10-Jan-2022 16:41:33

scip_solution_read_test():
  MATLAB/Octave version 9.8.0.1380330 (R2020a) Update 2
  scip_solution_read() reads a SCIP solution file.

scip_solution_read():
  Extract solution information from an SCIP solution file.
solution status: optimal solution found
objective value:                                    0
x1                                                  1 	(obj:0)
x7                                                  1 	(obj:0)
x12                                                 1 	(obj:0)
x18                                                 1 	(obj:0)
  File contains 6 lines
  Number of nonzero entries in X is 4

word =

    'x1'


text_rest =

    '                                                  1 	(obj:0)'


i =

     1

  x(1) = 1

word =

    'x7'


text_rest =

    '                                                  1 	(obj:0)'


i =

     7

  x(7) = 1

word =

    'x12'


text_rest =

    '                                                 1 	(obj:0)'


i =

    12

  x(12) = 1

word =

    'x18'


text_rest =

    '                                                 1 	(obj:0)'


i =

    18

  x(18) = 1

scip_solution_read():
  Extract solution information from an SCIP solution file.
solution status: optimal solution found
objective value:                               113.45
Dimes                                               2 	(obj:0.1)
Quarters                                           53 	(obj:0.25)
Dollars                                           100 	(obj:1)
Cu                                              999.8 	(obj:0)
Ni                                               46.9 	(obj:0)
Zi                                                 50 	(obj:0)
Mn                                                 30 	(obj:0)
  File contains 9 lines
  Number of nonzero entries in X is 7

word =

    'Dimes'


text_rest =

    '                                               2 	(obj:0.1)'


i =

     []

  x() = 1

word =

    'Quarters'


text_rest =

    '                                           53 	(obj:0.25)'


i =

     []

  x() = 1

word =

    'Dollars'


text_rest =

    '                                           100 	(obj:1)'


i =

     []

  x() = 1

word =

    'Cu'


text_rest =

    '                                              999.8 	(obj:0)'


i =

     []

  x() = 1

word =

    'Ni'


text_rest =

    '                                               46.9 	(obj:0)'


i =

   0.0000 + 1.0000i

  x(0) = 1