NAME: 5 MT263 problems Hand in your solutions, with this coversheet, for feedback in the Maths Office (McCrea 118) by 1pm on Monday 19th 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. Write down the dual of the following linear programming problems. Start by transforming the problems into standard form. (a) Minimise z=3x1+ 4x2 subject to x1+4x2 ?8 2x1+3x2 ? 12 2x1+x2 ? 6 x1, X2 ? 0
Solution: maximize 8y1 + 12y2 y1 4y1 + 3y2 y1 6y3 + + 2y2 + , y2 2y3 ?3 + 1y3 ? 4 , y3 ?0 (b) Maximise Z=3x1+2x2+5x3 + 7x4 subject to 3x1+2x2 + x3 ?8 5x1+ 2+ 2x3 + 4x4 = 7 4x1+x3-2x4 ? 12 x1, x2, 13, 14 ? 0 Solution: Transforming the linear program into standard form gives maximize 3x1 + 2x2 + 5x3 3x1 + 2x2 5.x1 + -5.1 ? 4x1 ?1 , x2 , ?3 + 7x4 + 23 ? 8 + 2x3 + 4x4 ? 7 x2 ? 2x3 ? 4x4 ? -7 + x3 ? 2x4 ? 12 , ?4 ? 0 And thus the dual is minimize 8x1 + 7 x2 ? 7x3 + 12x4 3x1 + 5x2 ? 5x3 + 4x4 ? 3 2x1 + ? 23 + ? 2 x1 + 2x2 ? 2x3 + ? X 4 5 4x2 ? 4x3 ? 2x4 ? 7 x2 , x3 , x4 ? 0 x1 , 2. Prove that the dual of a dual linear programming problem is the primal. That is, show that if you start with any LPP (the primal), and you form its dual and take the dual of that, you end up with the original LPP. Solution: Here is the solution in matrix form. You can also do this with the long-hand "sum notation". If the primal is maximise subject to CTx Ax < b x > 0 its dual is minimise subject to AT ATy > c bTy y > 0 Writing this in standard form gives maximise (-b)Ty subject to (-A)Ty ?(-c) y ?0 Then taking the dual gives minimise subject to (-c)Tz ((-A)T)Tz ? (-b) z ? 0
Recalling that (AT)T =