2011年數(shù)學(xué)建模詳細(xì)解釋腸衣問題

1、研究目的：本文研該題主要產(chǎn)天然腸衣及其搭配問題問題3、求解思路，使用線性優(yōu)化方法、編寫Lingo：生產(chǎn)天然腸 最優(yōu) 最大利-3.93.5378∞52表 原料描述3-3.5-4-4.5-5-5.5-6-7-7.5-8-8.5-9-12-13-16-17-20-21-224-25-060001為提高原料使用率，總長(zhǎng)度允許有±0.51147-13.57-13.5307～13.5表 成品規(guī)格132783∞51 2、所假設(shè)確實(shí)是建立數(shù)學(xué)模型所必需的，與建立模型無關(guān)的假設(shè)只從問題的性質(zhì)出發(fā)作出合乎的假設(shè)，或者由觀察所給數(shù)據(jù)的圖象，得到變量的函數(shù)形式，也可以參考其他資料由類推得到。對(duì)于后者應(yīng)參考文獻(xiàn)的

為某一規(guī)格中第ij

為第ij j 為第kk 為第k（1每捆成品總長(zhǎng)度允許有0.5的誤差。應(yīng)用線性規(guī)劃建立模型天然腸衣。88.5，得出針對(duì)長(zhǎng)度的最大捆數(shù)；用每種規(guī)格腸衣的總根數(shù)除以對(duì)應(yīng)規(guī)格每捆要求的數(shù)量dk(k1,2,3，nnkLkj

ajb

fk

,N},k1,2,3 ．fkLNajj 1853～6.57～13.514～25.5目標(biāo)函數(shù)為88x*89.5qi*y*89.5 qi*z*89.5 qi*

88.5*88.5*88.5*88x*20約束條件為8*y 5*z

19*7*4*

然后通過linggo

xyz表 原料剩余3-3.5-4-4.5-5-5.5-6-6.5-00000007-7.5-8-8.5-9-9.5-10-10.5-0000011-11.5-12-12.5-13-13.5-14-14.5-0000000015-15.5-16-16.5-17-17.5-18-18.5-0000000019-19.5-20-20.5-21-21.5-22-22.5-000000023-23.5-24-24.5-25-25.5-000000 LINGO使用且操作簡(jiǎn)單的工具是最好的選擇，本文運(yùn)用圖形用戶界面編程建立一[1]姜啟源《數(shù)學(xué)模型（第三版 ：高等教 1、lingomax=x+y+z;a1=33.544.555.56a2=77.588.599.51010.51111.51212.513a3=1414.51515.51616.51717.51818.51919.52020.52121.52222.52323.524.525b1=43593941272834b2=24242025212321183123225918b3=3529304228424549506452634935;2GlobaloptimalsolutionObjectiveObjectiveExtendedsolver0 ReducedXYZA1(A1(A1(A1(A1(A1(A1(A1(A2(A2(A2(A2(A2(A2(A2(A2(A2(A2(A2(A2(A2(A2(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(A3(B1(B1(B1(B1(B1(B1(B1(B1(B2(B2(B2(B2(B2(B2(B2(B2(B2(B2(B2(B2(B2(B2(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(B3(Q1(Q1(Q1(Q1(Q1(Q1(Q1(Q1(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q2(Q3(Q3(Q3(Q3(Q3(Q3(Q3(