lingo實(shí)驗(yàn)報(bào)告

1 實(shí)驗(yàn)名稱 推銷員指派問(wèn)題 二 實(shí)驗(yàn)?zāi)康募叭蝿?wù) 1 掌握 Lingo 軟件的使用方法 2 編寫簡(jiǎn)單的 Lingo 程序 3 解決 Lingo 中的最優(yōu)指派問(wèn)題 三 實(shí)驗(yàn)內(nèi)容 1 問(wèn)題描述 一個(gè)公司要分派 5 個(gè)推銷員去 5 個(gè)地區(qū)推銷某種產(chǎn)品 5 個(gè)推銷員在各個(gè) 地區(qū)推銷這種產(chǎn)品的預(yù)期利潤(rùn)如下表所示 若每個(gè)推銷員只能去一個(gè)地區(qū) 應(yīng) 如何分派這 5 個(gè)推銷員才能使公司的利潤(rùn)為最大 2 模型建立 決策變量 決策變量 設(shè) i j 1 2 3 4 5 個(gè)地區(qū)個(gè)人去第不指派第 個(gè)地區(qū)個(gè)人去第指派第 ji 0 ji 1 ij x 目標(biāo)函數(shù) 目標(biāo)函數(shù) 設(shè)總利潤(rùn)為 z 第 i 個(gè)人去第 j 個(gè)地區(qū)的利潤(rùn)為 Aij i j 1 2 3 4 5 假設(shè) Aij為指派矩陣 則 Max 5 1 5 1ij ijijx Az 約束條件 約束條件 1 第 j 個(gè)地區(qū)只有一個(gè)人去 j 1 2 3 4 5 1 5 1 i ij x 2 第 i 個(gè)人只去一個(gè)地區(qū) i 1 2 3 4 5 1 5 1 j ij x 由此得基本模型 Max 5 1 5 1ij ijijx Az S t j 1 2 3 4 5 1 5 1 i ij x i 1 2 3 4 5 1 5 1 j ij x i j 1 2 3 4 5 10或 ij x 3 Lingo 程序 一 常規(guī)程序 Lingo 輸入 model max 1 x11 8 x12 9 x13 2 x14 1 x15 5 x21 6 x22 3 x23 10 x24 7 x25 3 x31 10 x32 4 x33 11 x34 3 x35 7 x41 7 x42 5 x43 4 x44 8 x45 4 x 51 2 x52 6 x53 3 x54 9 x55 x11 x12 x13 x14 x15 1 x21 x22 x23 x24 x25 1 x31 x32 x33 x34 x35 1 x41 x42 x43 x44 x45 1 x51 x52 x53 x54 x55 1 x11 x21 x31 x41 x51 1 x12 x22 x32 x42 x52 1 x13 x23 x33 x43 x53 1 x14 x24 x34 x44 x54 1 x15 x25 x35 x45 x55 1 end Lingo 輸出 Global optimal solution found Objective value 45 00000 Infeasibilities 0 000000 Total solver iterations 8 Variable Value Reduced Cost X11 0 000000 7 000000 X12 0 000000 0 000000 X13 1 000000 0 000000 X14 0 000000 7 000000 X15 0 000000 8 000000 X21 0 000000 4 000000 X22 0 000000 3 000000 X23 0 000000 7 000000 X24 1 000000 0 000000 X25 0 000000 3 000000 X31 0 000000 7 000000 X32 1 000000 0 000000 X33 0 000000 7 000000 X34 0 000000 0 000000 X35 0 000000 8 000000 X41 1 000000 0 000000 X42 0 000000 0 000000 X43 0 000000 3 000000 X44 0 000000 4 000000 X45 0 000000 0 000000 X51 0 000000 4 000000 X52 0 000000 6 000000 X53 0 000000 3 000000 X54 0 000000 6 000000 X55 1 000000 0 000000 Row Slack or Surplus Dual Price 1 45 00000 1 000000 2 0 000000 9 000000 3 0 000000 10 00000 4 0 000000 11 00000 5 0 000000 8 000000 6 0 000000 9 000000 7 0 000000 1 000000 8 0 000000 1 000000 9 0 000000 0 000000 10 0 000000 0 000000 11 0 000000 0 000000 二 集合函數(shù)程序 Lingo 輸入 model sets person 1 5 area 1 5 assign person area A x endsets data A 1 8 9 2 1 5 6 3 10 7 3 10 4 11 3 7 7 5 4 8 4 2 6 3 9 enddata max sum assign A x for person i sum area j x i j 1 for area j sum person i x i j 1 for assign i j bin x i j end Lingo 輸出 Global optimal solution found Objective value 45 00000 Objective bound 45 00000 Infeasibilities 0 000000 Extended solver steps 0 Total solver iterations 0 Variable Value Reduced Cost A 1 1 1 000000 0 000000 A 1 2 8 000000 0 000000 A 1 3 9 000000 0 000000 A 1 4 2 000000 0 000000 A 1 5 1 000000 0 000000 A 2 1 5 000000 0 000000 A 2 2 6 000000 0 000000 A 2 3 3 000000 0 000000 A 2 4 10 00000 0 000000 A 2 5 7 000000 0 000000 A 3 1 3 000000 0 000000 A 3 2 10 00000 0 000000 A 3 3 4 000000 0 000000 A 3 4 11 00000 0 000000 A 3 5 3 000000 0 000000 A 4 1 7 000000 0 000000 A 4 2 7 000000 0 000000 A 4 3 5 000000 0 000000 A 4 4 4 000000 0 000000 A 4 5 8 000000 0 000000 A 5 1 4 000000 0 000000 A 5 2 2 000000 0 000000 A 5 3 6 000000 0 000000 A 5 4 3 000000 0 000000 A 5 5 9 000000 0 000000 X 1 1 0 000000 1 000000 X 1 2 0 000000 8 000000 X 1 3 1 000000 9 000000 X 1 4 0 000000 2 000000 X 1 5 0 000000 1 000000 X 2 1 0 000000 5 000000 X 2 2 0 000000 6 000000 X 2 3 0 000000 3 000000 X 2 4 1 000000 10 00000 X 2 5 0 000000 7 000000 X 3 1 0 000000 3 000000 X 3 2 1 000000 10 00000 X 3 3 0 000000 4 000000 X 3 4 0 000000 11 00000 X 3 5 0 000000 3 000000 X 4 1 1 000000 7 000000 X 4 2 0 000000 7 000000 X 4 3 0 000000 5 000000 X 4 4 0 000000 4 000000 X 4 5 0 000000 8 000000 X 5 1 0 000000 4 000000 X 5 2 0 000000 2 000000 X 5 3 0 000000 6 000000 X 5 4 0 000000 3 000000 X 5 5 1 000000 9 000000 Row Slack or Surplus Dual Price 1 45 00000 1 000000 2 0 000000 0 000000 3 0 000000 0 000000 4 0 000000 0 000000 5 0 000000 0 000000 6 0 000000 0 000000 7 0 000000 0 000000 8 0 000000 0 000000 9 0 000000 0 000000 1