310: Sensitivity Analysis
ASB and FJM
Spring 2016
1
Sensitivity Analysis
Once we solve an LP we should also consider potential errors in the model/parameters, i.e., what happens
if the RHS of the third constraint was higher/lower? what if a cost coefficie

124
With interval (5.894 —— 3.000) = 2.894 S 4 (mid)
point 5.431 is optimal to the speciﬁed accuracy.
13-23. (a) f1(3+A) = f(3) +Af’(3) = 33+20,\
(b) f2(3+A) = f(3)+/\f’(3)+%x\2f”(3) = 33+
20A + 6)?
(c)
ISO
100
[(x)
The approximations are accurate near

Chapter 12 Exercise Solutions
with best parent bound as node 4. Then depth
ﬁrst processing continues until node 5 is termi-
nated when we again seek the best parent bound.
Neither incumbent discovery produced any ter-
mination by parent bound.
12-39.
(a)

Chapter 12 Exercise Solutions
mal value 2 LP relaxation value 4(0) + 5(0) +
12(.533) +7(0) +6(0) = 6.396. (b) The LP opti-
mum is not ILP optimal because 0:3 violates inte-
grality requirements. Still, with all constraints _>_
and all LHS’s nonnegative, w

Chapter 11 Exercise Solutions
1/4(a:1+15+z2+8+23+20+z4+6),s.t. con-
straints of part (a) (d) x; = 16, 27* = 0, 2:; = 31,
1: = 10 (e) From principle , the same
schedule would be optimal for mean ﬂow time
and mean lateness. (f) min 2, s.t. z 2 x1 — 5,
z _>_

Chapter 11
Exercise Solutions
11-1. (a) Costs are proportional to variable
magnitude. (b) x* = (96.67,5,48.33) (c) Us-
ing the ﬁrst main constraint that the 23- should
sum to 150, we make a change of variables
to 31,- 2 mj/150. The result is model: min
22

106 Chapter 12
Exercise Solutions
12-1. (a) We enumerate all choices of the dis-
crete variables and optimize over the remaining
continuous variable as follows:
Enumerate Corres
Optimal
Value
b—‘HHHOOOO
Thus, the best feasible solution is x* = (

Chapter 10 Exercise Solutions 91
10-33. (a) Check ﬂow balance at every node and At t = 2, u[3][3] < 0 indicates a negative dicycle.
verify that all ﬂows are between 0 and capacity. Backtracking through corresponding d—labels re-
(b) covers dicycle 3 — 1 —

Chapter 8 Exercise Solutions
('0)
[ye optima
’ jective
x1
1 2 3 4
We ﬁrst minimize the second objective subject
to the given constraints. The result is alter-
native optima with x1 = 0, 2:2 2 4. Now
adding a constraint that $1 3 0 and minimizing

100
the time for i (b) More than one unit of time is
allocated by each decision. (c) 2713 = $313 =
933,1? = 132,1? = 933,3 = 933,1? = 1
11-18. (a) ziyjél if player i is assigned to
team j, = 0 otherwise; max 225:1 z§=1c,-,-z,-,-
(max total value of player

118 Optimization in Operations Research
(1,1,0)
(1,0,1)
The third gives:
(1,2,0)
(1,1,1)
5‘: = (1, 1,0) (e) Different starting points led to
diﬁ'erent local minima, but neither is a global
minimum. (f)
The best of the 3 local maxima is 5‘: = (0, 1, 1,

IEMS 310 - Solver in EXCEL
FJM and ASB
Spring 2016
SOLVER in EXCEL, is a built-in tool that uses Operations Research techniques to find optimal solutions to
all kinds of decisions problems. In this class we will use SOLVER to solve Linear, Integer and Mix

IEMS 310 HWl Yue Hu
Due Date: April 12, 2016
1
—1 —2 —3 3 6 9
(a)—A= —x1 —5 —6 (b) 3/1: 12 15 18 (c) A—Sb is undeﬁned (d) AT
_7 —8 —9 21 24 27
| 1
/'_"\
15
(g) bA is undeﬁned (h) (5%: ( 8 10 12 )
(i) A is singular because its determinant is O
4
(e)bT=(1 0

IEMS 310 - Operations Research
IEMS
Homework
2
:H
Northwestern University
due: April 19, 2016 (beginning of class)
1. True or False (justify your answers):
(a) Adding a constraint to a linear programming problem increases the size of the feasible region.

130
60% and 10% in each crop 2'), Where the m,- are
the mean returns and via- the covariances in the
given table. (b) x; = 0.2, 2:; = 0.6, so; = 0.2
14-11. (a) xjé volume fraction of ingredient j
5
in the blend; min ch-zj (minimize total cost),
j=1
s.t. 2

Chapter 13 Exercise Solutions 127
. _ —-1.003 —.0268 erywhere. (b) See Table 13.1.
W“ D ‘ ( —.0268 —.0267 and (c)
Ax = -D Vf(x(1) = (5.555, 0.151). (c) Check
that Dg = d for values in part (d) Check
that Vf(x(1) Ax > 0 for values in part
(e)
13-38. (a)

112
incumbent value 17 = —00. The ﬁrst relaxation
optimum (48.1, .20, .77, 1) is fractional on binary-
restricted 2:2 and (1:3. We branch on 32, which
is closest to integer, and make node 1 the = 0
child nearest to .20. The node 1 relaxation opti-
mum (42

Chapter 13
Exercise Solutions
13-1. (a) 13% number of units in a lot;
min 40(m/2) + 2000/(93/5)
(b)
5000
woo
4000
3500
E soon
0
5T
2? 2500
g 2300
1500
moo
500
on 5 1o 15 20 25 so 35 4o 45 so
buzz:
93* = 22.4
13-2. (a) aé the chosen parameter;
max 1'1le ae

64
8-16.
(a) We ﬁrst solve min d1, s.t. x2 — d1 S l,
5151 +3$2+d2 2 30, 21171 +3232 2 6, 231 S 5,
—x1 + :32 S 2, m1,x2,d1,d2 Z 0. Any feasible
solution with x2 S 1 is optimal. min d2, s.t.
d1 S 0 (or (Q S 1) and all constraints of the ﬁrst
‘LP. The resu

82
to locations #1 through #5. There are two op-
posed arcs between any pair of points, an uphill
one with the indicated energy and a downhill
one with half the energy. (b) The digraph is not
acyclic because each pair of opposed arcs forms
a dicycle. (c)

40
components are positive, i.e. no nonnegativ-
ity constraint is active. (a) No, because :62 =
0. (b) Yes, because all components are posi-
tive, 4(2) + 1(5) = 13, and 5(2) +5(1) = 15.
(c) Yes, because all components are positive,
4(1) + 1(9) = 13, and 5

Chapter 5
Exercise Solutions
W1
(b) W0) is boundary and extreme because it
forms a corner point of the feasible region. W(2)
is infeasible because it lies outside the feasible
region. w(3) is an interior point of the feasible
region. W(4) is a boundary