IEOR 240: Midterm Exam
November 19, 2019
5:00 PM - 8:00 PM
Name:
(please print)
SID:
• The use of any electronic device (such as smart phone, tablets, laptops,
calculators, etc.) is strictly prohibited. Please turn off all such devices.
• Any communication with other students during the exam (including showing, viewing or sharing any writing) is strictly prohibited.
• Any violation of the above will result in a score of 0 points for the exam,
and will be reported to the Fung Institution.
• Clearly state all assumptions.
• Answer each of the following questions in the space provided. If you need more space
to show major computations you performed to obtain your answer for a particular
problem, use the back of the preceding page.
• You can quote and use any general result stated in class notes or in the main body
of the textbook as well as well-known general mathematical results but no references
to other sources (including homework and examples - both in textbook and class - are
allowed).
• Present your work in an organized and neat fashion.
• Good luck!
Problem
Score
1 (25)
2 (25)
3 (25)
4 (25 +10)
Total (100+5)
1. [25 points] Suppose you are a production engineer in a factory. Now, you need to
decide how much to produce in each period for the upcoming 5 periods.
1 There is a demand at the end of each period, and it must be satisfied.
2 There are two costs involved:
• The first cost is a holding cost: for every unit of product held in inventory
from the end of period i to the beginning of period i + 1, you incur a holding
cost hi .
• The second cost is associated with changing production level: if you change
the quantity of product produced in i − 1 to a different amount in i, you incur
a cost per unit change ci (for example, if your order quantity from period 1
to period 2 increases or decreases by 5 units, the associated order change cost
is 5 × c2 ).
3 Assume you start with zero inventory and a production level of zero. There is no
restrictions on the level of inventory at the end of period 5.
4 The relevant data is given in following table:
Demand (di )
Holding cost (hi )
Production change cost (ci )
Period 1
50
0.5
3
Period 2
80
1.5
2
Period 3
60
1
1.5
Period 4
110
1
2
Period 5
90
0.5
2
(a) formulate a linear programming model that minimizes cost.
(b) Now, assume that you can choose to satisfy any portion (up to what specified
in the table), but that for each unit you satisfy, you earn a net profit of $1 per
unit. In addition, you are required to pay the cost associated with holding and
production changes. To do that, you are provided with a budget of $100 which
is available before the start of the first period. For each period, you can also
use money earned from previous periods. At the end of each period, you can
deposit whatever cash you have at hand to a bank which pays %1 interest per
period (you can withdraw money from the accumulated deposits at the beginning
of every period). Revise your formulation to model this new scenario to maximize
profit at the end of period 5..
Clearly define any variables and parameters you use and briefly explain
each of the constraints. In addition, clearly state any assumption you
make.
Solutions:
(a)
• zi : The change of production in period i
• xi : Production level in period i
• Ii : Inventory in period i
N
X
(hi Ii + ci zi )
minimize
i=1
subject to
(b)
Ii = Ii−1 + xi − di
zi ≥ xi − xi−1
zi ≥ xi−1 − xi
z1 ≥ x1
I1 = I1 − d1
Ii ≥ 0
xi ≥ 0
i = 2, . . . , N
i = 2, ..., N
i = 2, ..., N
i = 1, ..., N
i = 1, ..., N
• zi : The change of production in period i
• xi : Production level in period i
• Ii : Inventory in period i
• yi : Demand to meet in period i
• ki : Money you have in period i (right after you make change on the production
level)
maximize
N
X
(yi − hi Ii − ci zi ) + k5
i=1
subject to
Ii = Ii−1 + xi − yi
zi ≥ xi − xi−1
zi ≥ xi−1 − xi
z1 ≥ x1
I1 = I1 − y1
k1 = 100 − zi
ki = 1.01 ∗ ki−1 + yi−1 − zi
Ii ≥ 0
xi ≥ 0
yi ≥ 0
yi ≤ di
ki ≥ 0
i = 2, ..., N
i = 2, ..., N
i = 2, ..., N
i = 2, ..., N
i = 1, ..., N
i = 1, ..., N
i = 1, ..., N
i = 1, ..., N
i = 1, ..., N
2. [25 points] The NCAA is making plans for allocating tickets to the upcoming regional
basketball championship of the South-East. Up to 10,000 available seats will be allocated to the media, students of the two competing universities, and the general public,
subject to the following conditions:
• Media people are admitted free, and at least 1000 tickets must be reserved for the
media.
• The NCAA receives $45 per ticket sold to students of the two competing universities and $100 per ticket from tickets sold to the general public.
• At least half as many tickets should be allocated to students of the two competing
universities as to the general public.
With these restrictions, we formulate an LP so that the NCAA maximizes their revenue.
To model this problem algebraically, we let x1 , x2 and x3 (in thousands) represent the
numbers of tickets allocate to media, students and general public, respectively. The
linear programming formulation is as follows:
max 45x2 + 100x3
s.t.
x1 + x2 + x3 ≤ 10,
1
x2 − x3 ≥ 0,
2
x1 ≥ 1,
xi ≥ 0, i = 1, 2, 3.
Solving the above model using the AMPL solver, we obtained the following sensitivity
analysis report:
Variables
Name
Media Tickets
University Tickets
Public Tickets
Final
Value
1
3
6
Reduced Objective
Cost
Coefficients
0
0
0
45
0
100
Upper
Bound
81.667
100
1E+20
Lower
Bound
-1E+20
-200
45
Constraints
Final
Name
Value
Capacity Constraint
10
Univ./Public Constraint
0
Media Constraint
1
Shadow
Price
81.667
-36.663
-81.667
Constraint Upper
R.H.Side Bound
10
1E+20
0
9
1
10
Lower
Bound
1
-4.5
0
Consider each of the following questions independently. Be sure to explain (using the
sensitivity report) all your answers.
(a) What is the optimal value of the decision variables, and what is the optimal
revenue?
(b) Suppose it is possible to reconfigure the arena to have a total of 20,000 seats (so
that 20,000 tickets can be sold). How much additional revenue would be gained
from the expanded seating?
(c) In order to make the championship game more affordable for fans, it is proposed
to reduce the price of the general public tickets to $50. Will the same solution
be optimal if we implement this change? If the same solution (with the reduced
price) is optimal, how much revenue for NCAA would be lost?
(d) Suppose you don’t have the data to answer question (c), provide an upper bound
to the possible revenue loss from the price reduction as in (c). You should try to
get (and justify!) the lowest upper bound that you can.
(e) To accommodate the high demand from the students in the participating universities, the NCAA is considering marketing a new scrunch seat that costs $15
each and consumes only 50% of a regular seat (that is, each physical seat can
accommodate two students who purchase scrunch seat), but counts fully against
the “university ≥ half public” rule. Specifically, let x4 indicate the number of
scrunch seats (in thousands) allocated, then the problem can be reformulated as
follows:
max 45x2 + 100x3 + 30x4
s.t.
x1 + x2 + x3 + x4 ≤ 10,
1
x2 − x3 + 2x4 ≥ 0,
2
x1 ≥ 1,
xi ≥ 0, i = 1, 2, 3, 4.
Would selling tickets for such seats be profitable?
(f) Explain, in words, why the shadow price of the Media Constraint is the same as
the shadow price of the Capacity Constraint, but with opposite sign.
Solutions:
(a) The optimal value of the decision variables are (x1 , x2 , x3 ) = (1, 3, 6). The optimal
revenue is 735.
(b) The increase of 10 of the right-hand side of the first constraint is within the
allowable increase. So the shadow price remains the same and the additional
revenue gained from the expanded seating is 816.67.
(c) The decrease of 50 in the objective coefficient of x3 is within the allowable decrease.
This implies that the optimal solution is still optimal and the lost revenue for
NCAA will be 300.
(d) The current optimal solution will be at least feasible after we change the objective
coefficient of x3 to 50. It is clear that this feasible solution has the new objective
value of 45∗3+50∗6 = 435. By the optimality condition, the new optimal revenue
will be greater than or equal to this value. Therefore, the loss is not greater than
735 − 435 = 300.
(e) Adding one more decision variable in the primal problem is equivalent to adding
one more constraint to the dual problem. The feasible set of the dual problem
becomes smaller, so its optimal objective value becomes larger since it is a minimization problem. By the optimality condition, the optimal value of the dual
is the same as that of the primal problem, and the new optimal objective value
is greater than or equal to the current one. Therefore, we conclude that selling
tickets for such seats is profitable.
(f) If we increase the right-hand side of the Media constraint by 1, then in order to
maximize the objective value, the value of x1 will be set to 2 and x2 + x3 will
be equal to 8 in the new optimal solution. On the other hand, when we decrease
the right-hand side of Capacity constraint by one unit, again to maximize the
objective function, x1 remains 1 and x2 + x3 will be equal to 8 in the new optimal
solution. Therefore, increasing the right-hand side of the Media constraint will
change the optimal objective value in the way decreasing the right-hand side of
the Capacity constraint does.
3. [25 points] A factory works a 24 hour day, 7 day week in producing four products.
Since only one product can be produced at a time the factory operates a system where,
throughout one day, the same product is produced (and then the next day either the
same product is produced or the factory produces a different product). The rate of
production is:
Product
Number of units produced per hour worked
1
100
2
250
3
190
4
150
The only complication is that in changing from producing product 1 one day to producing product 2 the next day five working hours are lost (from the 24 hours available
to produce product 2 that day) due to the necessity of cleaning certain oil tanks.
To assist in planning the production for the next week the following data is available
that details the current Stock and Demand (units) for each day of the week .
Product
1
2
3
4
Current Stock(units)
5000
7000
9000
8000
1
1500
4000
2000
3000
2
1700
500
2000
2000
3
4
1900 1000
1000 3000
3000 2000
2000 1000
5
6
7
2000 500 500
500 1000 2000
2000 2000 500
1000 500 500
Product 3 was produced on day 0. The factory is not allowed to be idle (i.e. one of
the four products must be produced each day). Stockouts are not allowed. At the end
of day 7 there must be (for each product) at least 1750 units in stock.
If the cost of holding stock is $1.50 per unit for products 1 and 2 but $2.50 per unit
for products 3 and 4 (based on the stock held at the end of each day) formulate the
problem of planning the production for the next week as an integer program in which
all the constraints are linear.
• Formulate the problem as a linear integer programming problem.
Solutions: The solutions are as follows:
• Approach 1:
minimize
7
X
hi Ii
i=1
subject to
x3,0 = 1
4
X
xi,t
t = 1, . . . , 7
i=1
Ii,7 ≥ 1750
Ii,0 = si
Ii,t = Ii,t−1 + Pi,t−1 − di,t
zt ≥ x2,t + x1,t−1 − 1
zt ≥ (x2,t +x2 1,t−1 )
Tt = 24 − 5zt
Pi,t ≤ xi,t ∗ M
Pi,t ≤ Tt θi
Ii,t , Tt , Pi,t , zt ≥ 0
xi,t = {0, 1}
i = 1, . . . , 4
i = 1, . . . , 4
i = 1, . . . , 4, t = 1, . . . , 7
t = 1, . . . , 7
t = 1, . . . , 7
t = 1, . . . , 7
i = 1, . . . , 4, t = 1, . . . , 7
i = 1, . . . , 4, t = 1, . . . , 7
• Approach 2:
minimize
7
X
hi Ii
i=1
subject to
x3,0 = 1
4
X
xi,t
t = 1, . . . , 7
i=1
Ii,7 ≥ 1750
Ii,0 = si
Ii,t = Ii,t−1 + Pi,t−1 − di,t
zt ≥ x2,t + x1,t−1 − 1
zt ≥ (x2,t +x2 1,t−1 )
Pi,t = θi ∗ (24 ∗ xi,t − 5 ∗ zt )
Ii,t , Tt , Pi,t , zt ≥ 0
xi,t = {0, 1}
i = 1, . . . , 4
i = 1, . . . , 4
i = 1, . . . , 4, t = 1, . . . , 7
t = 1, . . . , 7
t = 1, . . . , 7
i = 1, . . . , 4, t = 1, . . . , 7
4. [25 points + 10 points bonus for question (f )] For each of the following claims,
determine whether it is true or false. Justify your answer by providing either a short
proof or a counterexample as appropriate.
No explanation - no credit
(a) For any linear program which is feasible, the dual program is also feasible.
(b) An integer program must either have no optimal solution, a unique optimal solution, or infinitely many optimal solutions.
(c) The optimal objective value of the linear programming
max c1 x1 + c2 x2 + · · · + cn xn
s.t. x1 + x2 + · · · + xn ≤ 1
xj ≥ 0, j = 1, 2, ..., n.
where ci > 0 for i = 1, . . . , n, is max{c1 , c2 , ..., cn }.
(d) Suppose the linear program
P : min c> x
s.t. Ax ≥ b, x ≥ 0.
is unbounded. Then changing the values of the bi ’s will never result in this linear
program having an optimal solution.
(e) Suppose two minimization problems (A and B) with linear objective and linear
constraints are identical, except that problem A requires to satisfy at least one
of the first two constraints, while problem B requires to satisfy both the first two
constraints. Given that problem A is feasible and is certified to be bounded with
a lower bound of objective value α for every feasible solution, the optimal solution
(if exists) of problem B has an objective function value of at least α.
(f) Let x̄1 , . . . , x̄n ; ȳ1 , . . . , ȳm be feasible solutions to the primal and dual linear program (in symmetric form), respectively. Also, let s̄1 , . . . , s̄m ; r̄1 , . . . , r̄n be the
slacks to the primal and dual feasible solutions above, respectively. Then, if
Pm
Pn
Pm
i=1 bi ȳi ≤ 0, then
j=1 r̄j x̄j +
i=1 s̄i ȳj is an upper bound on the optimal value
of the dual.
Solutions: The solutions are as follows,
(a) False. The LP
min
x
x≥1
s.t.
−x ≥ 0
x≥0
is infeasible, since x cannot be ≥ 1 and ≤ 0, yet its dual,
max
s.t.
y1 − 2y2
y1 − y2 ≤ 1
y1 , y2 ≤ 0
is feasible. Reverse the orders for a counterexample.
(b) False. The IP
max
s.t.
x+y
x+y ≤1
x, y ∈ {0, 1}
has exactly two optimal solutions, (1, 0) and (0, 1).
(c) False. This is true only if ci ≥ 0, ∀i. If all ci < 0, for example, we can choose
xi = 0, ∀i, giving us the optimal solution and value of 0. However, the optimal
objective value of the above LP is equal to max{0, c1 , c2 , . . . , cn }.
(d) True. We need to consider two cases. If P is still feasible after changing the
values of the bi ’s, we find that
c> z < 0, Az ≥ 0, z ≥ 0,
still has a solution since the original P is unbounded. This implies that the
certificate of unboundedness exists and P can not be optimal after changing the
values of the bi ’s. Otherwise, P is infeasible and hence not optimal after changing
the values of the bi ’s.
(e) True. The solutions of B are all solutions of A since x or y ⇒ x and y. So the
minimum is being taken over a (weakly) smaller set in B, so the min. of it will
be at least min(A) = α.
(f-1) True. Satisfying complementary slackness means being, in particular, feasible.
Then complementary slackness says that they are optimal solutions for their respective systems, then weak duality gives the desired equation.
(f-2) False. Choose c = 0, b = 0 so that all x, y satisfy this condition. Choose any
A, x, y such that Ax 6≥ 0, AT y 6≤ 0, then these choices are not feasible, hence don’t
satisfy complementary slackness conditions.