Due: Friday Nov. 14, beginning of class
Late submissions accepted with 10% penalty until Tues. Nov. 18 at 12:30pm.
I solved this problem with my program and the final dictionary was:
X7 = 440 + 0 X1 + 0 X4 - 1 X8 - 1 X9 + 1 X10+ 1 X11 X6 = 210 + 1 X1 - 1 X4 - 1 X8 - 1 X9 + 1 X10+ 0 X11 X5 = 20 - 1 X1 + 1 X4 + 1 X8 + 0 X9 - 1 X10+ 0 X11 X3 = 400 + 0 X1 - 1 X4 - 1 X8 + 0 X9 + 0 X10+ 0 X11 X2 = 40 - 1 X1 + 0 X4 + 1 X8 + 1 X9 - 1 X10- 1 X11 ------------------------------------------------- z = 4550 - 1 X1 - 1 X4 - 1 X8 - 2 X9 - 7 X10- 3 X11
(a) [2] Explain the meaning of each variable Xj, j=1, 2, 3, ..., 6.
(b) [2] What are the units for the primal problem.
(c) [6] State the dual problem and determine the units for Yi, i=1, 2, 3, 4, 5. Then give an economic interpretation for the solution for the dual.
(d) [4] What is the new values of the objective function if constraints 1-5 change by t1, t2, t3, t4, t5 respectively? (assume the ti's are small)
(e) [4] What is the new optimal solution if constraints 1-5 change by t1, t2, t3, t4, t5 respectively? (assume the ti's are small)
(f) [2] Consider t1, t2, t3, t4, and t5 one at a time. What limit is there to the increase of each before a new basis would result?
X5 = 3 + 1 X1 + 3 X2 + 5 X3 + 1 X4 X6 = -4 + 1 X1 + 2 X2 + 4 X3 - 4 X4 X7 = 2 + 1 X1 + 6 X2 + 2 X3 + 7 X4 X8 = -12 + 1 X1 - 3 X2 + 3 X3 + 3 X4 ----------------------------------------------- z = 0 - 10 X1 + 0 X2 - 24 X3 + 0 X4 The second last dictionary is: X5 = 23.00- 0.67 X1 + 8.00 X2 - 4.00 X4 + 1.67 X8 X6 = 12.00- 0.33 X1 + 6.00 X2 - 8.00 X4 + 1.33 X8 X7 = 10.00+ 0.33 X1 + 8.00 X2 + 5.00 X4 + 0.67 X8 X3 = 4.00- 0.33 X1 + 1.00 X2 - 1.00 X4 + 0.33 X8 ----------------------------------------------- z = -96.00- 2.00 X1 - 24.00 X2 + 24.00 X4 - 8.00 X8 The last dictionary is: X5 = 17.00- 0.50 X1 + 5.00 X2 + 0.50 X6 + 1.00 X8 X4 = 1.50- 0.04 X1 + 0.75 X2 - 0.12 X6 + 0.17 X8 X7 = 17.50+ 0.12 X1 + 11.75 X2 - 0.62 X6 + 1.50 X8 X3 = 2.50- 0.29 X1 + 0.25 X2 + 0.12 X6 + 0.17 X8 ----------------------------------------------- z = -60.00- 3.00 X1 - 6.00 X2 - 3.00 X6 - 4.00 X8
x | y |
---|---|
1 | 11 |
4 | 7 |
7 | 3.7 |
9 | 1 |
The adjacency matrix (with the degree of each vertex indicated):
0 (2): 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 (2): 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 (2): 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 (2): 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 4 (3): 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 5 (2): 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 6 (2): 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 7 (3): 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 8 (3): 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 9 (2): 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 10 (2): 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 11 (2): 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 12 (2): 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 13 (3): 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 14 (2): 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 15 (2): 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 16 (3): 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 17 (3): 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0
(a) [3] How many perfect matchings does this graph have?
Justify you answer by
showing all the perfect matchings of this benzenoid
on the
attached Perfect Matchings worksheet.
Print as many copies as you need to show all the perfect matchings.
(b) [2] For each edge, label the edge on
the attached Hadamard-Fries Algorithm worksheet
with the number
of perfect matchings of the graph containing the edge.
(c) [3] Describe a contrived linear programming problem that could
be solved to find the inverse of the adjacency matrix
for this graph.
There should be one variable per vertex.
(d) [2]
Solve your LP from (c) and explain how you
determine A-1 from the final dictionary.
(e) [3]
This matrix B is equal to the number of perfect matchings
times the inverse of the adjacency matrix:
0: 0 5 0 -5 0 2 0 -2 0 1 0 -1 0 1 0 -1 0 3 1: 5 0 3 0 -3 0 3 0 -1 0 1 0 -1 0 2 0 -2 0 2: 0 3 0 5 0 -2 0 2 0 -1 0 1 0 -1 0 1 0 -3 3: -5 0 5 0 3 0 -3 0 1 0 -1 0 1 0 -2 0 2 0 4: 0 -3 0 3 0 2 0 -2 0 1 0 -1 0 1 0 -1 0 3 5: 2 0 -2 0 2 0 6 0 -2 0 2 0 -2 0 4 0 -4 0 6: 0 3 0 -3 0 6 0 2 0 -1 0 1 0 -1 0 1 0 -3 7: -2 0 2 0 -2 0 2 0 2 0 -2 0 2 0 -4 0 4 0 8: 0 -1 0 1 0 -2 0 2 0 3 0 -3 0 3 0 -3 0 1 9: 1 0 -1 0 1 0 -1 0 3 0 5 0 -5 0 2 0 -2 0 10: 0 1 0 -1 0 2 0 -2 0 5 0 3 0 -3 0 3 0 -1 11: -1 0 1 0 -1 0 1 0 -3 0 3 0 5 0 -2 0 2 0 12: 0 -1 0 1 0 -2 0 2 0 -5 0 5 0 3 0 -3 0 1 13: 1 0 -1 0 1 0 -1 0 3 0 -3 0 3 0 2 0 -2 0 14: 0 2 0 -2 0 4 0 -4 0 2 0 -2 0 2 0 6 0 -2 15: -1 0 1 0 -1 0 1 0 -3 0 3 0 -3 0 6 0 2 0 16: 0 -2 0 2 0 -4 0 4 0 -2 0 2 0 -2 0 2 0 2 17: 3 0 -3 0 3 0 -3 0 1 0 -1 0 1 0 -2 0 2 0