Name:
Section:
ISyE 3833: Engineering Optimization
Final
May 3rd, 2016
6:00pm to 8:50pm
Remarks.
• This is a closed book exam.
• There are 5 problems in total.
• Write your name on every sheet.
Row:
1
Name:
Problem 1.
Section:
Row:
2
(18 pts) We are given a maximization integer programming problem (IP):
max
s.t.
cx
Ax ≤ b
x≥0
x integer
(a) The LP relaxation of the IP is obtained by
In the questions below circle the most appropriate answer. Note that always and sometimes
are mutually exclusive, i.e., if an event always happens, “sometimes” is an incorrect answer.
(b) The optimal value of the LP relaxation is
always
sometimes
never
strictly greater than the optimal value of the IP.
(c) If the LP relaxation is infeasible, then the IP is
always
sometimes
never
infeasible.
(d) If the LP relaxation has an optimal solution, then the IP
always
sometimes
never
has an optimal solution.
(e) If the LP relaxation has an optimal solution in which all of the variables have integer
values, then that solution is
always
sometimes
never
an optimal solution to the IP.
(f) Suppose we are solving the IP by a branch-and-bound algorithm. We solve the LP
relaxation at node i of the tree and the value of the optimal LP solution is larger than
the value of the best solution we have to the IP. Then it will
always
sometimes
never
be necessary to create children nodes from node i to solve the IP.
.
Name:
Section:
Row:
3
(g) Suppose we are solving the IP by a branch-and-bound algorithm. We solve the LP
relaxation at node i of the tree and the value of the optimal LP solution is less than
or equal to the value of the best solution we have to the IP. Then it will
always
sometimes
never
be necessary to create children nodes from node i to solve the IP.
(h) Suppose we are solving the IP by a branch-and-bound algorithm. We solve the LP
relaxation at node i of the tree. The optimal solution to the LP relaxation is given
by x1 = 3, x2 = 2.7, x3 = 0 and we need to create 2 children nodes from node i, say
nodes i + p and i + p + 1. Give any additional constraints needed to define the problem
at each child node:
node i + p
node i + p + 1
.
.
Name:
Problem 2.
Section:
Row:
(22 pts) In the questions below circle the most appropriate answer. We are given an LP:
max
cx
s.t. Ax ≤ b
x≥0
(a) The objective function
always
sometimes
never
strictly increases at each iteration of the simplex algorithm.
(b) If the LP has a feasible solution, its dual
always
sometimes
never
has an optimal solution.
(c) If the LP has an optimal solution, its dual
always
sometimes
never
has an optimal solution.
(d) If the LP is infeasible, its dual is
always
sometimes
never
infeasible.
(e) If the LP is unbounded, its dual is
always
sometimes
never
feasible.
(f) If the dual LP is unbounded, the primal LP is
always
feasible.
sometimes
never
4
Name:
Section:
Row:
5
(g) Suppose the optimal solution of the LP is unique and in the optimal solution, the
slack variable of the i-th constraint is positive. Then in an optimal solution to the
dual, the i-th dual variable is
always
sometimes
never
positive.
(h) Suppose the optimal solution of the LP is unique and in the optimal solution, the variable xj is positive. Then in an optimal solution to the dual, the j-th dual constraint is
always
sometimes
never
satisfied at equality.
(i) Suppose the optimal solution of the LP is unique and in the optimal solution, the
objective coefficient (reduced profit) of the slack variable of the i-th constraints equals
-12. It would be profitable to purchase some amount of the i-th resource if its cost
per unit was
more than
equal to
less than
12.
(j) Suppose the LP has 3 variables and 2 constraints, and that (x1 , x2 , x3 , s1 , s2 ) =
(2, 0, 0, 0, 3) is feasible to the LP problem and (y1 , y2 , t1 , t2 , t3 ) = (2, 2, 1, 0, 0) is feasible to its dual, where s is the vector of primal slacks, y is the vector of dual variables
and t is the vector of dual surplus (negative slack) variables. Then it is
always
sometimes
never
the case that both of these solutions are optimal.
(k) Suppose the LP has 3 variables and 2 constraints, and that (x1 , x2 , x3 , s1 , s2 ) =
(2, 0, 0, 0, 3) is feasible to the LP problem and (y1 , y2 , t1 , t2 , t3 ) = (2, 2, 1, 0, 0) is feasible to its dual, where s is the vector of primal slacks, y is the vector of dual variables
and t is the vector of dual surplus (negative slack) variables. Then it is
always
sometimes
never
the case that at least one of these solutions is optimal.
Name:
Problem 3.
Section:
Row:
6
(15 pts)
(a) (5 pts) Suppose we have two binary variables, x and y, and the nonlinear constraint
x 6= y. For binary variables there is one simple linear constraint for enforcing x 6= y,
i.e., excluding the points (x, y) = (0, 0) and (x, y) = (1, 1). That constraint is:
.
(b) (10 pts) Now suppose x and y are both integer variables between 0 and 9. Now it is
harder to enforce the constraint x 6= y, but it can be done with two constraints and
an additional binary variable z. One of the constraints is:
x ≤ y − 1 + 10z
which enforces x 6= y with z = 0 when x < y. Give the second constraint required in
the variables x, y and z:
.
Explain why the two constraints together guarantee x 6= y.
Name:
Problem 4.
Section:
Row:
(25 pts) Consider the LP:
max
5x1 + 2x2 + c3 x3
s.t.
x1 + 5x2 + 2x3 + s1
x1 − 5x2 − 6x3
= b1
+ s2 = b2
x1 , x2 , x3 , s1 , s2 ≥ 0
where c3 , b1 and b2 are constants. Suppose that the optimal set of equations is:
x1 = 30 + dx2 + ex3 − s1
s2 = 10 + f x2 + gx3 + s1
z = 150 + hx2 − 7x3 − 5s1
Note that the optimal basis matrix is:
AB =
1
1
0
1
with inverse:
A−1
B
1
=
−1
Determine:
(a) (6 pts) The values of c3 , b1 and b2 .
(b) (10 pts) The values of d, e, f, g, h.
(c) (5 pts) The dual of the LP.
(d) (4 pts) An optimal solution to the dual.
0
.
1
7
Name:
Problem 4.
Section:
Row:
8
Name:
Problem 5.
Section:
Row:
9
(20 pts) A company requires at least Di machines in month i for each of the next four
months, i = 1, 2, 3, 4. Depending on its quality, a machine is productive for 1, 2 or 3
months, i.e., it can have a service life of up to 3 months. A machine bought in month i
with a service life of k months costs cik dollars. Naturally a machine with longer service
life costs more so that cip < cis for p < s and all i. In addition, if any machines are bought
in month i, there is a delivery charge of fi > 0. Let xik be the number of machines bought
at the start of month i, i = 1, 2, 3, 4, with service life of k months, k = 1, 2, 3. Note that it
is undesirable, with respect to cost, to buy a machine whose service life goes beyond month
4.
(a) (10 pts) Formulate a linear integer program that will determine the number of machines of different service life to be bought each month to minimize total cost. Please
define any additional variables you add to the problem.
(b) (10 pts) Now suppose that if more than 5 machines of service life k are bought in
month i, i = 1, 2, 3, 4 and k = 1, 2, 3 then each of the additional machines of this type
and this month can be bought at a 10% discount. Formulate a linear integer program
that will determine the number of machines of different service life to be bought each
month to minimize total cost. Please define any additional variables you add to the
problem.
Name:
Problem 5.
Section:
Row:
10