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Mathematics, 17.07.2019 19:10, froyg1234
2. {10 marks} consider a generic lp (p) in sef. max{cx : ax = b, x > 03. we previously mentioned a certificate of unboundedness, that is, if there exist a feasible solution ī and a vector d such that ad=0,d > 0,c7d > 0, then (p) is unbounded. the goal of this question is to prove the converse of this, that is, if (p) is unbounded, then such a certificate must exist. we start by considering the following linear program (p') using d as the variable. max{cid : ad = 0,d > 0}. suppose (p) is unbounded. (a) write down the dual (d) of (p), and the dual (d') of (p'). (b) prove that (d) and (d') are both infeasible. (c) prove that (p') is unbounded. (d) prove that there exist a feasible ī and a vector d such that ad = 0,> 0, cd > 0.
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Mathematics, 22.06.2019 01:00, sdwhitneyhillis
Arrange the steps to solve this system of linear equations in the correct sequence. x + y = -2 2x – 3y = -9 tiles subtract 3x + 3y = -6 (obtained in step 1) from 2x – 3y = -9 (given) to solve for x. substitute the value of x in the first equation (x + y = -2) to get y = 1. the solution for the system of equations is (-3, 1). x = -15 the solution for the system of equations is (-15, 13). add 3x + 3y = -6 (obtained in step 1) to 2x – 3y = -9 (given), and solve for x. x = -3 substitute the value of x in the first equation (x + y = -2) to get y = 13. multiply the first equation by 3: 3(x + y) = 3(-2) 3x + 3y = -6.
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2. {10 marks} consider a generic lp (p) in sef. max{cx : ax = b, x > 03. we previously mentione...
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