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摘Lingo、等軟件進行編程,得出了滿足條件的汽車最優租賃調度方案。Lingo29次的迭代計算,得到了未來四周每天的29次的迭代計算,得到最佳調度方案以及最低的總轉運費和短缺損失費。規劃的最大利潤模型,Lingo29次的迭代計算,得到未來四周的根據問題三的模型,利用上半年每個點的日平均需求量和各點原始車輛數進行lingo軟件調出其價格和靈敏度表,得出840輛的購車方案。 汽車租賃行業早在70余年前就已經在迅速發展,時至今日,已經培養出、等行業巨頭。其中旗下用于汽車租賃業務的汽車已達一百五十萬以上,在全150大型國企等企業用戶。真正大規模投入個人業務的也是近幾情。隨著中國汽車產37920個1lingo程序(2lingo程序(3lingo程序(4lingo程序(問題一:在盡量滿足需求的條件下,建立合理的數學模型,設計出未來四問題二:考慮由于汽車數量不足而帶來的經濟損失,設計出未來四每天的問題四:在假定只一種車型的前提下,為了使年度總獲利最大,判斷是否 首先對給定坐標的 點作圖如CGJEHTQLDRISOPNFMBK 0圖1點位置分布LINGO編程求2針對未來四每個點的汽車需求量與各點汽車數量的初始值,假定各代LINGO進行編程求解。求相等的前提下,建立線性規劃模型,運用LINGO求解,計算出未來四公司的調模型,利用上半年的每個點的日平均需求量和各點原始車輛數進行利潤最大化的最優調運方案求解。并利用其結果進行價格分析,判斷是否購車。若需要購車,§3假定每個點即為車輛調入點,也為車輛調出點為了方便模型的設立,把點A,B,C…T轉化為點1,2,3,...200§4與符號說歐氏距離(euclideanmetric)是一個通常采用的距離定義,指在m中兩固定資產折舊指在固定資產使用內,按照確定的方法對應計折舊額進行1調運前第i個點第k日擁有的車輛2未來四第i個點第k日的汽車需求3第k日第i個點向第j個點轉運的車4從第i個點向第j個點調運的單位距離所需要的費5dij第i個點與第j個點之間的歐式距6從第i個點向第j個點調運需要的總費7第i個點一輛車一天的短缺損失8第i個點一輛車一天的租賃收9第i個點第k日的租賃總收D第i個點年平均需求§5①對于給定 點坐標,對其歐式距離進行求解2假設第i 點與第j個點的坐標分別為(xi,yi)(xj,yj),那2

)2(

yj

i②對一輛車從第i個點調運到第j 點的總運費cij求解ic1.2(xx)2(yy)2

③對于②中,當i

j時,即同 點向自身調運,顯然cij0設xij表示點i(供應地)向點j(需求地)調運的汽車數量,其i1,2,3,.....20;j Eik20i1j

xijk xijkxijk

iii

kkk其中:Eik——調運前第i個點第k日擁有的車輛 Dik——未來四第i個點第k日的汽車

xijk——第k日第i個點向第j個點轉cij——從第i個點向第j個點調運需要

①根據模型對未表1由表1表2741138943247431815451371362515255261337534625721 184255846541314112175412386226248239241852141134334141618641325511915473648114413352458598642321635581215891232724471787582136112267742464 32275623411511315361259176513315682241141223433368551374632921112111211412313747353251425945531411322212123253126851177237755514116872327212221365756914426241159183174322134第、、、日調運涉及到得點最多,車輛也較多。以第29天為例,調運方案有閑置車輛,一個點空缺車輛的情況,此時的短缺費用為0;當供給小于需求時,LINGO進行①假定在轉運費大于短缺費時,為了確保顧客需求,仍然進行調運。不出現一個②通過表格數據的計算得知,當k時,每天各點總需求量大于總供給量,當k6,7,12,13,14時,每天各點的總供①當 點的總需求大于總供給20 minZ(Dikxjik)vi

i1

j xijkj xjikj

ii

kk

i

k其中:Eik——調運前第i個點第k日擁有的車輛Dik——未來四第i個點第k日的汽車

xijk——第k日第i個點向第j個點轉運cij——從第i個點向第j個點調運需要的vi——第i個點一輛車一天的短缺損失費②當 點的總供給大于總需求

20

i1j

xijk xjik

i ki1,2,3...20k

i1,2,3...20k其中:Eik——調運前第i個點第k日擁有的Dik——未來四第i個點第k日的

xijk——第k日第i個點向第j個cij——從第i個點向第j個點調表3代表從A點運送到B點,后面的數字表示轉運車輛的數量。表443441536243156571142311917612651446712114321347429514171145112521895943826253271121628138417613813311542481117211254116142984118156184151449374823166345325532316855129531124272847178758213611226774246413751212412341151131136155912651731368524214122134563385513746329211121112114123135473132594255451314213222131252321851777177135511745231627182157129236136182553771227334421最少。第、、、、、日調運涉及到得點最多,車輛也較多。以第29天M點調運到點8點調運到點11點調運到D點3輛汽車,從G點調運到J點15輛汽車,從E點調運到J點4輛汽車,從I點調運到L點1輛點調運到點1點調運到M點5點調運到M點9從N點調運到M點1輛汽車,從S點調運到M點8輛汽車,從H點調運到O點3輛汽車,從S點調運到O點1輛汽車,從R點調運到P點7輛汽車,從E點調運到Q點4輛汽車,從T點調運到點2輛汽車。總的調運費為2.2439萬元。①假定租賃公司偏向于在盡可能多的滿足各點需求量的前提下,實現利潤最②通過表格數據的計算得知,當k時,每天各點總需求量大于總供給量,當k6,7,12,13,14時,每天各點的總供③由已經條件知,第i個點第k日的租賃總收入Yik j①當各點的總需求大于總供給20

maxZxjikqi(Dikxjik)vixijkcij

i1j

j

j

ii

i

其中:Eik——調運前第i個點第k日擁有的車輛Dik——未來四第i個點第k日的汽車

xijk——第k日第i個點向第j個點轉cij——從第i個點向第j個點調運需要

vi——第i個點一輛車一天的短缺損失qi——第i個點一輛車一天的租賃收②當各點的總供給大于總需求20

minZDikqixijkcij

i1j

j xijk xjik

ii

kk

i

k其中:Eik——調運前第i個點第k日擁有的車輛Dik——未來四第i個點第k日的汽車

xijk——第k日第i個點向第j個點轉運cij——從第i個點向第j個點調運需要的qi——第i個點一輛車一天的租賃收入表52345678-9-----表6kk3k4k5413934434213135185535815914312165572131461361181531755415kkkk8954372768263151176284227138412737662115382257kkkk8142238469161851381851119 5145169821433226 kkkk315968331345661415891161632651221381132214111385431215kkkk23121364648132413321111173717132211121453154361724642kkkk641727453336222198713433113521224158433557223423111111311549kkkk431845131349483694644282921335159224371911412913469732249162在29天的方案中我們發現依然是第19日需要調轉的車輛最少涉及到得點也最少。第、、、、日調運涉及到得點最多,車輛也較多。以第29天為例調運方案為從M點調運到點8點調運到點3點調運到點9輛汽車,從點調運到點3輛汽車,從J點調運到點6輛汽車,從J點調運到點8從I點調運到L點1輛汽車,從N點調運到L點11輛汽車,從E點調運到M點9輛汽車,從F點調運到M點14點調運到點9點調運到點1點調運到點3點調運到點7點調運到點48.588萬元。問題三的模型,以利潤最大化為目標,利用LINGO求解,并結合價格,進行靈①假定租賃公司為各點的車輛直接屬于各點,不考慮期間的調撥,②由于考慮到數量與價格幅度之間的關系,假設如果新車,只③由于公司規模的擴大和汽車租賃需求的增加,先假定需要進行車輛的購置來使第一年成本第二年總成本第一年成本第二年總成本第三年總成本第四年總成本第五年總成本第六年總成本第七年總成本第八年總成本年平均總成本123456789圖3每年的平均折舊額。從表中易知,第8種車型的年平均折舊最低,那么選擇第八種車型做購置準備。通過計算得,購置一輛第八種類型車輛,每天的單位總成本為0.014120

ji

i1j

j i

j js.txjij

ix x

i

xij——第k日第i個 點向第j個 cij——從第i個點向第j個點調運需要的總費用vi——第i個點一輛車一天的短缺損失費qi——第i個點一輛車一天的租賃收Di——第i個點年平均需求Ei——調運前第i個點擁有的車輛根據附件2表7各點年平均需求量與原始車點ABCDEFGHIG各點原始車點KLMNOPQRST各點原始車附件中計算年平均需求量均為分數,由于車輛必為整數,因此對數據進行“進一利用Lingo進行編程求解,調用出靈敏度表,去除虛擬變量導入excel表8價格與靈敏度分析RighthandSideSlackorDual1A0A10B0B10C0C20D0D20E0E20F0F20G0G20H0H10I0I10J0J20K0K20L0L30M0M0N0N30O0O30P0P30Q0Q20R0R30S0S30T0T20A0A01B0B01C0C02D0D02E0E02F0F02G0G02H0H01I0I01J0J02K0K02L0L03M0M0N0N03O0O03P0P03Q0Q02R0R03S0S03T0T02由表5知,當A~T為供給地時,以分析A點為例,當A地增加一輛車時,其單位所以A點可以購置1輛車。同理對剩下的19個點進行分析發現每個點增加一輛車的單位利潤都大于增加一輛車的單位成本,依據靈敏度分析表中購置車輛的范圍,得當點購置1輛,B點購置1輛C點購置2輛D點購置2輛E點購置2輛F點購置2輛,G點購置2輛,H點購置1輛,I點購置1輛,J點購置2輛,K點購置2輛L點購置3輛M點購置0輛N點購置3輛O點購置3輛,P點購置3輛Q點購置2輛點購置3輛S點購置3輛T點購置2輛。總共購置40輛。§6Lingo軟§70-1變量。0-1引入一個變量wi

cjivicji

,i①當各點的總需求大于總供給時為20

minZ(Dikwixjik)viwixijkciji1

j

j xijkj

i

k j

i

kxijk

ii

k[1].黃己立.數學建模[M].合肥;中國科技大 [2]張家善.線性規劃在產銷不平衡問題中的應用 [3].運籌優化在物資不平衡調運中的應用研究[J].物流技術,2013,15:150-[4]葉桂林.Lingo軟件在優化問題中的應用[J].現代商業[5]金晶晶.Lingo軟件在數學建模競賽中的應用[J].十堰學院學報,2010,04:85- 0.0339410.1233290.0656130.1445680.1081260.2543140.1022480.1139130.055887 0.0611880.0822220.0771160.098920.0949680.0966580.220010.177610.033941 0.179930.0483710.1728170.1241350.0926650.066270.1310730.1374090.0613910.0136820.0429140.0314870.051920.0785480.0837690.0594270.1566890.1233290.17993 0.1261230.1136370.4180120.1873120.348910.6595120.2236780.6965930.946580.6060640.6529181.8766070.477121.1621241.7325370.465370.0656130.0483710.126123 0.1139540.0858020.0434690.0395490.0739460.0585880.0723590.0336420.0253950.022710.039950.0638010.0272170.054110.0530060.1445680.1728170.1136370.113954 0.1448280.0251830.1241810.0533480.006480.1754980.065454 0.0695090.0912970.0337310.1372730.0796650.050090.1081260.1241350.4180120.0858020.144828 0.1344110.0458550.0470010.0323150.0983690.0319270.0360450.032020.0812590.0430340.0206460.0311520.0474210.2543140.0926650.1873120.0434690.0251830.134411 0.1084080.0556690.00979 0.0636450.1023440.108850.0845530.0705060.071010.0502360.1052520.1022480.066270.348910.0395490.1241810.0458550.108408 0.0443530.0664490.0826870.0909630.051930.032760.0872090.0330420.0250570.0278090.0123690.1139130.1310730.6595120.0739460.0533480.0470010.0556690.044353 0.0265630.0778130.0969790.0438790.071760.0496890.0447950.0378430.044010.0643990.2052990.1374090.2236780.0585880.006480.032315 0.0265630.1348330.2300060.1209020.0966320.1525650.1479370.0618470.0701950.0457450.0804670.0558870.1051710.825740.0169710.0967230.0158330.138480.08906 0.1348330.0909670.082606 0.0468750.0414610.055680.0421010.0496010.071051 0.0613910.6965930.0723590.1754980.098369 0.0826870.0778130.090967 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0.0145990.0141260.220010.0594271.7325370.054110.0796650.0311520.0502360.0278090.044010.0496010.0629720.025960.015480.0232330.0315880.0276910.014599 0.0309260.177610.1566890.465370.0530060.050090.0474210.1052520.0123690.0643990.0710510.1131680.0324180.0472260.0685310.0601940.0107330.0141260.030926 附件2 0.0000 0.033940.123330.065610.144570.108130.254310.102250.113910.05589 0.0611900.03394 0.179930.048370.061390.013680.0429100.123330.17993 0.126120.696590.946580.6060600.065610.048370.12612 0.072360.03364 00.144570.172820.113640.11395 0.06545 0.06951 00.108130.124140.41801 0.14483 0.098370.031930.036050.032020.0812600.254310.092670.187310.043470.025180.13441 0.063650.102340.108850.0845500.102250.066270.348910.039550.124180.045860.108410.082690.090960.051930.032760.087210.0330400.113910.131070.659510.073950.05335 0.055670.044350.077810.096980.043880.071760.04969 0.037840 0.137410.223680.058590.006480.032320.009790.066450.02656 0.23001 0.096630.152570.147940.06185 0.045750.055890.105170.825740.016970.096720.015830.138480.08906 0.134830.090970.08261 0.046880.041460.05568 0.0909700.061190.082610.0659600.082220.042910.096630.045240.0149300.077120.031490.046880.081420.390940.0174700.098920.051921.876610.03995 0.081260.041460.050620.012240.017130.03757 00.094970.078550.47712 0.033730.043030.055680.076270.063690.074930.070290.0990500.096660.083771.162120.027220.137270.020650.103010.058140.05904 0.018440.04001 00.220010.059431.732540.054110.079670.031150.050240.062970.025960.015480.023230.031590.02769 00 0000000000000000 00-0.705088000000 -0.54352- 00.01725600 00 0 0 -0.11288- 0 04.94E-074.94E-07-000000000000000000000 0.0339410.1233290.0656130.1445680.1081260.2543140.1022480.1139130.055887 0.0611880.0822220.0771160.098920.0949680.0966580.220010.177610.033941 0.179930.0483710.1728170.1241350.0926650.066270.1310730.1374090.0613910.0136820.0429140.0314870.051920.0785480.0837690.0594270.1566890.1233290.17993 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