Added examples/multiple_objective.py

Author: jajhall

examples/multiple_objective.py

@@ -0,0 +1,120 @@
+import numpy as np
+import highspy
+
+hscb = highspy.cb 
+
+h = highspy.Highs()
+h.setOptionValue("output_flag", False);
+
+inf = highspy.kHighsInf
+lp = highspy.HighsLp()
+lp.num_col_ = 2
+lp.num_row_ = 3
+lp.col_cost_ = np.array([0, 0], dtype=np.double)
+lp.col_lower_ = np.array([0, 0], dtype=np.double)
+lp.col_upper_ = np.array([inf, inf], dtype=np.double)
+lp.row_lower_ = np.array([-inf, -inf, -inf], dtype=np.double)
+lp.row_upper_ = np.array([18, 8, 14], dtype=np.double)
+lp.a_matrix_.start_ = np.array([0, 3, 6])
+lp.a_matrix_.index_ = np.array([0, 1, 2, 0, 1, 2])
+lp.a_matrix_.value_ = np.array([3, 1, 1, 1, 1, 2], dtype=np.double)
+h.passModel(lp)
+# LP constraints are
+#
+# 3x +  y <= 18
+#
+#  x +  y <= 8
+#
+#  x + 2y <= 14
+linear_objective0 = highspy.HighsLinearObjective()
+linear_objective0.weight = -1
+linear_objective0.offset = -1
+linear_objective0.coefficients = np.array([1, 1], dtype=np.double)
+linear_objective0.abs_tolerance = 0
+linear_objective0.rel_tolerance = 0
+linear_objective0.priority = 10
+
+linear_objective1 = highspy.HighsLinearObjective()
+linear_objective1.weight = 1e-4
+linear_objective1.offset = 0
+linear_objective1.coefficients = np.array([1, 0], dtype=np.double)
+linear_objective1.abs_tolerance = -1
+linear_objective1.rel_tolerance = -1
+linear_objective1.priority = 0
+
+h.addLinearObjective(linear_objective0)
+h.addLinearObjective(linear_objective1)
+# Objectives
+#
+# f0: x + y - 1 (weight -1)
+#
+# f1: x         (weight 1e-4)
+#
+# With objective min -f0 (since its weight is -1) the LP has nonunique
+# optimal solutions on the line joining (2, 6) and (5, 3)
+#
+# Blending -f0 + 1e-4 f1 minimizes x along the line joining (2, 6) and
+# (5, 3) to give optimal solution at (2, 6) with objective -7
+h.run()
+solution = h.getSolution()
+print(f"Solution: ({solution.col_value[0]}, {solution.col_value[1]}) for min -f0 + 1e-4 f1")
+
+# Switch to lexicographic optimization 
+h.setOptionValue("blend_multi_objectives", False);
+
+linear_objective0.coefficients = np.array([1.0001, 1], dtype=np.double)
+linear_objective0.abs_tolerance = 1e-5;
+linear_objective0.rel_tolerance = 0.05;
+linear_objective1.weight = 1e-3
+
+h.clearLinearObjectives()
+h.addLinearObjective(linear_objective0)
+h.addLinearObjective(linear_objective1)
+# Objectives
+#
+# f0 = 1.0001 x + y - 1 (with priority 10 and absolute tolerance 1e-5)
+#
+# f1 =        x         (with priority  0)
+#
+# Lexicographically: HiGHS 
+#
+# minimizes f0 (to give objective 7.0005) 
+#
+# adds a constraint that
+#
+# 1.0001 x + y - 1 >= 7.0005 - 1e-5 => 1.0001 x + y >= 8.00049
+#
+# to ensure that the initial objective is within 1e-5 of its optimal
+# value
+#
+# minimizes f1 to give optimal solution at (4.90000, 3.10000) - where
+# x + y = 8 and 1.0001 x + y = 8.00049
+
+h.run()
+solution = h.getSolution()
+print(f"Solution: ({solution.col_value[0]}, {solution.col_value[1]}) for min f1 and 1.0001 x + y >= 8.00049")
+
+linear_objective0.abs_tolerance = -1;
+h.clearLinearObjectives()
+h.addLinearObjective(linear_objective0)
+h.addLinearObjective(linear_objective1)
+# Objectives as before, but absolute tolerence for f0 now -1 (negative
+# => ignored) so relative tolerance of 0.05 is used
+#
+# Lexicographically: HiGHS 
+#
+# minimizes f0 (to give objective 7.0005) 
+#
+# adds a constraint that
+#
+# 1.0001 x + y - 1 >= 7.0005 * 0.95 => 1.0001 x + y >= 7.650475
+#
+# to ensure that the initial objective 95% of its optimal value
+#
+# minimizes f1 to give optimal solution at (1.30069, 6.34966) - where
+# x + y = 8 and 1.0001 x + y = 7.650475
+h.run()
+solution = h.getSolution()
+print(f"Solution: ({solution.col_value[0]}, {solution.col_value[1]}) for min f1 and 1.0001 x + y = 7.650475")
+
+