OR-MA-ST 706 HW #3 Due Thurs. March 31
Spring 2016
1.
Reiland
Let B‡ − \ œ ÖB − I 8 À 13 ÐBÑ !ß 3 œ "ß á ß 7à 24 ÐBÑ œ !ß 4 œ "ß á ß :× and define
^ " ÐB‡ Ñ œ œ ÖD − I 8 À D X f13 ÐB‡ Ñ !ß 3 − MÐB‡ Ñß D X f24 ÐB‡ Ñ œ !ß 4 œ "ß á ß :×
where MÐB‡ Ñ œ Ö3 À 13 ÐB‡ Ñ œ !×Þ
Show that ^ " ÐB‡ Ñ is a closed convex cone.
(Note: a set E © I 8 is closed if it contains its limit points, that is, if E is closed and ÖB5 × is a sequence of
points in E such that B5 Ä Bß then B − E).
If D − ^ " ÐB‡ Ñß then clearly αD − ^ " ÐB‡ Ñ for α !Þ Therefore ^ " ÐB‡ Ñ is a cone.
Suppose D " and D # are elements of ^ " ÐB‡ Ñà then for 3 − MÐB‡ Ñß D "X f13 ÐB‡ Ñ 0,
D #X f13 ÐB‡ Ñ 0ß and D "X f24 ÐB‡ Ñ œ !ß D #X f24 ÐB‡ Ñ œ ! for 4 œ "ß á ß :Þ
Then for 3 − MÐB‡ Ñß
ÐD "  D # ÑX f13 ÐB‡ Ñ œ D "X f13 ÐB‡ Ñ  D #X f13 ÐB‡ Ñ !
(1)
ÐD "  D # ÑX f24 ÐB‡ Ñ œ D "X f24 ÐB‡ Ñ  D #X f24 ÐB‡ Ñ œ !
Ð#Ñ
In addition,
(1) and (2) imply that D "  D # − ^ " ÐB‡ Ñ.
Since ^ " ÐB‡ Ñ is a cone and D "  D # − ^ " ÐB‡ Ñ when D " ß D # − ^ " ÐB‡ Ñß ^ " ÐB‡ Ñ is a convex
cone.
^ " ÐB‡ Ñ is closed:
Let ÖD 5 × be a sequence in ^ " ÐB‡ Ñ such that lim D 5 œ DÞ Since each D 5 − ^ " ÐB‡ Ñß we have
5X
‡
D65 `13 ÐB Ñ
`B6
6œ"
8
‡
D f13 ÐB Ñ œ
and
D 5X f24 ÐB‡ Ñ œ D65
8
6œ"
`24 ÐB‡ Ñ
`B6
5Ä∞
! for 3 − MÐB‡ Ñ
œ !ß 4 œ "ß á ß :Þ
We must show that D − ^ " ÐB‡ ÑÞ
For 3 − MÐB‡ Ñß
3 ÐB Ñ
3 ÐB Ñ
3 ÐB Ñ
5 `13 ÐB Ñ
D X f13 ÐB‡ Ñ œ D6 `1`B
œ  lim D65  `1`B
œ  lim D65 `1`B
 œ lim D6 `B6
6
6
6
8
6œ"
‡
8
6œ"
‡
5Ä∞
5
"
8
6œ"
‡
5Ä∞
‡
‡
‡
5Ä∞ 6œ"
œ lim D f13 ÐB Ñ ! since D − ^ ÐB Ñ for all 5Þ
5X
8
5Ä∞
For 4 œ "ß á ß :ß
D X f24 ÐB‡ Ñ œ D6
8
6œ"
`24 ÐB‡ Ñ
`B6
œ  lim D65 
8
6œ"
5Ä∞
`24 ÐB‡ Ñ
`B6
œ  lim D65
8
6œ"
5Ä∞
`24 ÐB‡ Ñ
`B6 
`24 ÐB Ñ
œ lim D65 `B
6
8
5Ä∞ 6œ"
‡
OR-MA-ST 706 HW #3 Solutions
page 2
œ lim D 5X f24 ÐB‡ Ñ œ ! since D 5 − ^ " ÐB‡ Ñ for all 5Þ
5Ä∞
Remark: Recall that we proved in class that HÐB‡ Ñ © ^ " ÐB‡ Ñ where HÐB‡ Ñ œ ÖD − I 8 À B œ B‡  ) D − \ for
) − Ð!ß $ Ñ for some $  !× is the set of feasible directions at B‡ . Since ^ " ÐB‡ Ñ is closed from the above
problem, then in fact we have HÐB‡ Ñ © ^ " ÐB‡ Ñ where HÐB‡ Ñ denotes the closure of HÐB‡ Ñ
Òthis latter set containment is true since if F − I 8 ß G − I 8 with G a closed set and F © G , then F © GÞ
Proof: Suppose B − Fà then there exists a sequence ÖB5 × © F such that B5 Ä B. ÖB5 × © F and F © G
implies ÖB5 × © Gà since G is closed, the limit of ÖB5 × is in G , that is, B − GÞ Therefore, F © G ].
2(a). Consider the constraints
1" ÐBÑ œ Ð"  B"  B# Ñ$ !
1# ÐBÑ œ B" !
1$ ÐBÑ œ B# !Þ
X
Prove that at the point B‡ œ  "# ß "#  , we have HÐB‡ Ñ § ^ " ÐB‡ Ñ, that is, show that HÐB‡ Ñ is properly
contained in ^ " ÐB‡ Ñ.
^ " ÐB‡ Ñ À
 $Ð"  B"  B# Ñ#
!
X
At B‡ œ  "# ß "#  , MÐB‡ Ñ œ Ö"×Þ f1" ÐB‡ Ñ œ 
 œ  !  so
 $Ð"  B"  B# Ñ#
^ " ÐB‡ Ñ œ ÖD − I # À D X f1" ÐB‡ Ñ !× œ ÖD − I # À !D"  !D# !× œ I # Þ
HÐB‡ Ñ À
HÐB‡ Ñ œ ÖD − I # À B‡  ) D − \ß for ) − Ð!ß $ Ñ for some $  !×Þ
"
 )D"
B‡  )D œ  #"
;
#  ) D#
1" ÐB‡  )DÑ œ Ð"  "#  )D"  "#  )D# Ñ$ œ Ð  )ÐD"  D# ÑÑ$ œ  )$ ÐD"  D# Ñ$ ! Ê ÐD"  D# Ñ$ Ÿ !
Ê D"  D# Ÿ ! Ê D# Ÿ  D"
D
‡
1# ÐB  )DÑ œ "#  )D" ! for ) − (!ß $ ÐD" ÑÑ for some $ ÐD" Ñ  ! for any D œ  "  − I #
D#
D
"
1$ ÐB‡  )DÑ œ #  )D# ! for ) − (!ß $ ÐD# ÑÑ for some $ ÐD# Ñ  ! for any D œ  "  − I #
D#
Therefore HÐB‡ Ñ œ ÖD − I # À D# Ÿ  D" ×Þ Since HÐB‡ Ñ is closed, HÐB‡ Ñ œ HÐB‡ ÑÞ
Therefore HÐB‡ Ñ œ ÖD − I # À D# Ÿ  D" × § ^ " ÐB‡ Ñ œ I # .
2(b). Show that the constraints
1% ÐBÑ œ Ð"  B"  B# Ñ !
1& ÐBÑ œ B" !
1' ÐBÑ œ B# !
define precisely the same feasible region as the constraints in 2(a). In addition, show that for any feasible
B, HÐBÑ œ ^ " ÐBÑÞ
Let \ " œ ÖB − I # À 13 ÐBÑ !ß 3 œ "ß #ß $×ß \ # œ ÖB − I # À 13 ÐBÑ !ß 3 œ %ß &ß '×.
i)
B − \ " Ê Ð"  B"  B# Ñ$ !ß B" !ß B# ! Ê Ð"  B"  B# Ñ !ß B" !ß B# ! Ê B − \ #
so \ " © \ # Þ
OR-MA-ST 706 HW #3 Solutions
page 3
ii) B − \ # Ê Ð"  B"  B# Ñ !ß B" !ß B# ! Ê Ð"  B"  B# Ñ$ !ß B" !ß B# ! Ê B − \ "
so \ # © \ " Þ
\ " © \ # and \ # © \ " Ê \ " œ \ # Þ So the feasible regions (shown below) are the same.
Show that for any feasible B, HÐBÑ œ ^ " ÐBÑÞ For an arbitrary feasible Bß it is not known which constraints are
active. There are 7 cases to consider.
Case i): if B is in the interior of the feasible region, that is, 13 ÐBÑ  ! for 3 œ %ß &ß 'ß then we can move a short
distance in any direction from B and remain feasible, so HÐBÑ œ HÐBÑ œ I # Þ In addition, since
MÐBÑ œ g, by definition ^ " ÐBÑ œ I # Þ So HÐBÑ œ ^ " ÐBÑÞ
For reference in what follows, note that for any feasible B,
f1% ÐBÑ œ 
"
"
!
ß f1& ÐBÑ œ  ß f1' ÐBÑ œ  Þ
 "
!
"
In addition, for any feasible Bß
HÐBÑ œ ÖD − I # À B  ) D − \ß for ) − Ð!ß $ Ñ for some $  !×Þ
so 1% ÐB  )DÑ œ "  B"  )D"  B#  )D# œ "  ÐB"  B# Ñ  ) ÐD"  D# Ñ !à
1& ÐB)DÑ œ B"  )D" !à
1' ÐB)DÑ œ B#  )D# !à
Case ii) MÐBÑ œ Ö%×ß that is "  B"  B# œ !ß B"  !ß B#  !:
1% ÐB  )DÑ œ  )ÐD"  D# Ñ ! for ) ! Ê D"  D# Ÿ ! Ê D# Ÿ  D"
so HÐBÑ œ HÐBÑ œ ÖD − I # À D# Ÿ  D" ×Þ
^ " ÐBÑ œ ÖD − I # À D X f1% ÐBÑ !× œ ÖD − I # À D# Ÿ  D" ×Þ
Therefore HÐBÑ œ ^ " ÐBÑÞ
Case iii) MÐBÑ œ Ö&×, that is "  B"  B#  !ß B" œ !ß B#  !:
1& ÐB)DÑ œ B"  )D" ! Ê D" !
so HÐBÑ œ HÐBÑ œ ÖD − I # À D" !×Þ
^ " ÐBÑ œ ÖD − I # À D X f1& ÐBÑ !× œ ÖD − I # À D" !×Þ
Therefore HÐBÑ œ ^ " ÐBÑÞ
Case iv) MÐBÑ œ Ö'×ß that is "  B"  B#  !ß B"  !ß B# œ !:
OR-MA-ST 706 HW #3 Solutions
1' ÐB)DÑ œ B#  )D# ! Ê D# !
so HÐBÑ œ HÐBÑ œ ÖD − I # À D# !×Þ
^ " ÐBÑ œ ÖD − I # À D X f1' ÐBÑ !× œ ÖD − I # À D# !×Þ
Therefore HÐBÑ œ ^ " ÐBÑÞ
Case v) MÐBÑ œ Ö%ß &×, that is, "  B"  B# œ !ß B" œ !ß B#  !:
from cases ii) and iii), HÐBÑ œ HÐBÑ œ ÖD − I # À D# Ÿ  D" and D" !×Þ
^ " ÐBÑ œ ÖD − I # À D X f1% ÐBÑ !ß D X f1& ÐBÑ !× œ ÖD − I # À D# Ÿ  D" and D" !×
Therefore HÐBÑ œ ^ " ÐBÑÞ
Case vi) MÐBÑ œ Ö%ß '×, that is, "  B"  B# œ !ß B"  !ß B# œ !:
from cases ii) and iv), HÐBÑ œ HÐBÑ œ ÖD − I # À D# Ÿ  D" and D# !×
^ " ÐBÑ œ ÖD − I # À D X f1% ÐBÑ !ß D X f1' ÐBÑ !× œ ÖD − I # À D# Ÿ  D" and D# !×
Therefore HÐBÑ œ ^ " ÐBÑÞ
Case vii) MÐBÑ œ Ö&ß '×, that is, "  B"  B#  !ß B" œ !ß B# œ !:
from cases iii) and iv), HÐBÑ œ HÐBÑ œ ÖD − I # À D" ! and D# !×
^ " ÐBÑ œ ÖD − I # À D X f1& ÐBÑ !ß D X f1' ÐBÑ !× œ ÖD − I # À D" ! and D# !×
Therefore HÐBÑ œ ^ " ÐBÑÞ
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