Game Theory

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By Greg Knowles (Eds.)

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You may assume the path traveled is a straight line, and neglect friction, the rotation of the earth, etc. The only stipulation is that the eggs break if the acceleration exceeds 100 ft/sec 2 . 4. Calculate the minimum time to transfer from the initial point (1, 0) to the origin (0, 0) for the system x+X=U, IU(t)1 ~ 1. What are the switching times of the optimal control? 5. Find the optimal trajectories and switching locus for the problem of reaching the origin in minimum time for the control system Xl = X z lUll ~ Iuzl s + Ut> Xz = -Xl + Uz , 1, 1.

The synthesizer can be defined by '¥(Xb X2 ) = { for x2>W(xdandonr_ for X2 < W(Xt) and on I' + - I +1 and the optimal trajectories are solutions of x + 2bx + k 2 x = t/J(x, x), x(O) = Xo, x(O) = Yo. 5. NUMERICAL COMPUTATION OF THE SWITCHING TIMES In this section we discuss a method for the numerical solution of time optimal control problems x = Ax + bu. (1) Suppose the optimal control u* transferring Xo to 0 in minimum time is bang-bang with r switches. Without loss of generality, suppose that the first action of u* is -1 (see Fig.

1 -11------ Fig. 16 30 II. The General Linear Time Optimal Problem In other words, multiplying both sides by A and integrating yields Ax o = _(e- At1b - b) + (e- At2b - e- At1b) + ... , Ax o = b - 2e- At1b + 2e- At2b + ... + (_l)'e-Atrb. (2) If matrix Ais normal (commutes with its adjoint), has eigenvalues AI" .. , .. , e-Atb = n L j= 1 e-Ajt(b' X)Xj . So we can expand (2) as n L Aj(XO • Xj)Xj = L [(b' X)Xj - 2e- Ajt1(b • X)Xj j= 1 + 2e- AjI2(b' X)Xj + .. : + Aj(Xo'Xj) = (b' xj)(l - 2e- Ajt, + 2e- Ajt2 - 2e Ajl3 + ...

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