Inequality constrained minimization: log-barrier method
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1 Inequality constrained minimization: log-barrier method We wish to solve min c T x subject to Ax b with n = 50 and m = 100. We use the barrier method with logarithmic barrier function m ϕ(x) = log( (a T i x b i )) i=1 and solve a sequence of smooth unconstrained problems x (t) = argmin x R n tc T x + φ(x) Objective functional augmented with the log-barrier function [f,g,h] = objective_barrier(t,x,a,b,c); [m,n] = size(a); d = A*x - b; D = diag(1./d); f = t*c *x - log(-d )*ones(m,1); g = t*c - A *D*ones(m,1); H = A *D^2*A; Problem parameters m = 100; n = 50; ALPHA =.1; BETA =.7; mu = 50; A = randn(m,n); b = 1 + abs(randn(m,1)); c = randn(n,1); Smooth unconstrained minimization Start with strictly feasible point x = 0 Terminates when t = 10 8 (Duality gap 10 6 ) Centering uses Newton method with backtracking
2 x = zeros(n,1); t = 1; histobj=[]; NTTOL = 1e-10; MAXITERS = 500; % stop inner iteration if lambda^2/2 < NTTOL while (t <= 1e8), % Outer loop niter = 0; for k=1:maxiters, % Inner loop [val,g,h] = objective_barrier(t,x,a,b,c); v = -H\g; lambda = g *v; % Newton step % Newton decrement % Perform backtracking line search along search direction s = 1; while (min([b - A*(x+s*v) ]) < 0), s = BETA*s; % first get feasible point... then search minimum while (objective_barrier_val(t,x+s*v,a,b,c) > val + ALPHA*s*lambda), s = BETA*s; x = x+s*v; niter = niter + 1; % Test if optimum achieved if (abs(lambda/2) < NTTOL), break; % decrement smaller than NTTOL? % Display progress obj = c *x; histobj=[[histobj],[obj;niter;m/t]]; % Bookkeeeping disp([ obj:,num2str(obj, %1.6e ), ; PDGap:, num2str(m/t, %1.2e ), ; number iterations:,int2str(niter)]); % Update t = mu*t;
3 Plot results (more Matlab commands to produce the nice plot below) PDGap = histobj(3,:); niter = histobj(2,:); total_iter = cumsum(niter); figure; semilogy(total_iter,pdgap, * ); mu = 2 mu = 50 mu = Duality gap Total Number of iterations Figure 1: Plot of duality gap vs. total number of Newton iterations for µ = 2, 50, 150
4 Total Number of iterations mu Figure 2: Trade-off between µ and the total number of Newton iterations needed to reduce the duality gap from 100 to The optimization problem is a moderately small inequality constrained LP, just as before. This shows that the method is not very sensitive to the value of µ provided µ 10
5 Figure 11.7 Progress of barrier method for three randomly generated standard form LPs of different dimensions, showing duality gap versus cumulative number of Newton steps. The number of variables in each problem is n = 2m. Here too we see approximately linear convergence of the duality gap, with a slight increase in the number of Newton steps required for the larger problems. The following figures are taken from our textbook (Boyd and Vandenberghe). 35 PSfrag replacements Newton iterations m Figure 11.8 Average number of Newton steps required to solve 100 randomly generated LPs of different dimensions, with n = 2m. Error bars show standard deviation, around the average value, for each value of m. The growth in the number of Newton steps required, as the problem dimensions range over a 100:1 ratio, is very small.
6 PSfrag replacements Interior-point methods duality gap duality gap µ = 50 µ = 200 µ = 2 PSfrag replacements Newton iterations Figure Progress of barrier method for an SOCP, showing duality gap versus cumulative number of Newton steps with xproblems R 50, m with = 50, generalized and A i inequalities R The problem instance was randomly 603 generated, in such a way that the problem is strictly primal and dual feasible, and has optimal value p = 1. We start with a point x (0) on the central path, with a duality gap of The barrier method is used to solve the problem, using the barrier function 120 m 100 φ(x) = log ( (c T i x + d i ) 2 A i x + b i 2 2). Newton iterations 80 i=1 The centering problems are solved using Newton s method, with the same algorithm 60 parameters as in the examples of : backtracking parameters α = 0.01, β = 0.5, and a stopping 40 criterion λ(x) 2 / Figure shows the duality gap versus cumulative number of Newton steps. The plot is very20similar to those for linear and geometric programming, shown in figures 11.4 and 11.6, respectively. We see an approximately constant number 0 of Newton steps required 0 40 per centering 80 step, 120and therefore 160 approximately 200 linear convergence of the duality gap. For this µ example, too, the choice of µ has little effectfigure on the11.16 totaltrade-off numberinofthe Newton choicesteps, of the provided parameter µ µ, isfor ataleast small10 SOCP. or so. As in the examples The vertical foraxis linear shows andthe geometric total number programming, of Newton steps a reasonable required to choice reduce of µ is in the duality gap from 100 to 10 the range , which results in 3, and the horizontal axis shows µ. a total number of Newton steps around 30 (see figure 11.16). with A small variable SDPx R 100, and F i S 100, G S 100. The problem instance was generated randomly, in such a way that the problem is strictly primal and dual feasible, Our next with example p = 1. is an The SDP initial point is on the central path, with a duality gap of 100. minimize c T x We apply the barrier method with logarithmic n barrier function (11.46)
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