ISM206 Lecture, April 26, 2005 Optimization of Nonlinear Objectives, with Non-Linear Constraints

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1 ISM206 Lecture, April 26, 2005 Optimization of Nonlinear Objectives, with Non-Linear Constraints Instructor: Kevin Ross Scribe: Pritam Roy May 0, 2005 Outline of topics for the lecture We will discuss nonlinear objective functions with non-linear constraints. The topics are: SUMT- Algorithm and example Sequential Linear Approximation Frank-Wolfe Algorithm Duality in Non-linear Programming Duality example from a game Strong duality, Weak duality Lagrangian Duality Convex duality For details also see the reference listed in the class notes, Linear and Nonlinear Programming, by Nash and Sopher.

2 2 SUMT We have seen in the previous lecture that Barrier method is a strictly feasible method. P (x, r) = f(x) r.b(x) () where, f is a maximizing function and B is a barrier function. Consider the following: Maximize f(x) subject to g i (x) b i, x i 0 B(x) = m i= n b i g i (x) + x j= j Note: if b i g i (x), then B(x). Hence the function does not hit the barrier. Idea of SUMT : Solve the unconstrained min P(x,r) over a sequence of times. Each sequence reduces r i.e. allows us to move closer to boundary. Usually go until some error tolerance is reached. If x = max(p (x, r), x max f(x)), then f(x ) f(x ) f(x )+r.b(x ). Step : r new = θr, say θ = 0.. where the stopping condition is : r.b(x) ɛ. 2. Algorithm for SUMT. Let x = intial solution (not on boundary) k =, θ = 0.0 and r = 2. At k-th iteration, x k = starting point, we can use multivariate unconstrained technique e.g. gradient search method to find the local maximum/minimum of P (x, r) = f(x) r.[ 3. Stop if x k x k ɛ. m i= 4. else k = k + and r = θr; go to step 2 n b i g i (x) + ] (2) x j Note: if we had equality constraints i.e. g i (x i ) = b i, then instead of use (b i g i (x)) 2 r. 2 j= r b i g i (x) we would

3 2.2 Example of SUMT Maximize f(x) = 5x x 2 + 8x 2 2x 2 2 subject to 3x + 2x 2 6 x 0, x 2 0 Let us construct the barrier function for the problem. B(x) = 6 3x 2x 2 + x + x 2 (3) and we have to maximize P(x,r) = f(x) - r.b(x). Let us start with r = and θ = 0.. Initial trial solution x =, x 2 =. After the first iteration by gradient search we obtain x = (x, x 2 ) = (0.85,.329) In the second iteration, we take r = 0. and in the third and fourth iteration we value of r as 0.0 and 0.00 respectively. After fourth iteration, we obtain x = (,.5) which is close enough to barrier, so that we can stop the iteration. 3 Sequantial Linear Approximation We already know that LP s can be solved well.for solving NLP we can use the fact. From Taylor s series, n f(x) = f(x δf(x ) ) + (x x ) = f(x ) + f(x )(x x ) (4) δx j j= We know f(x ) and f(x ).x, so our goal is to maximize g(x) = f(x ).x, where f(x ) is a fixed, known constant. Therefore, g(x) = j c jx j, where c j = δf(x) δx j at x. The original problem reduces to solving a linear programming problemmaximize g(x) subject to Ax b, x 0. 4 Frank Wolfe Algorithm It is evident that Linear approximation is VERY much an approximation. The solution x LP, obtained from the linear problem, may not be optimal solution for the 3

4 original problem. We can obtain the optimal solution as a point on line segment [x, x LP ] by a search along the line for the best point of original f(x). h(t) = f(x) on line x k + t.(x LP x k ) (5) We optimize for t to obtain, x k = x k + t.(x LP x k ) (6) 4. Example Maximize f(x) = 32x x 4 + 8x 2 x 2 2 subject to, x x 2 3x + x 2 7 x 0, x 2 0 Let us start with x = (0,0). δf δx (0, 0) = {32 4x 3 }(0, 0) = 32 δf δx 2 (0, 0) = {08 2x 2 }(0, 0) = 08 We now maximize g(x) = 32x + 8x 2 subject to original constraints which yields a solution x = (2, ) and g(2, ) = 72. x = (0, 0) + t[(2, ) (0, 0)] = (2t, t)for0 t.h(t) = f(2t, t) = 72t t 2 6t 4 for0 t. Optimal(maximum) value of t i.e. t is, hence x = (2, ). We get t = and x = (.695,.94) in the next iteration. 4.2 Stopping condition x k x k < ɛ f(x k ) f(x k ) < ɛ 4

5 5 Duality in Non-linear Programming Let us recall from the LP duality results the following facts. The objective of Primal problem is to minimize a function then dual problem has a maximizing objective and vice versa. Dual of dual is primal. objective values give bounds to each other (weak duality). if one has an optimal solution, then so does the other and objective values are the same (strong-duality). We can enlist some differences in duality theory in NLP from LP this is not straight forward as their linear counterpart. lagrangian multipliers are used as dual variables. a problem has to be solved for lambda-values. We care for duality of an optimization problem because dual problem can give an estimate of lagrangian multiplier. good multiplier estimates leads to good solution estimates for the primal problem sometimes dual problems can be solved directly. 6 Duality Example From a Game Let us assume that there are two players P and P2 playing a two player game, where P chooses from X and P2 chooses from Y at the same time. If one player wins, then the other looses i.e. zero-sum game. We assume that both players are rational and want to minimize the worst possible outcome. Let P pays F (x, y) to P2, so worst for P = best for P 2 = F (x) = max y Y F (x, y) (7) P wants this function to have as small value as possible. So the function P want to solve is, min x X max y Y F (x, y) (8) Similarly, P2 will like to solve, max y Y min x X F (x, y) (9) 5

6 We clearly note that two players are solving the dual problems. problems satisfy weak duality, Moreover, two max y Y min x X F (x, y) min x X max y Y F (x, y) (0) 6. Example ( ) 4 Let A = payoff matrix =. Let P chooses rows and P2 chooses columns. 2 3 Here optimal solution for both problems are same. But we can produce some counter example where ( two solutions ) are different. 4 If A = then max-min problem has optimal solution whereas minmax problem has optimal solution Strong Duality, Weak Duality and Convex Duality We have seen in the previous section that weak duality holds in the NLP problems. For strong duality, let us define a term saddle point. Saddle Point: {(x, y ) x X, y Y } is a saddle point if F (x, y) F (x, y ) F (x, y ) () for all x X, y Y. The idea is that x minimizes F (x, y) at y = y and y minimizes F (x, y) at x = x. Theorem of Strong Duality: max y Y min x X F (x, y) = min x X max y Y F (x, y) holds if and only if there exists a pair (x, y ) satisfying the saddle point condition. 8 Lagrangian Duality Min-max duality is the basis for more general Lagrangian Duality. Minimize f(x) subject to, g(x) 0 for x X. Let us define, L(x, λ) = f(x) λ T.g(x) (2) 6

7 The primal problem is, L (x) = max λ 0 L(x, λ) = max λ 0 (f(x) λ T.g(x)) with constraints g(x) 0 for x X. Hence we have, L (x) = f(x) if g(x) 0 (feasible for x) = otherwise Now the primal min-max problem is min x X L (x) (3). Hence, the dual problem will be L (λ) = min x X L(x, λ) (4) So the corresponding max-min problem would be max λ 0 L (λ) i.e. max λ 0 min x X (f(x) λ T.g(x)) (5) 8. Weak Duality Theorem Let x be a feasible solution to primal problem and (x, λ) be the feasible solution pair to dual problem. Then,f(x) λ T.g(x) f(x) leads to, max λ 0 {L (λ)} min x X {f(x) : g(x) 0} (6) Duality Gap: Difference between solutions of primal and dual problem. 9 Convex Duality If f is convex in minimize f(x) subject to g(x) 0 and g is are all concave and assume all functions are continously differentiable and x is a regular point of constraints. Let λ = Lagrangian Multiplier of x, then (x, λ ) is dual feasible, and primal and dual objective values are equal. 7