Solutions |

MATP6640/ISYE6770

Linear Programming

Spring 2016

Midterm Exam, Friday, April 8, 2016.

Please do all three problems. Show all work. No books or calculators allowed. You may use any result from class, the homeworks, or the texts, except where stated. You may use one sheet of handwritten notes. The exam lasts 110 minutes.

Q1 | /30 | |

Q2 | /35 | |

Q3 | /35 | |

Total | /100 |

- (30 points; each part is worth 10 points)

Let A be an n × n matrix satisfying:- a
_{ij}≥ 0 if i≠j, for 1 ≤ i ≤ n and 1 ≤ j ≤ n, - and ∑
_{i=1}^{n}a_{ ij}= 0 for j = 1,…,n.

Show the following:

- Let ∈ ℝ
^{n}. Assume the smallest component of is_{ k}, so_{k}≤_{i}for i = 1,…,n. Show ∑_{i=1}^{n}a_{ ik}_{i}≥0. - Show the linear program
is infeasible, where e denotes the vector of ones.

- What can you say about the set {x ∈ ℝ
^{n}: Ax = 0,x ≥ 0}?

Solutions:

- We have
- For any y ∈ ℝ
^{n}, let y_{ k}be the smallest component of y. We have^{T }y)_{ k}⁄≤-1, so A^{T }y ⁄≤-e. - The dual LP to the infeasible problem in part (b) is
This problem is feasible (take x = 0), so it is unbounded. So there exists a nonzero x in the given set.

- a
- (35 points)

In the single commodity network flow problem in the figure, nodes s and t are supply nodes and nodes u and v are demand nodes. The supplies available are b_{s}= 3 and b_{t}= 2. The demands are d_{u}= 1 and d_{v}= 4. The unit shipping costs for each edge are indicated in the figure.An initial basic feasible solution can be obtained by taking x

_{su}, x_{ut}, and x_{tv}to be basic, and x_{st}, x_{uv}to be nonbasic.- (5 points) What is the initial basic feasible solution?
- (15 points) What are the primal and dual linear programs? Find a dual solution that satisfies complementary slackness. Show your dual solution is not dual feasible.
- (15 points) Make a simplex pivot to improve your primal solution. What is the updated dual solution? Is the new solution optimal?

Solutions:

- x
_{su}= 3, x_{ut}= 2, x_{tv}= 4, x_{st}= x_{uv}= 0. - Total supply and total demand are both equal to 5, so we can use equality constraints in
the primal LP. Primal LP:
Dual LP:

From complementary slackness, need y to satisfy

One solution is y

_{v}= 0, y_{t}= 3, y_{u}= 8, y_{s}= 13. This violates the second and fourth dual constraints, each by one unit. - We use network simplex to make the pivot. We choose edge (s,t) to enter the basis
(using the edge (u,v) instead is similar). The fundamental cycle also includes edges
(s,u) and (u,t). The values of the variables with different choices Δ for the incoming
basic variable is indicated in the figure:
Largest possible value of Δ = 2, and edge (u,t) leaves the basis. Get new bfs as indicated in the figure:

From complementary slackness, need y to satisfy:

One solution is y

_{v}= 0, y_{t}= 3, y_{s}= 12, y_{u}= 7, which is dual feasible. Therefore, the new bfs is primal optimal, and the given dual solution is dual optimal.

- (35 points)

The numbers on the edges in the following graph represent the probability of failure of an edge. Edges fail independently of one another.For a price, we can reinforce an edge so that its probability of failure drops to zero. We want to ensure that a path exists between s and t with high probability. More precisely, we want to ensure that, for each cutset,

(1) A cutset consists of all edges leading from P to Q, where P and Q form a partition of the vertices with s ∈ P and t ∈ Q, so

(2) Two possible partitions are:

- P
_{1}= {s}, Q_{1}= {t,u,v,w}. Edges from P_{1}to Q_{1}are (s,u), (s,v). Probability both edges in the cutset fail is 0.2 × 0.1 = 0.02 ≥ 0.01, so at least one edge in the cutset must be reinforced. - P
_{2}= {s,u,v,w}, Q_{2}= {t}. Edges from P_{2}to Q_{2}are (v,t), (w,t). Probability both edges in the cutset fail is 0.2 × 0.1 = 0.02 ≥ 0.01, so at least one edge in the cutset must be reinforced.

We introduce variables

(3) In order to ensure the solution x satisfies the constraint (1) for the cutset given by P

_{1}and Q_{1}, we can impose the constraintSimilarly, we can impose the constraint

corresponding to the other cutset. For a general graph, there may be a large number of possible cutsets, so we add the corresponding constraints as needed. The initial LP relaxation is as follows, where the objective function coefficients correspond to the costs of reinforcing a particular edge:

(4) - (5 points) Show that the solution to the initial relaxation is x
_{sv}= x_{wt}= 1, with all other x_{e}= 0. - (10 points) Find a cutset that leads to a valid inequality that is violated by the solution in part 3a.
- (5 points) Show that x
_{su}= x_{wt}= 1 with all other x_{e}= 0 is feasible after the new valid inequality is added to (4). - (15 points) Construct the dual LP to the new problem obtained by adding the constraint to (4). Find a dual solution that satisfies complementary slackness. Is your solution optimal?

Solution:

- Trivial.
- Choose P = {s,v}, Q = {t,u,w}. The edges going from P to Q are (s,u), (v,t)
and (v,w).
The product of the probabilities of failure of the three edges in the cutset is

so we must reinforce at least one edge in this cutset in order to satisfy constraint (1). This gives the valid constraint

(5) which is violated by the solution given in part (a).

- The updated primal LP is
(6) and the given solution is feasible in this LP, with value 6.

- The dual LP is
From complementary slackness, we need

This system has multiple solutions, including y = (2, 3, 1) and (1, 3, 2), both of which are feasible in the dual problem. Since we have primal and dual feasibility as well as complementary slackness, we are optimal.

The general set of dual optimal solutions has the form y = (1 + z, 3, 2 - z) for 0 ≤ z ≤ 1.

- P