Spring 2016
OR-MA-ST 706 HW #2
Reiland
Sketches of Solutions
1.
Solve the following problem using the method of Lagrange multipliers:
Min 0 ÐBÑ œ B#"  ÐB#  "Ñ#
=Þ >Þ
 #B#"  B# œ %
Now solve by eliminating the variable B" . Why is the incorrect answer obtained?
SOLUTION:
Lagrange multipliers gives B œ  !% ß - œ 'Þ
%B#
$
‡
#
#
eliminate B" À B#" œ %B
# à 0 ÐB# Ñ œ #  ÐB#  "Ñ is minimized at B# œ % à which implies
"$
#
B" œ  ) .
When we substitute for B" we lose information, that is, we don't take account of the fact that
#
B#" œ %B
# should be !.
2.
If the Lagrange multiplier associated with the kth constraint 15 ÐBÑ œ ,5 of an equality constrained problem
has a value of zero, is it necessarily true that the optimal value of the objective function would be unaffected
by small changes in the value of ,5 ? Suggest two reasons why a Lagrange multiplier might have a value of
zero, and devise a problem in which a zero-valued Lagrange multiplier occurs.
SOLUTION:
No.
2 reasons:
i) the constrained minimum of 0 subject to the constraints is the same as the unconstrained minimum
of 0 .
ii) a constraint is redundant.
Simple example: 738 0 ÐB" ß B# Ñ œ B#"  B## =?,4/-> >9 1ÐB" ß B# Ñ œ B"  B# œ ! has minimum at
B œ  !! , 0 ÐBÑ œ !ß with - œ !Þ Even though - œ !, the optimal value of the objective function is
affected by a small change in the RHS of the constraint. If the constraint is changed to 1ÐB" ß
B# Ñ œ B"  B# œ %, then the constrained minimum is B œ  #%  and 0 ÐBÑ œ
%
#
3.
%#
#Þ
Determine by inspection the optimal solution to the following problem:
Min 0 ÐBÑ œ B#"  B## , B % I $
=Þ >Þ B" B#  B$ œ "
and B"  B# B$ œ !Þ
Now use the Lagrange multiplier first-order necessary conditions to compute the associated values of the
Lagrange multipliers. Are they in fact equal to the rate of change of the optimal objective function value
with respect to changes in the right-hand-side constants? If not, why not?
SOLUTION:
By inspection B œ Ò!ß !ß 1ÓX . When use Lagrange mulitpliers, B œ Ò!ß !ß 1ÓX , with -" œ !ß -# œ ! Þ
-# œ ! is not the rate of change of the minimal value of 0 with respect to the RHS of constraint 2. In
general if the constraints are 1" ÐBÑ œ B" B#  B$ œ ," and 1# ÐBÑ œ B"  B# B$ œ ,# , then B œ Ò,# ß
!ß ," ÓX ß f1" ÐBÑ œ Ò!ß ,# ß "ÓX ß f1# ÐBÑ œ Ò"ß ," ß !ÓX ß f0 ÐBÑ œ Ò#,# ß !ß !ÓX and 0 ÐBÑ œ ,## Þ The
Lagrange muliplier conditions f0 ÐBÑ œ -" f1" ÐBÑ  -# f1# ÐBÑ yield the equations
#,# œ -# ß ! œ -" ,#  -# ," ß and ! œ -" , or
! œ -# ,"
and
#,# œ -# ;
if ," Á !ß this system has a solution (-# œ !Ñ only when ,# œ !à when ," Á ! and ,# Á !, the system
# ÑÑ
has no solution. So in general .0 ÐBÐ,
Á -# because 0 ÐBÐ,# ÑÑ is not a well-behaved function.
.,#
OR-MA-ST 706 HW #2
page 2
Derive a formula for the shortest distance . from the fixed point B % I 8 to the hyperplane D œ - X BÞ Now
assume that -  œ "; show that . # œ ÐD  - X BÑ# .
SOLUTION:
738 . # œ mB  Bm# =?,4/-> >9 - X B œ D
B − I8
4.
 #ÐB"  B" Ñ 
 -" 
ã
f0 ÐBÑ  -f1ÐBÑ œ ! Ê
- ã
œ ! Ê B3 œ -# -3  B3 ß 3 œ "ß ÞÞÞß 8Þ Ð‡Ñ
 #ÐB8  B8 Ñ 
 -8 
D œ - X B œ  Ò -# -3#  -3 B3 Ó œ -#  -3#  - X B Ê -# m-m# œ D  - X B Ê - œ
get
B3 œ
8
8
3œ"
3œ"
ÐD- X BÑ
m-m# -3
 B3 ß so B3  B3 œ
ÐD- X BÑ
m-m# -3
If -  œ " then . # œ ÐD  - X BÑ# .
#ÐD- X BÑ
à
m-m#
X
B
 -3# œ
and 738 . # œ  Dm-m# 
# 8
3œ"
plug into Ð‡Ñ to
ÐD- X BÑ#
m-m# Þ
5.
Assume that we wish to maximize 0 ÐBÑ =Þ >Þ 13 ÐBÑ œ ,3 ß 3 œ "ß á ß 7ß B % I 8 ß 7  8Þ Suppose that 0 ÐBÑ
has a local maximum at B subject to the constraints and that <ÐN ÐBÑÑ œ 7Þ If f0 ÐBÑ œ !, what are the
values of the Lagrange multipliers? What is the simple interpretation of such a situation?
SOLUTION:
Since <ÐN ÐBÑÑ œ 7, B is a regular point, that is, the gradients fg3 ÐBÑß 3 œ "ß #ß ÞÞÞß 7 are linearly
independent, so -3 œ !, 3 œ "ß #ß ÞÞÞß 7Þ The point B is an unconstrained local or global minimum,
local or global maximum, or a saddlepoint.
6.
Consider the problem
7+B 9< 738 0 ÐBÑ œ BX EB
=Þ >Þ B œ "ß
where B %I 8 and E is 8 ‚ 8 and symmetric. Determine the equations whose solutions contain the points at
which 0 takes on its maximum and minimum values. Do the same for the problem that results when the
constraint HB œ ! is added to the above problem, where H is 7 ‚ 8ß 7  8ß and <ÐHÑ  7.
SOLUTION:
#
f0 ÐBÑ œ #EB Ð2A #"Ñà " œ B Í " œ B œ B X B œ BX MBß so fgÐBÑ œ #MB œ #Bà so
f0 ÐBÑ  -fgÐBÑ œ ! Ê EB  -B œ !Þ With the constraint B œ ", this is the (unit)
eigenvector, eigenvalue problemÞ ! œ EB  -B œ ÐE  -MÑB has nonzero solutions B when
ÐE  -MÑ is singular, that is, when detÐE  -MÑ œ !, which results in an 8>2 degree polynomial
equation that will have 8 (not necessarily unique) roots -" Ÿ -# Ÿ ÞÞÞ Ÿ -8 Ðthese are the
eigenvalues, the corresponding B's are the eigenvectors). The B that corresponds to -8 , the maximum
eigenvalue, gives the maximum of 0 ÐBÑ œ BX EBÞ The B that corresponds to -" , the minimum
eigenvalue, gives the minimum of 0 ÐBÑ œ BX EBÞ
Add the constraint HB œ !Þ
The resulting Lagrange multiplier equations are
#EB  #-B  .H œ !ß mBm œ "ß HB œ !Þ
Note that ÖB − I 8 À HB œ !× is the null space of D, so we want the unit eigenvectors of A that are in the
null space of H. The projection matrix T into the null space of H is
T8‚8 œ M  HX ÐHHX Ñ H, where ÐHHX Ñ denotes a psuedoinverse of HHX à that is, for
C − I 8 ß T C is in the null space of H.
7.
In using Lagrange multipliers to solve the linear program
Q +B D œ - X B =Þ >Þ EB œ ,ß B % I 8 ,
OR-MA-ST 706 HW #2
page 3
how would (a) unboundedness and (b) infeasibility be discovered?
SOLUTION:
Lagange multiplier necessary conditions are -  EX - œ ! Ê EX - œ - .
(a) Unboundedness would be indicated by not being able to find - − I 7 satisfying EX - œ - (duality
theorem of LP: dual problem is min , X - =Þ>Þ EX - œ - ; if dual problem is infeasible, then primal is
unbounded).
(b) Infeasibility If there are many -'s satisfying EX - œ - such that , X - is unbounded, then Q +B
D œ - X B =Þ >Þ EB œ , is not feasible.