一、實驗?zāi)康募耙?/p>

1、分別用excel、matlab和lingo求解線性規(guī)劃、運輸問題和0-1整數(shù)規(guī)劃問題

2、撰寫實驗報告2

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二、實驗設(shè)備（環(huán)境）及要求

Microsoft Excel2016、matlab2017、lingo17

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三、實驗內(nèi)容

線性規(guī)劃

用excel求解

①加載宏—規(guī)劃求解加載項

②根據(jù)分析題目填寫約束條件、目標函數(shù)

D2=B5

D3=C5

D4=B5+2*C5

F6= =2*B5+5*C5

③添加線性求解

④求解結(jié)果

?

matlab

• ?

max S=2x1+5x2

? ? ? ? ? x1 ≤4

? ? ? ? ? x2≤3

s.t.???? x1+ 2x2≤8

? ? ? ? ? ?x1≥0

? ? ? ? ? ?x2≥0

?

將目標函數(shù)轉(zhuǎn)化為求函數(shù)-S的最小值,根據(jù)目標函數(shù)和約束條件，可以得出目標函數(shù)系數(shù)矩陣f=[2;5]，不等式約束系數(shù)矩陣A =[1 ?0;0? 1;1? 2 ]，不等式約束常向量b=[4; 3; 8], 調(diào)用MATLAB中l(wèi)ingprog函數(shù)求出-S的最小值，其相反數(shù)就是MaxS，如下圖所示:

結(jié)果:

>> f=[2;5];

A =[1? 0;0? 1;1? 2 ];

b=[4; 3; 8];

[x,fmin]=linprog(-f,A,b)

?

Optimal solution found.

?

?

x =

?

??? 2.0000

??? 3.0000

?

?

fmin =

?

?? -19

則MaxS=19

?

?

lingo

代碼:

max=2*X1+5*X2;

X1<=4;

X2<=3;

X1+2*X2<=8;

結(jié)果:

Global optimal solution found.

? Objective value:????????????????????????????? 19.00000

? Infeasibilities:????????????????????????????? 0.000000

? Total solver iterations:???????????????????????????? 1

? Elapsed runtime seconds:????????????????????????? 0.53

?

? Model Class:??????????????????????????????????????? LP

?

? Total variables:????????????????????? 2

? Nonlinear variables:????????????????? 0

? Integer variables:??????????????????? 0

?

? Total constraints:????? ??????????????4

? Nonlinear constraints:??????????????? 0

?

? Total nonzeros:?????????????????????? 6

? Nonlinear nonzeros:?????????????????? 0

?

?

?

??????????????????????????????? Variable?????????? Value??????? Reduced Cost

???????????????????????????????? ?????X1??????? 2.000000??????????? 0.000000

????????????????????????????????????? X2??????? 3.000000??????????? 0.000000

?

???????????????????????????????????? Row??? Slack or Surplus????? Dual Price

?????????????????????????????????????? 1??????? 19.00000 ???????????1.000000

?????????????????????????????????????? 2??????? 2.000000??????????? 0.000000

?????????????????????????????????????? 3??????? 0.000000??????????? 1.000000

?????????????????????????????????????? 4??????? 0.000000??????????? 2.000000

?

?

運輸問題

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Let xij = the number of truckloads to ship from cannery i to warehouse j
?????? (i = 1, 2, 3; j = 1, 2, 3, 4)
Minimize Cost = $464x11 + $513x12 + $654x13 + $867x14

????????????????????????? + $352x21 + $416x22+ $690x23 + $791x24

????????????????????????? + $995x31 + $682x32 + $388x33 + $685x34
subject to
?????? Cannery 1:????? x11 + x12 + x13 + x14 = 75
?????? Cannery 2:????? x21 + x22 + x23 + x24 = 125
?????? Cannery 3:????? x31 + x32 + x33 + x34 = 100
?????? Warehouse 1:? x11 + x21 + x31 = 80
?????? Warehouse 2:? x12 + x22 + x32 = 65
?????? Warehouse 3:? x13 + x23 + x33 = 70
?????? Warehouse 4:? x14 + x24 + x34 = 85
and?????????? xij ≥ 0 (i = 1, 2, 3;? j = 1, 2, 3, 4)

?

用excel求解

①輸入問題的數(shù)據(jù)，

• 根據(jù)分析題目填寫約束條件、目標函數(shù)

C15=C10+C11+C12

D15=D10+D11+D12

E15=E10+E11+E12

F15=F10+F11+F12

?

I10=C10+D10+E10+F10

I11=C11+D11+E11+F11

I12=C12+D12+E12+F12

?

I15=C3*C10+D3*D10+E3*E10+F3*F10+C4*C11+D4*D11+E4*E11+F4*F11+C5*C12+D5*D12+E5*E12+F5*F12

?

• 求解結(jié)果

?

?

?

Matlab

函數(shù)Min Cost的值,根據(jù)目標函數(shù)和約束條件，可以得出目標函數(shù)系數(shù)矩陣f=[464 513 654 867 352 416 690 791 995 682 388 685]，不等式約束系數(shù)矩陣A=[0 1 0 1 0 0 0 0 0 0 0 0

0 0 0 0 1 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 1

0 0 0 0 1 0 0 0 0 0 0 0

0 1 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 0

0 0 0 1 0 0 0 0 0 0 0 1]，不等式約束常向量b=[75 125 100 80 65 70 85], 調(diào)用MATLAB中l(wèi)ingprog函數(shù)求出Min Cost的值，如下圖所示:

結(jié)果:

>> f=[464 513 654 867 352 416 690 791 995 682 388 685]

A=[0 1 0 1 0 0 0 0 0 0 0 0

0 0 0 0 1 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 1

0 0 0 0 1 0 0 0 0 0 0 0

0 1 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 0

0 0 0 1 0 0 0 0 0 0 0 1]

b=[75 125 100 80 65 70 85]

c=[0 0 0 0 0 0 0 0 0 0 0 0 ];

[v,e]= linprog(f,[],[],A,b,c)

x=reshape(v,4,3);

x=x';

?

?

f =

?

?? 464?? 513?? 654?? 867?? 352?? 416?? 690?? 791?? 995?? 682?? 388?? 685

?

?

A =

?

???? 0???? 1???? 0???? 1???? 0???? 0???? 0???? 0???? 0???? 0???? 0???? 0

???? 0???? 0???? 0???? 0???? 1???? 1???? 0???? 0???? 0???? 0???? 0?? ??0

???? 0???? 0???? 0???? 0???? 0???? 0???? 0???? 0???? 0???? 0???? 1???? 1

???? 0???? 0???? 0???? 0???? 1???? 0???? 0???? 0???? 0???? 0???? 0???? 0

???? 0???? 1???? 0???? 0???? 0???? 1???? 0???? 0???? 0???? 0???? 0???? 0

???? 0???? 0???? 0???? 0???? 0?? ??0???? 0???? 0???? 0???? 0???? 1???? 0

???? 0???? 0???? 0???? 1???? 0???? 0???? 0???? 0???? 0???? 0???? 0???? 1

?

?

b =

?

??? 75?? 125?? 100??? 80??? 65??? 70??? 85

?

?

Optimal solution found.

?

?

v =

?

???? 0

??? 20

???? 0

??? 55

??? 80

??? 45

???? 0

???? 0

??? ?0

???? 0

??? 70

??? 30

?

?

e =

?

????? 152535

?

?

?

Lingo

代碼:

min=464*X11+513*X12+654*X13+867*X14

????????????????????????? +352*X21+416*X22+690*X23+791*X24

????????????????????????? +995*X31+682*X32+388*X33+685*X34;

X11+X12+X13+X14=75;
X21+X22+X23+X24=125;
X31+X32+X33+X34=100;
X11+X21+X31=80;

X12+X22+X32=65;
X13+X23+X33=70;
X14+X24+X34=85;
結(jié)果:

Global optimal solution found.

? Objective value:????????????????????????????? 152535.0

? Infeasibilities:????????????????????????????? 0.000000

? Total solver iterations:???????????????????????????? 7

? Elapsed runtime seconds:????????????????????????? 0.11

?

? Model Class:??????????????????????????????????????? LP

?

? Total variables:???????????????????? 12

? Nonlinear variables:????????????????? 0

? Integer variables:??????????????????? 0

?

? Total constraints:?? ?????????????????8

? Nonlinear constraints:??????????????? 0

?

? Total nonzeros:????????????????????? 36

? Nonlinear nonzeros:?????????????????? 0

?

?

?

??????????????????????????????? Variable?????????? Value??????? Reduced Cost

????????????????????????????? ???????X11??????? 0.000000??????????? 15.00000

???????????????????????????????????? X12??????? 20.00000??????????? 0.000000

???????????????????????????????????? X13??????? 0.000000??????????? 84.00000

???????????????????????????????????? X14??????? 55.00000??????????? 0.000000

???????????????????????????????????? X21??????? 80.00000??????????? 0.000000

???????????????????????????????????? X22??????? 45.00000??????????? 0.000000

???????????????????????????????????? X23??????? 0.000000??????????? 217.0000

?? ??????????????????????????????????X24??????? 0.000000??????????? 21.00000

???????????????????????????????????? X31??????? 0.000000??????????? 728.0000

???????????????????????????????????? X32??????? 0.000000??????????? 351.0000

??????????????????????????? ?????????X33??????? 70.00000??????????? 0.000000

???????????????????????????????????? X34??????? 30.00000??????????? 0.000000

?

???????????????????????????????????? Row??? Slack or Surplus????? Dual Price

?????????????????????????????????????? 1??????? 152535.0?????????? -1.000000

?????????????????????????????????????? 2??????? 0.000000?????????? -513.0000

?????????????????????????????????????? 3??????? 0.000000?????????? -416.0000

?????????????????????????????????????? 4??????? 0.000000?????????? -331.0000

?????????????????????????????????????? 5??????? 0.000000??????????? 64.00000

?????????????????????????????????????? 6??????? 0.000000??????????? 0.000000

?????????????????????????????????????? 7??????? 0.000000?????????? -57.00000

???????????????????????? ??????????????8??????? 0.000000?????????? -354.0000

?

?

0-1整數(shù)規(guī)劃問題

Maximize Z=9x1+5x2+6x3+4x4

?

Subject to

6x1+3x2+5x3+2x4≤10

x3+x4≤1

-x1????? + x3?? ≤0

-x1????????? +x4≤0

xj≤1

xj≥0

and

? xj ?is integer,? for j=1,2,3,4

? xj? is binary?? for j=1,2,3,4

?

用excel求解

①輸入問題數(shù)據(jù)

②分析題目添加約束條件、填寫目標函數(shù)

F2 =B2*B7+C2*C7+D2*D7+E2*E7

F3 =D3*D7+E3*E7

F4 =B4*B7+D4*D7

F5 =C5*C7+E5*E7

?

B7=X1;C7=X2;D7=X3;E=X4

?

H8 =B8*B7+C8*C7+D8*D7+E8*E7

?

③求解結(jié)果:

?

?

Matlab

將目標函數(shù)轉(zhuǎn)化為求函數(shù)的f=-1*[9 5 6 4];根據(jù)目標函數(shù)和約束條件，可以得出目標函數(shù)系數(shù)矩陣A=[6 3 5 2

0 0 1 1

-1 0 1 0

0 -1 0 1]

b=[10 1 0 0];不等式約束常向量b=[10 1 0 0]; 調(diào)用MATLAB中intlinprog函數(shù)求出f的值，則-1*fval則為最終結(jié)果，如下圖所示:

結(jié)果:

>> f=-1*[9 5 6 4];

intcon=[1 2 3 4];

A=[6 3 5 2

0 0 1 1

-1 0 1 0

0 -1 0 1]

b=[10 1 0 0];

lb=zeros(4,1);

ub=ones(4,1);

[x,fval,exitflag,output]=intlinprog(f,intcon,A,b,[],[],lb,ub);

x

-1*fval

?

A =

?

???? 6???? 3???? 5???? 2

???? 0???? 0???? 1???? 1

??? -1???? 0???? 1???? 0

???? 0??? -1???? 0???? 1

?

LP:??????????????? Optimal objective value is -16.500000.??????????????????????????????????????????

?

Cut Generation:? ??Applied 1 implication cut.??????????????????????????????????????????????????????

?????????????????? Lower bound is -14.000000.??????????????????????????????????????????????????????

?????????????????? Relative gap is 0.00%.?????????????????????????????????????????????????????????

?

?

Optimal solution found.

?

Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value,

options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer within tolerance, options.IntegerTolerance = 1e-05

(the default value).

?

?

x =

?

???? 1

???? 1

???? 0

???? 0

?

?

ans =

?

14

?

則結(jié)果為 14

?

3、lingo

代碼:

max=9*x1+5*x2+6*x3+4*x4;

6*x1+3*x2+5*x3+2*x4<=10;

x3+x4<=1;

-x1+x3<=0;

-x1+x4<=0;

?

結(jié)果:

Global optimal solution found.

? Objective value:????????????????????????????? 16.66667

? Infeasibilities:????????????????????????????? 0.000000

? Total solver iterations:???????????????????????????? 0

? Elapsed runtime seconds:????????????????????????? 0.44

?

? Model Class:??????????????????????????????????????? LP

?

? Total variables:????????????????????? 4

? Nonlinear variables:????????????????? 0

? Integer variables:??????????????????? 0

?

? Total constraints:??????????????????? 5

? Nonlinear constraints:??????????????? 0

?

? Total nonzeros:????????????????????? 14

? Nonlinear nonzeros:?????????????????? 0

?

?

?

??????????????????????????????? Variable?????????? Value??????? Reduced Cost

????????????????????????????????????? X1??????? 0.000000??????????? 0.000000

????????????????????????????????????? X2??????? 3.333333??????????? 0.000000

????????????????????????????????????? X3??????? 0.000000??????????? 2.333333

????????????????????????????????????? X4??????? 0.000000?????????? 0.3333333

?

???????????????????????????????????? Row??? Slack or Surplus????? Dual Price

?????????????????????????????????????? 1??????? 16.66667??????????? 1.000000

???????????????????????? ??????????????2??????? 0.000000??????????? 1.666667

?????????????????????????????????????? 3??????? 1.000000??????????? 0.000000

?????????????????????????????????????? 4??????? 0.000000??????????? 0.000000

?????????????????????????????????????? 5??????? 0.000000??????????? 1.000000

?

結(jié)果為14

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四、實驗結(jié)果

線性規(guī)劃:

Excel:

MATLAB:

LINGO：

運輸問題:

Excel:

MATLAB:

LINGO：

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0-1整數(shù)規(guī)劃:

Excel:

?

MATLAB:

LINGO：

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五、分析與討論

這次實驗分別用excel、matlab和lingo求解線性規(guī)劃、運輸問題和0-1整數(shù)規(guī)劃問題，相比第一次實驗，這次實驗學(xué)習(xí)了lingo這個計算機軟件工具，用不同的工具去求解問題。? 線性規(guī)劃問題是使用linprog函數(shù)將目標函數(shù)求Max值轉(zhuǎn)化為求函數(shù)-S的最小值；運輸問題調(diào)用matlab中l(wèi)ingprog函數(shù)求出Min Cost的值；求解0-1整數(shù)問題時，在使用matlab時，使用了intlinprog函數(shù)。

本次實驗實踐過程較為輕松，首先要明確問題類型，分析題目，理解題目的約束條件、目標函數(shù)，熟悉Excel、matlab、lingo軟件的公式或函數(shù)，就能輕松把結(jié)果求解出來。

總結(jié)

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