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final_sol_07

# final_sol_07 - EE263 Dec 78 or Dec 89 2007 Prof S Boyd...

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EE263 Prof. S. Boyd Dec. 7–8 or Dec. 8–9, 2007. Final exam solutions 1. Optimal initial conditions for a bioreactor. The dynamics of a bioreactor are given by ˙ x ( t ) = Ax ( t ), where x ( t ) R n is the state, with x i ( t ) representing the total mass of species or component i at time t . Component i has (positive) value (or cost) c i , so the total value (or cost) of the components at time t is c T x ( t ). (We ignore any extra cost that would be incurred in separating the components.) Your job is to choose the initial state, under a budget constraint, that maximizes the total value at time T . More specifically, you are to choose x (0), with all entries nonnegative, that satisfies c T x (0) B , where B is a given positive budget. The problem data ( i.e. , things you know) are A , c , T , and B . You can assume that A is such that, for any x (0) with nonnegative components, x ( t ) will also have all components nonnegative, for any t 0. (This occurs, by the way, if and only if the off-diagonal entries of A are nonnegative.) (a) Explain how to solve this problem. (b) Carry out your method on the specific instance with data A = 0 . 1 0 . 1 0 . 3 0 0 0 . 2 0 . 4 0 . 3 0 . 1 0 . 3 0 . 1 0 0 0 0 . 2 0 . 1 , c = 3 . 5 0 . 6 1 . 1 2 . 0 , T = 10 , B = 1 . Give the optimal x (0), and the associated (optimal) terminal value c T x ( T ). Give us the terminal value obtained when the initial state has equal mass in each component, i.e. , x (0) = α 1 , with α adjusted so that the total initial cost is B . Compare this with the optimal terminal value. Also give us the terminal value obtained when the same amount, B/n , is spent on each initial state component ( i.e. , x (0) i = B/ ( nc i )). Compare this with the optimal terminal value. Solution. (a) We have c T x ( T ) = c T e tA x (0) = b T x (0), where we define b = ( e TA ) T c , so our problem is to maximize b T x (0), subject to x (0) 0 (this means all its entries are nonnegative), and c T x (0) B . You can think of c i as the cost of investing in a unit of component i , and b i as the payoff received. Thus, the gain is b i /c i . The 1

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solution to this problem is to invest everything ( i.e. , the whole budget B ) in any component that has maximum gain. More formally, we choose any k for which b k /c k = max { b 1 /c 1 , . . . , b n /c n } , and then set x (0) = Be k . (Recall that we assume b i 0 and c i > 0 here.) We didn’t require a completely formal proof that this is the optimal strategy. But here is one, just so you know what one looks like. Suppose that x (0) satisfies x (0) 0, c T x (0) B . Then we have b T x (0) = n summationdisplay i =1 ( b i /c i )( c i x (0) i ) parenleftbigg max i =1 ,...,n ( b i /c i ) parenrightbigg parenleftBigg n summationdisplay i =1 c i x (0) i parenrightBigg B max i =1 ,...,n ( b i /c i ) . This shows that no feasible choice of x (0) can yield terminal value c T x ( T ) = b T x (0) more than B max i =1 ,...,n ( b i /c i ). But the choice described above yields this value of b T x (0), and so must be optimal.
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