By Hugo. Rossi
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Extra info for Advanced calculus. Problems and applications to science and engineering
Aji+1 .. . Ajk+1 .. . Ajm ]. Let E be the . . ˆ. . matrix E = [Aj1 .. . Aji+1 .. . Ajm ]. Then [D(z k+1 )]−1 = P E −1 where the matrix P permutes the columns of D(z k+1 ) such that E = D(z k+1 )P . Next, if ˆ Aj m = =1 to check that E −1 = M [D(z k )]−1 where γj Aj , it is easy 1 1 . .. 1 M = −γ j1 γ˜j 1 γ˜j −γ jm γ˜j 1 . . 1 ↑ ith column Then [D(z k+1 )]−1 = P M [D(z k )]−1 , so that these inverses can be easily computed. 4 is striking. The basic variables at z k correspond to the variables wk and non-basic variables correspond to uk .
Ii) Each activity combines the k inputs in fixed proportions into the m outputs in fixed proportions. Furthermore, each activity can be conducted at any non-negative intensity or level. Precisely, the jth activity is characterized completely by two vectors Aj = (a1j , a2j , . . , amj ) and B j = (bij , . . , bkj ) so that if it is conducted at a level xj ≥ 0, then it combines (transforms) the input vector (a1j xj , . . , amj xj ) = xj Aj into the output vector (b1j xj , . . , bkj xj ) = xj B j .
We now describe how to obtain an initial basic feasible solution. Phase I: Step I. 24) by −1 if necessary, we can assume that b ≥ 0. 32) involving the variables x and y: m Maximize − yi i=1 subject to ail x1 + . . + ain xn + yi = bi , 1 ≤ i ≤ m , xj ≥ 0 , y i ≥ 0 , 1 ≤ j ≤ n , 1 ≤ i ≤ m . 32) CHAPTER 4. LINEAR PROGRAMMING 42 Go to step 2. Step 2. 32). 32) starting with this solution. 32) lies between − bi and 0. Go to Step 3. i=1 y∗ x∗ y∗ Step 3. 24). 24) has no feasible solution. 24) has a feasible solution iff y ∗ = 0.