Unformatted text preview: Linear Programming and Integer Programming
Mid-term Examination
Solution 1.
(a) True. Hint: Recall the definitions of polyhedra set and convexity.
(b) False. Hint: Let the obj equal 0. Then every BFS is optimal, and in general every BFS is clearly not
adjacent.
(c) False. Hint: Introduce two auxiliary nonnegative variables to reformulate the LP equivalently.
(d) True. Hint: Let yi = |xi |, ∀i and a linear optimization problem in y follows. We can then form an
optimal solution by taking x = y, but any x with xi = ±yi , ∀i will be a solution.
(e) True. Hint: This means that improvements to this variable will increase the objective function at a
nonzero constant rate and no upper bound is imposed on the variable.
(f) True. Hint: There exists at least one feasible point (0, b).
(g) False. Hint: Consider the case where the obj is a constant.
2.
(a) True. According to certain Theorem.
(b) False. Hint: The origin may not be a BFS solution, in which case Big M or two-phase methods will
select the origin as an initial BFS, but this does not correspond to any BFS solution.
(c) True. According to certain Theorem.
(d) True. According to certain Theorem.
(e) False. Hint: According to definition, λ should be in [0, 1]
(f) False. 1 3. MATH
3205
/ MATH
2230
MATH
3205
/ MATH
2230
MATH
3205
/Programming
MATH
2230 / OR
Linear
andand
Integer
I I
Linear
Integer
Programming
/ OR
Linear and Integer Programming / OR I
Mid-term
solution
Mid-term
solution
Mid-term solution October
24,24,
2014
October
2014
October 24, 2014 1. 1.
1.
(a)(a)
4. (a)
max
900x
+ 1000x
+ 1200x
max 900x
1 +1 1000x
2 +2 1200x
3, 3,
xmax
x21,x
,x32 ,x900x
1 ,x
3
1 + 1000x2 + 1200x3 ,
x1 ,x2 ,x3
s.t.s.t.2x12x
+1 2x
+2 x+3 x≤3 180,
+ 22x
≤ 180,
s.t. 2x1 + 2x2 + x3 ≤ 180,
x1 x
+1 2x
+2 3x
+ 22x
+ 33x≤3 120,
≤ 120,
x1 + 2x2 + 3x3 ≤ 120,
x1 x
+1 x+2 x
+2 2x
+ 32x≤3 160,
≤ 160,
x1 + x2 + 2x3 ≤ 160,
x1 ,xx12,,xx23, x≥3 0.
≥ 0.
x1 , x2 , x3 ≥ 0.
Z
Z1
10
00
00
0
(b)(b)
(a)
(b) ) 1 ) (x2(x
) 2 ) (x3(x
) 3 ) x4 x4 x5 x5 x6 x6 RHS
Z (x1(x
RHS
(x1 ) (x
(x3 ) x40 x50 x60 RHS
2)
1 -900
-900-1000
-1000-1200
-1200 0
0
0 0 0
-900 -1000
0
2 2 -1200
1 1 01 1 00 0 00 0 180
0 2 2
180
21
22
13
10 01 00 180
0
1
2
3
0
1
0 120120
1
21
32
0
1
0
120
0 1 1
1
2 0 0 0 0 1 1 160160
1
1
2
0
0
1
160
R3 R3
R3 R
, 3,
R3 → →
3→
3 R3 ,
3− 2R + R → R ,
− 2R
3 3 +4 R4 →4 R4 ,
− 2R3 + R4 → R4 ,
−R
+3 R
R2 R
, 2,
−3 R
+2 R→
2 →
− R3 + R2 → R2 ,
1200R
R1 R
, 1,
1200R
+1 R→
3 +3 R
1 →
(b) 1200R3 + R1 → R1 , Z
Z1
10
00
00
0 ) 1 ) (x2(x
) 2 )x3 x3 x4 x4 (x5(x
) 5 )x6 x6 RHS
Z (x1(x
RHS
(x1 ) (x
x
x
(x5 ) x60 RHS
2)
1 -500
-500-200
-200 30 0 40 0 400
400
0 48000
48000
-500 -200 0
0
400
0 48000
0 5/35/3 4/34/3 0 0 1 1 -1/3
-1/3 0 0 140140
5/3
4/3
0
1
-1/3
0
140
0 1/31/3 2/32/3 1 1 0 0 1/31/3 0 0 40 40
1/3
2/3
1
0
1/3
0
40
0 1/31/3 -1/3
-1/3 0 0 0 0 -2/3
-2/3 1 1 80 80
1/3 -1/3 0
0 -2/3 1
80
3R3R
2 2
R2 R
, 2,
3R2 → →
5 →
5 R2 ,
5 R2 R
2
−R−
R3 R
, 3,
+3 R→
2 +R
3 →
− 3 +3 R3 → R3 ,
3R2 R
2
−R−
R4 R
, 4,
+4 R→
22 + R
4 →
− 3 +3 R4 → R4 ,
3 +R →R ,
500R
500R
2 2 +1 R1 →1 R1 ,
500R2 + R1 → R1 ,
1 1 Z
1
0
0
0 (x1 )
-500
5/3
1/3
1/3 (x2 )
-200
4/3
2/3
-1/3 x3
0
0
1
0 x4
0
1
0
0 (x5 )
400
-1/3
1/3
-2/3 x6
0
0
0
1 RHS
48000
140
40
80 3R2
→ R2 ,
5
R2
+ R3 → R3 ,
−
3
R2
−
+ R4 → R4 ,
3
500R2 + R1 → R1 ,
Z
1
0
0
0 x1
0
0
0
0 (x2 )
200
4/5
2/5
-3/5 x3 (x4 )
0 1 300
0
3/5
1 -1/5
0 -1/5 (x5 )
300
-1/5
2/5
-3/5 x∗ = (84, 0, 12, 0, 0, 52),
x∗ = (84, 0, 12, 0, 0, 52), 5. x6
0
0
0
1 RHS
90000
84
12
52 z ∗ = 900000.
z ∗ = 900000. (c) (a) Optimal Solution: (x∗1 , x∗2 , x∗3 ) = (0, 60, 0)and Z ∗ = 120
1 0.8 0 0.6 −0.2 0 1 −0.2 0.4 0 B −1 A = . 0 0.4
Bas Eq
Coefficient
of
Right
¯
¯
0
−0.6
0
−0.2
−0.6
1
Var No Z
X1
X2
X3
X4
Side
X5
X6
-5M
-4M
-8M
∵ B −1 A = [B −1 A B −1 ],
Z
0 1
-3
-2
-4
M
0
0 -180M
∴ ¯
X5
1 0
2
1
3
0
60
0.6 0 −0.2 0 1
¯6 X
2 0
3
3
5
-1
0
1
120 B −1 = −0.2 0.4 0 .
Coefficient of
Right
Bas Eq
−0.2 −0.6 1
¯
¯
Var No Z
X1
X2
X3
X4
X5
X6
Side
(d)
M/3 - 4M/3
8M/3
- 20M
0
0
0 1
- 1/3
- 2/3
M
+ 4/3
+ 80
Z
C = 900 1000 1200 0 0 0 ,
X3
1 0
2/3
1/3
1
0
1/3
0
20
CB = 900 1200 0 ,
¯6
X
2 0
-1/3
4/3
-1
-5/3
1
20 0 2 2 1 1 0 0 Bas Eq
Coefficient
of
Right
1 2 3 0 1 0 A=
,
¯
¯ Var No Z
X1
X2
X3
X4
X5
X6
Side
1 1 2 0 0 1 M
M −0.2 +0 1/2
Z
0 1
-1/2
0
0 0.6 -1/2
+ 1/2
90 X3
1 0
3/4
0 B −1 = 1 −0.2 1/4
3/4
-1/4
15
0.4 0 , X2
2 0
-1/4
1
0
-3/4
-5/4
3/4
15
−0.2 −0.6 1
−1
Bas Eq
Right
CCoefficient
A − Cof≥ 0.
BB
¯5
¯6
Var No Z
X1
X2
X3
X4
X
X
Side
Let 900 + ε be the new price for the low grade papa, B remains
optimal
if
M
M
Z
X1
X2
i.e., 0
1
2 1
0
0 0
1
0 0 ε 12002/30]B −1 A-1/3
1
+
1/3 0 0 100
[900 +
− [900 + ε+ 1000
1200
0]
0
4/3
1/3
1
-1/3
20
=[0 200 + 0.8ε 0 300 + 0.6ε 300 − 0.2ε 0]
1/3
-2/3
-1
2/3
20
≥0, 1 3
200 + 0.8ε ≥ 0 ε ≥ −250
300 + 0.6ε ≥ 0
ε ≥ −500 ⇒ −250 ≤ ε ≤ 1500.
⇒ 300 − 0.2ε ≥ 0
ε ≤ 1500 Bas
Var Eq
No Z X1 X2 Z
X4
X2 0
1
2 1
0
0 1
3
2 0
0
1 ¯6
X Right
Side 0
1
0 ¯5
X
M
+2
3
1 M
-1
0 120
60
60 (b) Optimal Solution: (x∗1 , x∗2 , x∗3 ) = (0, 60, 0) and Z ∗ = 120
Phase 1:
Bas Eq
Coefficient of
Var No Z
X1
X2
X3
X4
Z
0 1
-5
-4
-8
1
¯
X5
1 0
2
1
3
0
¯6
X
2 0
3
3
5
-1 ¯5
X
0
1
0 ¯6
X
0
0
1 Right
Side
-180
60
120 X2
-4/3
1/3
4/3 Coefficient of
X3
X4
0
1
1
0
0
-1 ¯5
X
8/3
1/3
-5/3 ¯6
X
0
0
1 Right
Side
-20
20
20 X2
0
0
1 Coefficient of
X3
X4
0
0
1
1/4
0
-3/4 ¯5
X
1
3/4
-5/4 ¯6
X
1
-1/4
3/4 Right
Side
0
15
15 X2
-2
0
1 Coefficient of
X3
X4
-4
0
1
1/4
0
-3/4 ¯5
X ¯6
X ¯5
X ¯6
X ¯5
X ¯6
X Bas
Var
Z
X3
¯6
X Eq
No
0
1
2 Bas
Var
Z
X3
X2 Eq
No
0
1
2 Phase 2:
Bas Eq
Var No
Z
0
X4
1
X2
2 Z
1
0
0
Z
1
0
0 Z
1
0
0 X1
1/3
2/3
-1/3
X1
0
3/4
-1/4 X1
-3
3/4
-1/4 Coefficient of
X3
X4
2
4
3 Bas
Var
Z
X3
X2 Eq
No
0
1
2 Z
1
0
0 X1
-1/2
3/4
-1/4 X2
0
0
1 Coefficient of
X3
X4
0
-1/2
1
1/4
0
-3/4 Bas
Var
Z
X1
X2 Eq
No
0
1
2 Z
1
0
0 X1
0
1
0 X2
0
0
1 Coefficient of
X3
X4
2/3
-1/3
4/3
1/3
1/3
-2/3 4 Right
Side
0
15
15
Right
Side
90
15
15
Right
Side
100
20
20 Bas
Var
Z
X4
X2 Eq
No
0
1
2 Z
1
0
0 X1
1
3
2 X2
0
0
1 Coefficient of
X3
X4
2
0
4
1
3
0 ¯5
X ¯6
X Right
Side
120
60
60 6. 4.6-17.
Method 1:
Reformulation:
maximize ^ œ %B" &B# $B$ subject to B" B# #B$ B% B( œ #!
"&B" 'B# &B$ B& œ &!
B" $B# &B$ B' œ $!
B" ß B# ß B$ ß B% ß B& ß B' ß B( ! Since this is the optimal tableau for Phase 1 and the artificial variable x¯7 = 5 ≥ 0, the
problem
Since thisis isinfeasible.
the optimal tableau for Phase 1 and the artificial variable B( œ & !, the
problem is infeasible.
Method 2 ( Use Big M method): 5 Reformulation:
max Z = 4x1 + 5x2 + 3x3 − M x¯7 s.t. x1 + x2 + 2x3 − x4 + x¯7 = 20
15x1 + 6x2 − 5x3 + x5 = 50
x1 + 3x2 + 5x3 + x6 = 30
x1 , x2 , x3 , x4 , x5 , x6 , x¯7 ≥ 0. Bas Eq
Var No Z
Z
X5
X6
¯7
X 0
1
2
3 1
0
0
0 Bas Eq
Var No Z
Z
X5
X6
¯7
X 0
1
2
3 1
0
0
0 Bas Eq
Var No Z
Z
X5
X6
X3 0
1
2
3 1
0
0
0 X1 X2 -4
1
15
1 -5
1
6
3 X1
-M
-4
1
15
1 X1
- 3M/5
- 17/5
3/5
16
1/5 X2
-M
-5
1
6
3 X2
M/5
- 16/5
-1/5
9
3/5 Coefficient of
X3
X4
-3
2
-5
5 0
-1
0
0 Coefficient of
X3
X4
- 2M
-3
M
2
-1
-5
0
5
0
Coefficient of
X3
X4
0
0
0
1 M
-1
0
0 6 Right
Side X5 X6 ¯7
X 0
0
1
0 0
0
0
1 M
1
0
0 0
20
50
30 X5 X6 ¯7
X Right
Side 0
0
1
0 0
0
0
1 0
1
0
0 -20M
20
50
30 X5
0
0
1
0 X6
2M/5
+ 3/5
-2/5
1
1/5 Right
¯7
X
Side
18
0 - 8M
1
8
0
80
0
6 (a) B& enters.
(b) B% leaves.
(c) Ð%ß #ß %ß !ß #ß !ß !Ñ
5.1-21.
Bas Eq
Var No Z
Z
X5
X1
X3 0
1
2
3 X1 1
0
0
0 0
0
1
0 Coefficient of
X3
X4 X2
43M/80
- 103/80
-43/80
9/16
39/80 X5
X6
3M/80
7M/16
M + 17/80 + 13/16
-1
-3/80
-7/16
0
1/16
1/16
0
-1/80
3/16 0
0
0
1 Right
Side
35
0 - 5M
1
5
0
5
0
5 ¯7
X Since this is the optimal tableau for Big M method, Z = 35 − 5M = −∞, and the artificial
variable x¯7 = 5 ≥ 0, the problem is infeasible. 5.2-1.
7. Î B$ Ñ
(a) Optimal Solution: B" œ F " , œ
Ï B& Ò
^ œ -B œ a ) % ' (b) Shadow prices: -F F " œ - œ a& ) ( % ' ! "
Iteration 0: F œ F " œ Œ
!
-F œ a ! " ÑÎ ")! Ñ Î &! Ñ
$
#(! œ $!
"! ÒÏ ")! Ò Ï &! Ò Î "" $
' *
Ï # $ " Ñ Î "Þ$$ Ñ
$ œ
"
Ò
Ï
"!
#Þ'( Ò Î $! Ñ
Ð ! Ó
Ð
Ó
* bÐ &! Ó
œ **!
Ð
Ó
!
Ï &! Ò $ 5.2-2. Î "" $
' *
Ï # $ "
#( "
#( a ' *b ) ! b, E œ Œ #
$ $
& $
% #
# #
% !
B
"
, BF œ Œ ' œ Œ "
B(
! ! b, - œ a & "
Revised B# coefficients: Œ
! ) ( % "
! !
#!
,,œŒ "
$! !
#!
#!
œŒ Œ "
$!
$!
' ! ! b, so B# enters. !
$
$
œ Œ , so B( leaves.
Π"
&
& 5-10 7 ...
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