NAME: 2 MT263 problems Hand in your solutions, with this coversheet, for feedback to the Maths Office (McCrea 118) by 1pm on Monday 28th Jan. 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. Construct initial feasible dictionary for maximise 5x1+ 5x2 subject to <1 2x1 + 4x2 -2x1+3x2 <0 4x1 - X2 < 2 x1, x2 ?0. Solution: x3 = x4 = x5 = 2 ? z = 1 2x1 ? ? 4x1 + 5x1 + 4x2 3x2 ? 2x1 x2 5.x2
2. Is a feasible dictionary for = 2x1 + 2x2 = + x2 2 = x1 + 2x2 maximise x1+2x2 subject to 2x1 + 4x2 ?1 x1+ 22 x1, X2 < - 2 ? ?0 Solution: No, as x1 = ^2 = 0 is not a feasible solution to the LPP (0 + 0 > -2). 3. Write down the LPP problem and the solution corresponding to the dictionary x3 = 2 + ?4 = 4 ? ?1 + + 3x2 z = ?1 + 2x2 Solution: maximise subject to 21+2x2 -x1-3x2 <2 x1-X2 ?4 x1, x2 ?0. The solution is x1 =0, x2 =0, x3 = 2, 04 = 4 4. Each of the following dictionaries appears as a step in the simplex method for some LPP. For each dictionary, complete the next step of the simplex method (i.e., do one pivot, or if the dictionary is final solve the LPP.) = 2 + x1 + (a) ?4 = 4 ? ?1 z = 3x2 ? + x2 + 2x2 ?1 Solution: x2 is the entering variable. 0 ? x3 = 2 +3x2 and 0 ? x4 = 4 + x2 put no bounds on 22 so the problem is unbounded. (b) x4 = 1 ? 2x5 x3 = 3 ? 2x1 + 4x2 ? 6x5 x6 = 2 + x1 ? 3x2 ? 4x5 z = -13 ? 2x1 ? ? 805 Solution: The dictionary is optimal, giving the optimal solution x1 = 0, x2 = 0, x3 = 3, x4 = 1, x5 = 0, x6 = 2, and optimal value z = - 13. (c) ?4 = 1 ? 2x3 = 3 ?