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Be sure this exam has 5 pages. THE UNIVERSITY OF BRITISH COLUMBIA Sessional Examination - December 18, 2010 MATH 340: Linear Programming Instructor: Dr. R. Anstee, Section 101
Special Instructions: No calculators. You must show your work and explain your answers. Quote names of theorems used as appropriate. Time: 3 hours Total marks: 100
- [12 marks]
a) [10pts] Solve the following linear programming problem, using our standard two phase method and using Anstee’s rule.
Maximize 5 x 1 +x 2 −x 3 3 x 1 +x 2 −x 3 ≤ − 2 x 1 −x 2 − 2 x 3 ≤ − 3 x 1 ≤ 2
x 1 , x 2 , x 3 ≥ 0
b) [2 marks] Give an optimal dual solution. How can you verify it is optimal?
- [12 marks] Consider the following linear program:
Maximize 10 x 1 +14x 2 +20x 3 3 x 1 +x 2 +3x 3 ≤ 45 2 x 1 +2x 2 ≤ 20 x 1 +2x 2 +5x 3 ≤ 15
x 1 , x 2 , x 3 ≥ 0
a) [2 marks] Give the Dual Linear Program of the above Linear Program.
b) [2 marks] State the Theorem of Complementary Slackness including a description of the conditions of complementary slackness.
c) [6 marks] You are given that an optimal primal solution has x∗ 1 = 10, x∗ 2 = 0, x∗ 3 = 1. Determine an optimal dual solution (without pivoting), stating which theorems you have used.
d) [2 marks] Does the dual solution determined in c) remain optimal if we replace the first two inequalities of the primal by 3x 1 − x 2 + 3x 3 ≤ 45 and 2x 1 + 3x 2 ≤ 20? Explain.
- [8 marks] Given A, b, c, current basis (and B−^1 for your computational ease), use our Revised Simplex method and Anstee’s rule to determine the next entering variable (if there is one), the next leaving variable (if there is one), and the new basic feasible solution after the pivot (if there is both an entering and leaving variable). The current basis is {x 7 , x 3 , x 4 }.
x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 5 1 3 1 1 1 0 0 x 6 − 1 4 1 2 0 1 0 x 7 − 1 − 1 0 − 1 0 0 1
b x 5 5 x 6 7 x 7 − 1
B− (^1) =
x 5 x 6 x 7 x 7 − 1 1 1 x 3 2 − 1 0 x 4 − 1 1 0
( x^1 x^2 x^3 x^4 x^5 x^6 x^7 c 1 9 2 3 0 0 0
)
- [25 marks] A manufacturer wishing to maximize profit can make three possible house types made from the three available resources according to the following table. We are not concerned with integrality in this question and allow fractional houses (we can think of our answer as an optimal product mix not a specific set of houses)
house 1 house 2 house 3 total available wood 1 2 1 6 labour 2 5 3 15 capital 3 6 4 19
$ profit 4 9 5
Let xi denote the amount of house i to produce and let x3+i denote the ith slack for i = 1, 2 , 3. The final dictionary is:
x 1 = 1 − 2 x 4 +2x 5 −x 6 x 2 = 2 −x 4 −x 5 +x 6 x 3 = 1 +3x 4 −x 6 z = 27 − 2 x 4 −x 5
B−^1 =
x 4 x 5 x 6 x 1 2 − 2 1 x 2 1 1 − 1 x 3 − 3 0 1
NOTE: All questions are independent of one another.
a) [2 marks] Give the marginal values for each of the resources wood, labour and capital.
b) [3 marks] Consider a new house (say house 4) with requirements 2 units of wood, 2 units of labour and 3 units of capital and profit of $7. Are you interested in producing this new house. Explain.
c) [5 marks] Give the range on c 3 (profit for house 3) so that the current solution remains optimal.
d) [5 marks] Give the range on b 2 (resource availability for labour) so that the current basis remains optimal. Also give the profit as a linear function of b 2 in that range. Hint for e),f): You need not complete all of the very final dictionary, merely the variables in the basis and the constants and all the entries in the z row.
e) [5 marks] Given resource availabilities of
, obtain (using the Dual Simplex method) the
new optimal solution as well as the new marginal values.
f) [5 marks] Consider adding a new constraint x 1 + x 2 ≤ 2 to our original problem. Solve using the Dual Simplex method. Report the new solution as well as the new marginal values.
The following is the output from LINDO:
OBJECTIVE FUNCTION VALUE
VARIABLE VALUE REDUCED COST
STRAT1 1.500000 0.
STRAT2 0.000000 5.
STRAT3 3.500000 0.
ROW SLACK OR SURPLUS DUAL PRICES
CAPITAL) 325.000000 0.
LABOUR) 140.000000 0.
SPACE) 6.000000 0.
HABITAT) 0.000000 26.
POLITICS) 0.000000 -63.
NO. ITERATIONS= 0
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
STRAT1 15.000000 63.000000 INFINITY
STRAT2 21.000000 5.000000 INFINITY
STRAT3 26.000000 INFINITY 5.
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
RHS INCREASE DECREASE
CAPITAL 1000.000000 INFINITY 325.
LABOUR 600.000000 INFINITY 140.
SPACE 23.000000 INFINITY 6.
HABITAT 8.000000 1.272727 3.
POLITICS 1.500000 1.166667 0.
- [5 marks] Consider a two person zero sum game whose payoff matrix for player 1 (the ‘row’ player) is A =
( 5 3 1 7 9 12
)
Give explicitly the LP that determines the optimal strategy for player 2 (the ‘column’ player). Do not solve. What does your objective function measure?
- [10 marks] Let A be an m × n matrix, C be an p × n matrix, b be an m × 1 vector and d be a p × 1 vector. Show that either
i) There exists an x with Ax ≤ b, Cx = d or
ii) There exists y, z with AT^ y + CT^ z = 0 , y ≥ 0 and b · y + d · z < 0
but not both.
Name theorems used as you use them.
- [10] Let A, b be given. Consider the following LP with c not yet specified:
LP (c) :
max c · x Ax ≤ b x ≥ 0
Assume that when c = d we have that B yields an optimal basis and that x∗^ is the optimal solution for LP(d) in this case.
a) [2 marks] Show that x∗^ is still an optimal solution for LP(1. 02 × d) namely when c = 1. 02 × d (this might correspond to an inflation of 2%).
b) [8 marks] Let e, f be two vectors such that B is still an optimal basis when c = e and also when c = f. Show that x∗^ is an optimal solution for LP(2 × e + 3 × f), namely when c = 2 × e + 3 × f.
- [5 marks] Consider a ‘battleship’ type game played on a 4×4 board. Player 1 secretly chooses a location for a domino (there are 24 possibilities but that is not so crucial to answering this question). Player 2 secretly chooses a position (among the 16 different possibilities). Player 1 wins $1 (and player 2 loses $1) if the domino does not occupy a position chosen by player 2 else player 2 wins $1 (and player 1 loses $1). One would quess that the value of the game is 12/16. Give a proof of this fact. Explicitly considering the 24 × 16 payoff matrix would probably be unproductive but you can use properties of the payoff matrix.
