[SCIP] set/vbc/dispsols
Gotzes, Uwe Dr.
Uwe.Gotzes at open-grid-europe.com
Fri Dec 16 12:02:08 CET 2016
Hi,
I have a very simple problem and tried to tell scip to apply plain lp-based branch and bound to it.
I have set vbc/dispsols = TRUE to visualize the nodes where solutions are found.
This option results in the two lines 00:00:00.00 P 6 14 and 00:00:00.00 P 5 14 in the vbc-file. (P int int Number of the node, number of the nodes colour.)
I have two questions:
1. Is it possible to figure out the value of the solution found in the respective node from the vbc-file?
2. What is the meaning of the minus sign in the "frac"-column of node 2?
Scip output is:
time | node | left |LP iter|LP it/n| mem |mdpt |frac |vars |cons |cols |rows |cuts |confs|strbr| dualbound | primalbound | gap
0.0s| 1 | 0 | 0 | - | 166k| 0 | - | 2 | 3 | 2 | 0 | 0 | 0 | 0 | -- | -- | Inf
(node 1) LP relaxation is unbounded (LP 0)
0.0s| 1 | 0 | 2 | - | 167k| 0 | - | 2 | 3 | 2 | 2 | 2 | 0 | 0 | -- | -- | Inf
(node 1) LP relaxation is unbounded (LP 1)
0.0s| 1 | 0 | 3 | - | 168k| 0 | 1 | 2 | 3 | 2 | 2 | 3 | 0 | 0 |8.428571e+000 | -- | Inf
0.0s| 1 | 2 | 3 | - | 168k| 0 | 1 | 2 | 3 | 2 | 2 | 3 | 0 | 0 |8.428571e+000 | -- | Inf
0.0s| 2 | 1 | 3 | 0.0 | 169k| 1 | - | 2 | 3 | 2 | 2 | 3 | 0 | 0 |8.000000e+000 | -- | Inf
0.0s| 3 | 2 | 5 | 1.0 | 169k| 1 | 1 | 2 | 3 | 2 | 2 | 3 | 0 | 0 |7.500000e+000 | -- | Inf
0.0s| 4 | 3 | 6 | 1.0 | 170k| 2 | 1 | 2 | 3 | 2 | 2 | 3 | 0 | 0 |7.000000e+000 | -- | Inf
* 0.0s| 5 | 2 | 7 | 1.0 | 170k| 3 | - | 2 | 3 | 2 | 2 | 3 | 0 | 0 |7.000000e+000 |4.000000e+000 | 75.00%
* 0.0s| 6 | 0 | 8 | 1.0 | 170k| 3 | - | 2 | 3 | 2 | 2 | 3 | 0 | 0 |7.000000e+000 |7.000000e+000 | 0.00%
The content of the generated vbc-file:
#TYPE: COMPLETE TREE
#TIME: SET
#BOUNDS: SET
#INFORMATION: STANDARD
#NODE_NUMBER: NONE
00:00:00.00 N 0 1 3
00:00:00.00 I 1 \inode:\t1\idepth:\t0\nvar:\t-\nbound:\t-100000000000000000000.000000
00:00:00.00 P 1 11
00:00:00.00 N 1 2 3
00:00:00.00 I 2 \inode:\t2\idepth:\t1\nvar:\t-\nbound:\t-8.428571
00:00:00.00 I 2 \inode:\t2\idepth:\t1\nvar:\tt_x1 [-0,1e+020] <= 2.000000\nbound:\t-8.000000
00:00:00.00 N 1 3 3
00:00:00.00 I 3 \inode:\t3\idepth:\t1\nvar:\t-\nbound:\t-8.428571
00:00:00.00 I 3 \inode:\t3\idepth:\t1\nvar:\tt_x1 [-0,1e+020] >= 3.000000\nbound:\t-8.428571
00:00:00.00 I 1 \inode:\t1\idepth:\t0\nvar:\t-\nbound:\t-8.428571\nnr:\t1
00:00:00.00 P 1 2
00:00:00.00 I 3 \inode:\t3\idepth:\t1\nvar:\tt_x1 [3,1e+020] >= 3.000000\nbound:\t100000000000000000000.000000\nnr:\t2
00:00:00.00 P 3 2
00:00:00.00 I 3 \inode:\t3\idepth:\t1\nvar:\tt_x1 [3,1e+020] >= 3.000000\nbound:\t100000000000000000000.000000\nnr:\t2
00:00:00.00 P 3 4
00:00:00.00 N 2 4 3
00:00:00.00 I 4 \inode:\t4\idepth:\t2\nvar:\t-\nbound:\t-7.500000
00:00:00.00 I 4 \inode:\t4\idepth:\t2\nvar:\tt_x2 [-0,3] <= 0.000000\nbound:\t-7.500000
00:00:00.00 N 2 5 3
00:00:00.00 I 5 \inode:\t5\idepth:\t2\nvar:\t-\nbound:\t-7.500000
00:00:00.00 I 5 \inode:\t5\idepth:\t2\nvar:\tt_x2 [-0,3] >= 1.000000\nbound:\t-7.000000
00:00:00.00 I 2 \inode:\t2\idepth:\t1\nvar:\tt_x1 [-0,2] <= 2.000000\nbound:\t-7.500000\nnr:\t3
00:00:00.00 P 2 2
00:00:00.00 N 4 6 3
00:00:00.00 I 6 \inode:\t6\idepth:\t3\nvar:\t-\nbound:\t-6.000000
00:00:00.00 I 6 \inode:\t6\idepth:\t3\nvar:\tt_x1 [-0,2] <= 1.000000\nbound:\t-4.000000
00:00:00.00 N 4 7 3
00:00:00.00 I 7 \inode:\t7\idepth:\t3\nvar:\t-\nbound:\t-6.000000
00:00:00.00 I 7 \inode:\t7\idepth:\t3\nvar:\tt_x1 [-0,2] >= 2.000000\nbound:\t-6.000000
00:00:00.00 I 4 \inode:\t4\idepth:\t2\nvar:\tt_x2 [-0,0] <= 0.000000\nbound:\t-6.000000\nnr:\t4
00:00:00.00 P 4 2
00:00:00.00 I 6 \inode:\t6\idepth:\t3\nvar:\tt_x1 [-0,1] <= 1.000000\nbound:\t-4.000000\nnr:\t5
00:00:00.00 P 6 2
00:00:00.00 P 6 14
00:00:00.00 U -4.000000
00:00:00.00 I 5 \inode:\t5\idepth:\t2\nvar:\tt_x2 [1,3] >= 1.000000\nbound:\t-7.000000\nnr:\t6
00:00:00.00 P 5 2
00:00:00.00 P 5 14
00:00:00.00 I 7 \inode:\t7\idepth:\t3\nvar:\tt_x1 [-0,2] >= 2.000000\nbound:\t-6.000000\nnr:\t6
00:00:00.00 P 7 4
00:00:00.00 U -7.000000
Thanks,
Uwe
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