Learning more about APOPT solver

Question

I have a few questions about how APOPT solves MINLPs.

• What nonlinear programming method APOPT uses (interior point, trust region, etc.)?
• How does APOPT deal with mixed integers (B&B, outer approximation, generalized benders decomposition,etc)?

Answer 1

APOPT is an active-set Sequential Quadratic Programming (SQP) solver that uses Branch and Bound. APOPT uses a warm-start method to speed up successive Nonlinear Programming (NLP) solutions. There is more information about APOPT from Wikipedia, APMonitor documentation, and APOPT.com. There is benchmark information from a 2013 INFORMS presentation and in the 2014 APMonitor CACE paper.

1. Hedengren, J.D., Mojica, J.L., Lewis, A.D. and Nikbakhsh, S., MINLP with Combined Interior Point and Active Set Methods, INFORMS Annual Meeting, Minneapolis, MN, Oct 2013.
2. Hedengren, J. D. and Asgharzadeh Shishavan, R., Powell, K.M., and Edgar, T.F., Nonlinear Modeling, Estimation and Predictive Control in APMonitor, Computers and Chemical Engineering, Volume 70, pg. 133–148, 2014, doi: 10.1016/j.compchemeng.2014.04.013.

Here is a sample MINLP problem solved with Python Gekko after getting the package with pip install gekko

from gekko import GEKKO
m = GEKKO() # Initialize gekko
m.options.SOLVER=1  # APOPT is an MINLP solver

# optional solver settings with APOPT
m.solver_options = ['minlp_maximum_iterations 500', \
                    # minlp iterations with integer solution
                    'minlp_max_iter_with_int_sol 10', \
                    # treat minlp as nlp
                    'minlp_as_nlp 0', \
                    # nlp sub-problem max iterations
                    'nlp_maximum_iterations 50', \
                    # 1 = depth first, 2 = breadth first
                    'minlp_branch_method 1', \
                    # maximum deviation from whole number
                    'minlp_integer_tol 0.05', \
                    # covergence tolerance
                    'minlp_gap_tol 0.01']
# Initialize variables
x1 = m.Var(value=1,lb=1,ub=5)
x2 = m.Var(value=5,lb=1,ub=5)
# Integer constraints for x3 and x4
x3 = m.Var(value=5,lb=1,ub=5,integer=True)
x4 = m.Var(value=1,lb=1,ub=5,integer=True)
m.Equation(x1*x2*x3*x4>=25)
m.Equation(x1**2+x2**2+x3**2+x4**2==40)
m.Obj(x1*x4*(x1+x2+x3)+x3) # Objective
m.solve(disp=False) # Solve
print('x1: ' + str(x1.value))
print('x2: ' + str(x2.value))
print('x3: ' + str(x3.value))
print('x4: ' + str(x4.value))
print('Objective: ' + str(m.options.objfcnval))

The iteration summary gives more information about the branch and bound process to find the solution.

 ----------------------------------------------
 Steady State Optimization with APOPT Solver
 ----------------------------------------------
Iter:     1 I:  0 Tm:      0.00 NLPi:    7 Dpth:    0 Lvs:    3 Obj:  1.70E+01 Gap:       NaN
--Integer Solution:   1.75E+01 Lowest Leaf:   1.70E+01 Gap:   3.00E-02
Iter:     2 I:  0 Tm:      0.00 NLPi:    5 Dpth:    1 Lvs:    2 Obj:  1.75E+01 Gap:  3.00E-02
Iter:     3 I:  0 Tm:      0.00 NLPi:    6 Dpth:    1 Lvs:    2 Obj:  1.75E+01 Gap:  3.00E-02
--Integer Solution:   1.75E+01 Lowest Leaf:   1.70E+01 Gap:   3.00E-02
Iter:     4 I:  0 Tm:      0.00 NLPi:    6 Dpth:    2 Lvs:    1 Obj:  2.59E+01 Gap:  3.00E-02
Iter:     5 I:  0 Tm:      0.00 NLPi:    5 Dpth:    1 Lvs:    0 Obj:  2.15E+01 Gap:  3.00E-02
 No additional trial points, returning the best integer solution
 Successful solution

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :   1.609999999345746E-002 sec
 Objective      :    17.5322673012512     
 Successful solution
 ---------------------------------------------------

x1: [1.3589086474]
x2: [4.5992789966]
x3: [4.0]
x4: [1.0]
Objective: 17.532267301