NAME: 3 MT263 problems Hand in your solutions, with this coversheet, for feedbacko the Maths Office (McCrea 118) by 1pm on Monday 4th Feb. I will keep a record of how many questions you made a decent attempt on. I can provide you with additional feedback if you want it. Many students don't collect their feedback so if you don't want additional feedback, please tick the relevant box below. Otherwise, if you make a decent attempt of N questions, you can request additional feedback on [+] questions (note the use of the floor function here). Please indicate which questions you'd like feedback on in the relevant box below (if you don't indicate what feedback you want, I'll assume you don't want any). • Put a tick in this box if you don't want additional written feedback on your solutions. · I attempted questions. Please provide me with feedback on questions YOUR FEEDBACK TO THE LECTURER (e.g., what was hard, interesting, fun, etc. this week): FEEDBACK ON YOUR SOLUTIONS FROM THE MARKER: 1. Consider maximise -2x1- 12 subject to -x1 + 22 ? -1 -x1 - 2x2 ?-2 x2 <1 > 0. x1, X2 (a) Write down the auxiliary problem. (b) Find a feasible dictionary for the auxiliary problem. (c) Use this to find a feasible dictionary for the original problem. (If at some stage you have a choice between x1 and x2 for entering variable, choose x1.) (d) Solve the original problem.
(For this question, you should obtain an optimal value of -3.) Solution: The auxiliary problem is maximise subject to -x0 -x1+ 22 - 20 -x1- 2x2 - 20 22-20 xo, x1, X2 <- 1 < - 2 <1 ? 0. The corresponding (infeasible) dictionary is x3 = -1 = -2 2 5 = 1 + w = + + x1 + + x1 ? ? x2 + 2x2 ? X2 Pivoting x0 (entering) and x4 (leaving) gives a feasible dictionary for the auxiliary problem: = 1 = 2 ? x5 = 3 ? ?1 w = -2 + x1 ? 3x2 + ? 2x2 + x 4 ? 3x2 + We choose x1 as entering variable. For the leaving variable we get x0. Pivoting gives the dictionary = 2 ? ? 2x2 + X4 ? + x3 = 1 ? 3x2 + x5 = 1 + ? x2 w = ? Since x0 is now a nonbasic variable, the original LPP is feasible. Setting x0 = 0 and substituting @1 = 2 - 2x2 + x4 into the objective function z = - 2x1 - 22 gives z = - 4 + 3x2 - 2x4. The initial feasible dictionary for the original LPP is thus ?1 = 2 2x2 ? x3 = 1 3x2 ? x5 = ? 1 -4 z = + 3x2 + ?4 + ? 2x4 Using the ordinary simplex method, we now get the following optimal dictionary x2 = 0.33 ?1 = x5 = z = ? 0.33x3 1.33 + 0.67 + -3 ?