X5 = 12.00+ 2.00 X1 + 3.00 X2 + 1.00 X3 - 4.00 X4 X6 = 8.00+ 1.00 X1 - 8.00 X2 + 4.00 X3 - 4.00 X4 X7 = 6.00+ 2.00 X1 - 6.00 X2 + 2.00 X3 - 3.00 X4 -------------------------------------------------------- z = 17.00- 10.00 X1 + 6.00 X2 + 2.00 X3 + 10.00 X4Which variables enter and exit if
subject to 1.00 X1 - 4.00 X2 - 2.00 X3 <= -4 1.00 X1 + 1.00 X2 + 1.00 X3 <= 10 -1.00 X1 + 2.00 X2 - 1.00 X3 <= -8 -1.00 X1 - 2.00 X2 + 4.00 X3 <= -2 X1, X2, X3 >= 0
(a) [5] Set up the initial phase 1 dictionary and show the dictionary resulting from the first pivot.
(b) [5] The final dictionary is this one:
X1 = 7.38- 0.04 X4 - 0.46 X5 + 0.32 X6 + 0.18 X7 X2 = 1.41+ 0.14 X4 - 0.14 X5 - 0.08 X6 + 0.08 X7 X8 = 1.11+ 0.20 X4 + 0.30 X5 + 0.38 X6 + 0.12 X7 X3 = 2.32+ 0.11 X4 - 0.11 X5 + 0.14 X6 - 0.14 X7 ----------------------------------------------- z = -1.11- 0.20 X4 - 0.30 X5 - 0.38 X6 - 0.12 X7what does this tell you about the original problem?
X5 = 6- 1 X1 + 1 X2 + 0 X3 + 0 X4 X6 = 6- 1 X1 + 1 X2 - 1 X3 + 1 X4 X7 = 2+ 1 X1 - 1 X2 - 1 X3 + 1 X4 X8 = -4+ 0 X1 + 0 X2 + 1 X3 - 1 X4 X9 = 4+ 1 X1 - 1 X2 + 1 X3 - 1 X4 ----------------------------------------------- z = 0- 1 X1 + 1 X2 - 1 X3 + 1 X4
The initial Phase 1 dictionary looks like this: Phase 1: Before pivoting to make feasible. X5 = 6- 1 X1 + 1 X2 + 0 X3 + 0 X4 + 1 X0 X6 = 6- 1 X1 + 1 X2 - 1 X3 + 1 X4 + 1 X0 X7 = 2+ 1 X1 - 1 X2 - 1 X3 + 1 X4 + 1 X0 X8 = -4+ 0 X1 + 0 X2 + 1 X3 - 1 X4 + 1 X0 X9 = 4+ 1 X1 - 1 X2 + 1 X3 - 1 X4 + 1 X0 ----------------------------------------------- z = 0+ 0 X1 + 0 X2 + 0 X3 + 0 X4 - 1 X0This standard form problem S was created from another problem P that was not in standard form because:
X5 = 4.00+ 0.00 X2 + 0.00 X4 + 0.75 X6 - 0.25 X7 + 0.50 X8 X1 = 2.00+ 1.00 X2 + 0.00 X4 - 0.50 X6 + 0.50 X7 + 0.00 X8 X3 = 4.00+ 0.00 X2 + 1.00 X4 - 0.25 X6 - 0.25 X7 + 0.50 X8 X0= 0.00+ 0.00 X2 + 0.00 X4 + 0.25 X6 + 0.25 X7 + 0.50 X8 X9 = 10.00+ 0.00 X2 + 0.00 X4 - 0.50 X6 + 0.50 X7 + 1.00 X8 ----------------------------------------------- z = -0.00+ 0.00 X2 + 0.00 X4 - 0.25 X6 - 0.25 X7 - 0.50 X8Which variable should enter and which one should leave before starting phase 2?
X5 = 4.00+ 0.00 X2 + 0.00 X4 - 1.00 X7 - 1.00 X8 X1 = 2.00+ 1.00 X2 + 0.00 X4 + 1.00 X7 + 1.00 X8 X3 = 4.00+ 0.00 X2 + 1.00 X4 + 0.00 X7 + 1.00 X8 X6 = -0.00+ 0.00 X2 + 0.00 X4 - 1.00 X7 - 2.00 X8 X9 = 10.00+ 0.00 X2 + 0.00 X4 + 1.00 X7 + 2.00 X8 ----------------------------------------------- z = -6.00+ 0.00 X2 + 0.00 X4 - 1.00 X7 - 2.00 X8What are the values for xi for i from 1 to 9, and for z that correspond to this dictionary?
(a) [5] Set up the initial dictionary for the phase 1
problem then take the first pivot to make the
current solution feasible.
(b) [2] Show the dictionary that results after one more
pivot that has x1 entering
and x4 leaving.
(c) [5] Note that X0 is still in the basis after the pivot from (b) yet the solution is optimal.
Do what is needed to prepare the dictionary
to start phase 2, explaining what you are doing
and why.
(d) [3] Finish solving the problem either by hand
or using your program and explaining what you
are doing.
Y4 = 2.0+ 0.0 Y1 + 0.5 Y5 + 0.5 Y6 Y2 = 2.0+ 1.0 Y1 - 0.5 Y5 + 0.5 Y6 Y3 = 5.0+ 0.0 Y1 + 0.0 Y5 + 1.0 Y6 ----------------------------------------------- z =-21.0+ 0.0 Y1 - 1.0 Y5 - 4.0 Y6 The optimal solution: -21 Y1=0, Y2=2, Y3=5, Y4=2, Y5=0, Y6=0How can you determine what the primal solution should be from the information in this final dual dictionary? That is, explain how to compute x1, x2, and x3 and z for the primal from this last dual dictionary. Hint: you can use this to double check your work from the previous question.
Maximize 2 X1 - 1 X2 subject to 2 X1 - 2 X2 <= 2 -3 X1 + 3 X2 <= -6 X >= 0(a) [3] Find the solution to the primal problem either by hand or by using your computer program.
For the middle school classes, he has 4 students per class and each one pays a fee of $15 per hour. These classes require 2 tutors. He pays his tutors the minimum wage- $8 per hour. For the high school classes, Tom has 4 students per class each paying $10 per hour, but only one tutor is required for these. For the adult classes he offers one on one tutoring (one adult and one tutor) and the fee per student is $30 because of this individual attention. He has 14 tutor hours available per week.
In one hour of each of the middle school and high school classes, Tom uses 8 pages of worksheets per student for each of the 4 students so a total of 32 worksheet pages per hour for each of these classes. The adults are expected to get their course resources off the web and print them themselves. But he must photocopy the worksheets for the other students and he has a photocopying quota of 256 pages per week.
(a) [2] State the problem of maximizing Tom's profit per week as a linear programming problem.
(b) [4] Tom's optimal solution involves classes for all three groups of students. Find the inverse of the basis matrix for this situation. Show your work.
(c) [4] Use your result from (b) to determine the final dictionary for the Simplex method (including the z row).