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By Richard Bellman

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1) is −1 1 1 −1 0 1 Aδ = u = −1 1 0 1 1 1 Adding two times the last row to 1 3 1 2 Bδ = r = 1 3 1 1 27 equations Aδ = u corresponding to 0 δ1 0 1 δ2 = 0 . 12) the other rows, the system becomes δ1 2 3 2 δ2 2 3 3 = . 13) 2 3 δ3 2 δ4 1 1 1 The matrix B and the vector r are now positive. 13). The MART iteration is as follows. , let δjk+1 = δjk ri (Bδ k )i m−1 i Bij . 2)T . 1, equals g(t∗ ). 2 , so that, after a little arithmetic, we discover that v(δ ∗ ) = g(t∗ ) = 100; the lowest cost is one hundred dollars.

Solving the DGP Problem . . . . . . . . . . . . . . . . . . . . 1 The MART . . . . . . . . . . . . . . . . . . . . . . . 2 MART I . . . . . . . . . . . . . . . . . . . . . . . . . 3 MART II . . . . . . . . . . . . . . . . . . . . . . . . . 4 Using the MART to Solve the DGP Problem . . . . . . Constrained Geometric Programming . . . . . . . . . . . . . . Exercises . .

The Cauchy Inequality then tells us that (bT s)2 = (bT CC −1 s)2 ≤ [bT CC T b][sT (C −1 )T C −1 s], with equality if and only if the vectors C T b and C −1 s are parallel. It follows that b = α(CC T )−1 s = αQ−1 s, for any constant α. 9 1 Q−1 s. sT Q−1 s An Inner Product for Square Matrices The trace of a square matrix M , denoted trM , is the sum of the entries down the main diagonal. Given square matrices A and B with real entries, the trace of the product B T A defines the trace inner product, that is A, B = tr(B T A), 12 A First Course in Optimization where the superscript T denotes the transpose of a matrix.