(wù) 員連續(xù)工作8h （即四個時段）為一班。現(xiàn)要求安排服務(wù)員的工作時間，使服務(wù)部門服 務(wù)員總數(shù)最少。時段12345678服務(wù)員最少數(shù)目10891113853(

22、1)給出原始代碼；(2)計算結(jié)果(決策變量求解結(jié)果粘貼)model:sets :time/x1.x8/: required,start;endsetsdata :!每天所需的最少職員數(shù)；required = 10 8 9 11 13 8 5 3;enddata!最小化每周所需職員數(shù)；min=sumtime: start);for(time (J):sumtime(I) | I #le# 4:start( wrap(J+I+2,8) >= required(J);end結(jié)果Global optimal solution found.Objective value:23.00000Total

23、 solver iterations:3Variable Value Reduced CostREQUIRED( X1)10.000000.000000REQUIRED( X2)8.0000000.000000REQUIRED( X3)9.0000000.000000REQUIRED( X4)11.000000.000000REQUIRED( X5)13.000000.000000REQUIRED( X6)8.0000000.000000REQUIRED( X7)5.0000000.000000REQUIRED( X8)3.0000000.000000START( X1)13.000000.0

24、00000START( X2)0.0000000.000000START( X3)0.0000000.000000START( X4)2.0000000.000000START( X5)8.000000START( X6)0.000000START( X7)0.000000START( X8)0.0000000.0000000.0000000.0000000.000000RowSlack or SurplusDual Price123.00000-1.00000020.000000-1.00000030.0000000.00000044.0000000.00000052.0000000.000

25、00060.000000-1.00000077.0000000.00000085.0000000.00000097.0000000.0000004、指派問題:已知如下效率矩陣，求極大化指派問題BB2RBB5A4871512A79171410A691287A6714610A6912106(1)給出原始代碼；(2)計算結(jié)果(決策變量求解結(jié)果粘貼)model:!5個工人，5個工作的分配問題；sets :workers/w1.w5/;jobs/j1.j5/;links(workers,jobs): cost,volume;endsets!目標函數(shù)；min=surminks: cost*volume);

26、!每個工人只能有一份工作；for(workers(I):sumjobs(J): volume(I,J)=1;);!每份工作只能有一個工人；for(jobs(J):sumworkers(I): volume(I,J)=1;);data :cost= 4 8 7 15 127 9 17 14 106 9 12 8 76 7 14 6 106 9 12 10 6 enddata endGlobal optimal solution found.Objective value:34.00000Total solver iterations:10VariableValue Reduced CostCOS

27、T( W1, J1)4.0000000.000000COST( W1, J2)8.0000000.000000COST( W1, J3)7.0000000.000000COST( W1, J4)15.000000.000000COST( W1, J5)12.000000.000000COST( W2, J1)7.0000000.000000COST( W2, J2)9.0000000.000000COST( W2, J3)17.000000.000000COST( W2, J4)14.000000.000000COST( W2, J5)10.000000.000000COST( W3, J1)

28、6.0000000.000000COST( W3, J2)9.0000000.000000COST( W3, J3)12.000000.000000COST( W3, J4)8.0000000.000000COST( W3, J5)7.0000000.000000COST( W4, J1)6.0000000.000000COST( W4, J2)7.0000000.000000COST( W4, J3)14.000000.000000COST( W4, J4)6.0000000.000000COST( W4, J5)10.000000.000000COST( W5, J1)6.0000000.

29、000000COST( W5, J2)9.0000000.000000COST( W5, J3)12.000000.000000COST( W5, J4)10.000000.000000COST( W5, J5)6.0000000.000000VOLUME( W1, J1)0.0000003.000000VOLUME( W1, J2)0.0000005.000000VOLUME( W1, J3)1.0000000.000000VOLUME( W1, J4)0.00000013.00000VOLUME( W1, J5)0.00000011.00000VOLUME( W2, J1)0.000000

30、0.000000VOLUME( W2, J2)1.0000000.000000VOLUME( W2, J3)0.0000004.000000VOLUME( W2, J4)0.0000006.000000VOLUME( W2, J5)0.0000003.000000VOLUME( W3, J1)1.0000000.000000VOLUME( W3, J2)0.0000001.000000VOLUME( W3, J3)0.0000000.000000VOLUME( W3, J4)0.0000001.000000VOLUME( W3, J5)0.0000001.000000VOLUME( W4, J

31、1)0.0000001.000000VOLUME( W4, J2)0.0000000.000000VOLUME( W4, J3)0.0000003.000000VOLUME( W4, J4)1.0000000.000000VOLUME( W4, J5)0.0000005.000000VOLUME( W5, J1)0.0000000.000000VOLUME( W5, J2)0.0000001.000000VOLUME( W5, J3)0.0000000.000000VOLUME( W5, J4)0.0000003.000000VOLUME( W5, J5)1.0000000.000000Row

32、 Slack or Surplus Dual Price134.00000-1.00000020.000000-1.00000030.000000-7.00000040.000000-6.00000050.000000-5.00000060.000000-6.00000070.0000000.00000080.000000-2.00000090.000000-6.000000100.000000-1.000000110.0000000.0000005、一維資源分配問題：某工業(yè)部門根據(jù)國家計劃的安排，擬將某種高效率的設(shè)備五臺，分配給所屬的甲、 乙、丙三個工廠，各工廠若獲得這種設(shè)備之后，可以為國家

33、提供的盈利如下表所示。問：這五臺設(shè)備如何分配給各工廠，才能使國家得到的盈利最大。二:工廠 設(shè)備臺數(shù)甲a乙b丙c00 n0015542151526340 140404806045_ _5907050(1)給出原始代碼；(2)計算結(jié)果(決策變量求解結(jié)果粘貼)代碼sets :R/1.6/:z;L/1.3/;c(R,L):x,y;endsetsdata :X=0 0 05 5 415 15 2640 40 4080 60 4590 70 50;z=0 1 2 3 4 5;enddatamax=sur(c(i,j):X(i,j)*y(i,j);for(l(i):sumc(j,k)|k#eq# 1:y(j

34、,k)=1);sumc(i,j):y(i,j)*z(i)=5;for(c(i,j):B in(y(i,j);end答案Global optimal solution found.Objective value:90.00000Extended solver steps:0Total solver iterations:0VariableValueReduced CostZ( 1)0.0000000.000000Z( 2)1.0000000.000000Z( 3)2.0000000.000000Z( 4)3.0000000.000000Z( 5)4.0000000.000000Z( 6)5.00

35、00000.000000X(1, 1)0.0000000.000000X( 1,2)0.0000000.000000X( 1,3)0.0000000.000000X( 2,1)5.0000000.000000X( 2, 2)5.0000000.000000X( 2, 3)4.0000000.000000X( 3, 1)15.000000.000000X( 3, 2)15.000000.000000X( 3, 3)26.000000.000000X( 4,1)40.000000.000000X( 4, 2)40.000000.000000X( 4, 3)40.000000.000000X( 5,

36、1)80.000000.000000X( 5, 2)60.000000.000000X( 5, 3)45.000000.000000X( 6,1)90.000000.000000X( 6, 2)70.000000.000000X( 6, 3)50.000000.000000Y( 1,1)0.0000000.000000Y( 1,2)0.0000000.000000Y( 1,3)0.0000000.000000Y( 2, 1)0.000000-5.000000Y( 2, 2)0.000000-5.000000Y( 2, 3)0.000000-4.000000Y( 3, 1)0.000000-15

37、.00000Y( 3, 2)0.000000-15.00000Y( 3, 3)0.000000-26.00000Y( 4,1)0.000000-40.00000Y( 4, 2)0.000000-40.00000Y( 4, 3)0.000000-40.00000Y( 5, 1)0.000000-80.00000Y( 5, 2)0.000000-60.00000Y( 5, 3)0.000000-45.00000Y( 6,1)1.000000-90.00000Y( 6, 2)0.000000-70.00000Y( 6, 3)0.000000-50.00000RowSlack or SurplusDu

38、al Price190.000001.00000020.0000000.00000030.0000000.00000040.0000000.00000050.0000000.0000006、最短路問題：求從V1至V。的最短路。l 1,2=6,l 1,3 =5,l 2,4 =3,l 2,5=6,l 2,6=9,l 3,4 =7,l 3,5 =5,l 3,6 =11,l 4,7=9, l 4,8=1,l 5,7=8,l 5,8 =7,l 5,9=5,l 6,8 =4,l 6,9=10, l 7,10 =5, l 8,10=7, l 9,10=9(l i,j表示V到V之間的權(quán)重)(1) 給出原始代碼；(2) 計算結(jié)果 ( 決策變量求解結(jié)果粘貼)! 最短路問題;model:data : n=10;enddata sets :cities/1.n/: F;!10 個城市 ;roads(cities,cities)/ 1,2 1,3 2,4 2,5 2,6 3,4 3,5 3,6 4,7 4,8 5,7 5,8 5,9 6,8 6,9 7,10 8,10 9,10/: D, P; endsets data :D= 6 5 3 6 9 7 5 11 9 1 8 7 5 4 10 5 7 9;enddataF(n)=0;for(cities(i) |