作者xainbo (阿伯)
看板NCCU06_LAND
标题Re: [资班] 公设课程所需的软体程式
时间Tue Nov 4 11:19:54 2008
P CENTER
min D
subject to
x1 + x2 + x3 + x4 = 2
y11 + y12 + y13 + y14 = 1
y21 + y22 + y23 + y24 = 1
y31 + y32 + y33 + y34 = 1
y41 + y42 + y43 + y44 = 1
y11 - x1 <= 0
y12 - x2 <= 0
y13 - x3 <= 0
y14 - x4 <= 0
y21 - x1 <= 0
y22 - x2 <= 0
y23 - x3 <= 0
y24 - x4 <= 0
y31 - x1 <= 0
y32 - x2 <= 0
y33 - x3 <= 0
y34 - x4 <= 0
y41 - x1 <= 0
y42 - x2 <= 0
y43 - x3 <= 0
y44 - x4 <= 0
D - 5 Y11 + 10 Y12 + 8 Y13 + 12 Y14 >= 0
D - 10 Y21 + 5 Y22 + 6 Y23 + 8 Y24 >= 0
D - 8 Y31 + 6 Y32 + 5 Y33 + 12 Y34 >= 0
D - 12 Y41 + 8 Y42 + 12 Y43 + 5 Y44 >= 0
End
INT X1 X2 X3 X4
INT Y11 Y12 Y13 Y14
INT Y21 Y22 Y23 Y24
INT Y31 Y32 Y33 Y34
INT Y41 Y42 Y43 Y44
LP OPTIMUM FOUND AT STEP 9
OBJECTIVE VALUE = 0.000000000E+00
FIX ALL VARS.( 3) WITH RC > 0.000000E+00
SET Y31 TO >= 1 AT 1, BND= -8.000 TWIN= 0.0000E+00 19
NEW INTEGER SOLUTION OF 8.00000000 AT BRANCH 1 PIVOT 19
BOUND ON OPTIMUM: 0.0000000E+00
FLIP Y31 TO <= 0 AT 1 WITH BND= 0.00000000E+00
SET Y11 TO <= 0 AT 2, BND= 0.0000E+00 TWIN= -5.000 25
NEW INTEGER SOLUTION OF 0.000000000E+00 AT BRANCH 2 PIVOT 25
BOUND ON OPTIMUM: 0.0000000E+00
DELETE Y11 AT LEVEL 2
DELETE Y31 AT LEVEL 1
ENUMERATION COMPLETE. BRANCHES= 2 PIVOTS= 25
LAST INTEGER SOLUTION IS THE BEST FOUND
RE-INSTALLING BEST SOLUTION...
OBJECTIVE FUNCTION VALUE
1) 0.0000000E+00
VARIABLE VALUE REDUCED COST
X1 1.000000 0.000000
Y11 0.000000 0.000000
Y21 0.000000 0.000000
Y31 0.000000 0.000000
Y41 0.000000 0.000000
D 0.000000 1.000000
X2 0.000000 0.000000
X3 0.000000 0.000000
X4 1.000000 0.000000
Y12 0.000000 0.000000
Y13 0.000000 0.000000
Y14 1.000000 0.000000
Y22 0.000000 0.000000
Y23 0.000000 0.000000
Y24 1.000000 0.000000
Y32 0.000000 0.000000
Y33 0.000000 0.000000
Y34 1.000000 0.000000
Y42 0.000000 0.000000
Y43 0.000000 0.000000
Y44 1.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
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) 1.000000 0.000000
8) 0.000000 0.000000
9) 0.000000 0.000000
10) 0.000000 0.000000
11) 1.000000 0.000000
12) 0.000000 0.000000
13) 0.000000 0.000000
14) 0.000000 0.000000
15) 1.000000 0.000000
16) 0.000000 0.000000
17) 0.000000 0.000000
18) 0.000000 0.000000
19) 1.000000 0.000000
20) 0.000000 0.000000
21) 0.000000 0.000000
22) 0.000000 0.000000
23) 12.000000 0.000000
24) 8.000000 0.000000
25) 12.000000 0.000000
26) 5.000000 0.000000
NO. ITERATIONS= 25
BRANCHES= 2 DETERM.= 1.000E 0
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