To start from [0 0]' and allow a maximum of 25 linesearches to minimize the function, do:

>> [x fx c] = minimize([0 0]', 'rosenbrock', 25) Linesearch 20; Value= 2.724035e-30 x = 1.00000000000000 1.00000000000000 fx = 0.78251060908545 0.39556533940410 0.21891183735928 0.08594362786359 0.08167279250615 0.04803799828332 0.03385004042904 0.01810891772958 0.01747706436156 0.01719294050541 0.00647710313861 0.00317008743069 0.00002390064486 0.00000859309602 0.00000859309029 0.00000076626706 0.00000000009216 0.00000000000000 0.00000000000000 0.00000000000000 c = 22this shows that the minimum (at x = [1 1]') was found after c = 22 iterations and fx contains the function value after each iteration. If instead we want to limit the search to 25 function evaluations:

>> [x fx c] = minimize([0 0]', 'rosenbrock', -25) Function evaluation 23; Value= 1.810892e-02 x = 0.86905063456945 0.75343426564357 fx = 0.78251060908545 0.39556533940410 0.21891183735928 0.08594362786359 0.08167279250615 0.04803799828332 0.03385004042904 0.01810891772958 c = 25where the function value has been reduced to 0.018. Note that in this example each iteration (linesearch) used an average of around 3 function evaluations.

>> x0 = zeros(100, 1); >> a = fminunc('rosenbrock', x0, optimset('GradObj', 'on', 'MaxFunEval', 20)); >> disp([rosenbrock(x0) rosenbrock(a)]) 99.00000000000000 88.38645133447895reduces the function from 99 to around 88 using 2121 function and gradient evaluations (don't ask me why it calls the function 2121 times when MaxFunEval is set to 20). Using minimize:

>> x0 = zeros(100, 1); >> a = minimize(x0, 'rosenbrock', -2121); Function evaluation 2120; Value= 6.500827e-06 >> disp([rosenbrock(x0) rosenbrock(a)]) 99.00000000000000 0.00000650082744showing considerably more progress in the same number of function evaluations. For larger problems, the difference is even more severe: for the 1000 dimensional Rosenbrock problem, minimize converges in 19640 function evaluations to a value of 5.552907e-26, whereas fminunc reduces the function value from 999 (at zero) to 980 in 21021 function evaluations (if you want to replicate this experiment, be aware that for some reason matlab's fminunc is computationally extremely slow for the 1000 dimensional problem).

>> randn('seed', 21); >> checkgrad('rosenbrock', randn(3,1), 1e-5) 1.0e+02 * 9.91926964244770 9.91926964286449 -3.76300059587095 -3.76300059622281 0.36091809447947 0.36091809448635 ans = 2.569411416671008e-11to verify that the partial derivatives and finite differences approximation have very close values at some random point.