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Be sure this exam has 5 pages. THE UNIVERSITY OF BRITISH COLUMBIA Sessional Examination - December 6, 2011 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 x 1 −x 2 +x 3 2 x 1 −x 2 ≤ − 3 x 1 −x 2 +x 3 ≤ − 2 − 2 x 2 −x 3 ≤ − 5
x 1 , x 2 , x 3 ≥ 0
b) [2 marks] Give two optimal solutions.
- [12 marks] Consider the following linear program:
Maximize 4 x 1 +5x 2 +5x 3 2 x 1 +3x 2 +x 3 ≤ 9 2 x 1 −x 2 +3x 3 ≤ 7 2 x 1 −x 2 +2x 3 ≤ 7
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 = 0, x∗ 2 = 2, x∗ 3 = 3. Determine an optimal dual solution (without pivoting), stating which theorems you have used.
d) [2 marks] Consider replacing the first inequality of the primal by 2x 1 + 3x 2 + x 3 ≤ 10. Does the primal solution x∗ 1 = 0, x∗ 2 = 2, x∗ 3 = 3 and the optimal dual solution determined in c) remain optimal to their new LP’s? 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 5 , x 2 , x 4 }.
x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 5 3 2 1 6 1 0 0 x 6 1 1 2 2 0 1 0 x 7 − 2 − 1 1 − 1 0 0 1
b x 5 4 x 6 1 x 7 − 1
B− (^1) =
x 5 x 6 x 7 x 5 1 − 4 − 2 x 2 0 − 1 − 2 x 4 0 1 1
( x^1 x^2 x^3 x^4 x^5 x^6 x^7 cT^5 1 2 1 0 0
)
- [25 marks] An electronic greeting card company wishes to maximize profit by considering what products to make. They can make three possible products made from the three available resources (hours of available staff time) according to the following table. We seek an optimal product mix. We are not concerned with integrality in this question and allow fractional products (we can think of our answer as an optimal product mix not a specific set of products)
product 1 product 2 product 3 total hours available artist hours per unit 1 2 1 2 tech hours per unit 2 3 1 3 sales hours per unit 2 1 3 7
$ profit per unit 4 6 3
Let xi denote the amount of product i to produce and let x3+i denote the ith slack for i = 1, 2 , 3. The final dictionary is:
x 1 = 1 −x 2 +x 4 −x 5 x 3 = 1 −x 2 − 2 x 4 +x 5 x 6 = 2 +4x 2 +4x 4 −x 5 z = 7 −x 2 − 2 x 4 −x 5
B−^1 =
x 4 x 5 x 6 x 1 − 1 1 0 x 3 2 − 1 0 x 6 − 4 1 1
NOTE: All questions are independent of one another.
a) [3 marks] Give the marginal values for each of the resources: artist hours, tech hours and sales hours.
b) [2 marks] Give the range on c 2 (profit for product 2) so that the current solution remains optimal.
c) [5 marks] Give the range on c 3 (profit for product 3) so that the current solution remains optimal.
d) [5 marks] Find the range on p so that we may change the availability of each labour resource by p units (i.e. change to 2 + p artist hours, 3 + p tech hours and 7 + p sales hours) and still have the same optimal basis {x 1 , x 3 , x 6 }. Also give the profit as a linear function of p 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] Change the resource availabilities to 3 for artist hours, 4 for tech hours and 6 for sales hours. Determine the new optimal solution using the Dual Simplex method. Report the new solution as well as the new marginal values.
f) [5 marks] Consider adding a new constraint 2x 1 + x 2 + x 3 ≤ 2 to our original problem. Solve using the Dual Simplex method. Report the new solution as well as the new marginal values.
LP OPTIMUM FOUND AT STEP 0
OBJECTIVE FUNCTION VALUE
VARIABLE VALUE REDUCED COST
X1 1.500000 0.
X2 6.000000 0.
X3 0.000000 9.
D1 6.500000 0.
D2 0.000000 4.
D3 0.000000 11.
ROW SLACK OR SURPLUS DUAL PRICES
SERVICE1) 0.000000 -5.
SERVICE2) 0.000000 -0.
SERVICE3) 0.000000 6.
NO. ITERATIONS= 0
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
X1 20.000000 5.000000 1.
X2 25.000000 3.500000 6.
X3 38.000000 9.916667 INFINITY
D1 -5.000000 4.760000 24.
D2 -5.000000 4.166667 INFINITY
D3 -5.000000 11.250000 INFINITY
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
RHS INCREASE DECREASE
SERVICE1 50.000000 INFINITY 6.
SERVICE2 63.000000 1.800000 9.
SERVICE3 72.000000 8.666667 2.
- [12 marks]
a) [2 marks] Let 1 denote the n × 1 vector of 1’s. Let x ≥ 0 be an n × 1 vector. Show that 1 · x > 0 if and only if x 6 = 0.
b) [10 marks] Let A be a given m × n matrix, let b be a m × 1 vector. Show that either
i) There exist an x 6 = 0 with Ax = 0 , x ≥ 0 , or ii) There exists a y with AT^ y ≥ 1 , but not both. You may use the result from a) even if you did not prove it.
- [4 marks] Let A be a matrix satisfying A > 0 which is the same as saying each entry of A is strictly greater than 0. Let b ≥ 0 and c ≥ 0 be some given vectors. Then show that the LP
max c · x Ax ≤ b x ≥ 0
always has an optimal solution.
- [10 marks] Let A = (aij ) be an m × n matrix representing the payoff to the row player where aij is the payoff to the row player if the row player plays strategy i and the column player plays strategy j. A mixed strategy for the row player can be given as x = (x 1 , x 2 ,... , xm)T^.
a) [5 marks] Assume that for some given k, that the kth row of A is at most theth row of A, namely akj ≤ a`j for j = 1, 2 ,... , n. Show that there is an optimal strategy x for the row player with xk = 0.
b) [5 marks] Assume for some given k, that the kth row of A is strictly less than theth row of A, namely akj < a`j for j = 1, 2 ,... , n. Show that any optimal strategy x for the row player has xk = 0.
- [5 marks] Consider an LP in the following standard form:
max c · x Ax ≤ b x ≥ 0
Let bi denote the ith entry of b. We have found the optimal solution and optimal basis B using LINDO. We have discovered by considering the LINDO output that the same basis is optimal if we increase b 1 by 4 and also the same basis is optimal if we increase b 2 by 6. Show that the same basis is optimal if we increase b 1 by 2 and simultaneously increase b 2 by 3. You might note that
