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Exercises for Part V (Px means exercise on page x)
P18 ex The problem is
Min − y 1 (^) − y 2 (^) − y 3
St (^1 2 3 4 )
y + y + y + y = t
1 2 3 5 2
y + y + y + y = t
2 3 6 3
y + y + y = t
2 3 7 4
y + y + y = t
3 8 5
y + y = t
3 9 6
i^0
y y t
y
t 1 (^) = t 2 (^) = 300, t 3 (^) = t 4 (^) = 100, t 5 (^) = t 6 = 50
From previous exercise, we know that the final basic variables are Y4,Y1,Y2,Y7,Y8,Y9.
Then we have
T^1 A
− = [ 1 2/7 5/7 0 0 0
0 5/7 2/7 0 0 0 0 0 2/3 0 0 0 0 0 1/3 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1]^(-1) =[ 1.0000 -0.4000 -0.9000 0 0 0 0 1.4000 -0.6000 0 0 0 0 0 1.5000 0 0 0 0 0 -0.5000 1.0000 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 1.0000]
T^1 T A A
− (^) ′ =[0 0 0 1.0000 -0.4000 -0.9000 0 0 0
1.0000 0 0.6667 0 1.4000 -0.6000 0 0 0 0 1.0000 0.3333 0 0 1.5000 0 0 0 0 0 0.3333 0 0 -0.5000 1.0000 0 0 0 0 0.6667 0 0 0 0 1.0000 0 0 0 0.3333 0 0 0 0 0 1.0000] =
-2/5 -9/10 0 7/5 -3/5 2/ 0 3/2 1/ 0 -1/2 1/ 0 0 2/
0 0 1/
1 4 1 2 7 8 9
T f c c c c c c A
∗ − = b = -510 (min problem)
[ ] ( )
1 4 1 2 7 8 9
T (^) T y y y y y y A
− = b =90.0000 >
360.0000 > 150.0000 > 50.0000 > 50.0000 > 50.0000 > other variables are zero.
1 3 5 6 4 1 2 7 8 9 3 5 6
T c c c c c c c c c A
− − a a a =[0.0000 1.4000 0.9000] ≥ 0
∇ b (^) f =[0 -1.4000 -0.9000 0 0 0]
∇ c (^) f =[90.
50.0000]
∇ (^) d f =
Exercises for Part VI (Px means exercise on page x)
P13 ex
± -9 5 3 -f
ξ 1 3 ± -2 ± -1 X
ξ 2 2 ± -1 ± -1 X
Y3 3 ± -3 ± -1 η 3
-g Y1 Y
Notice: this solution should be correct, there was a small mistake in this problem in previous
solutions (Solutions I)
Y4 Y 8800 40 4 -f X1 60 -2 3/25 Y X4 20 1 -1/10 Y
X5 100 0 1/10 Y -g X2 X Primal solution (20,100,60,0,0) Dual solution(0 40 4 0 0) e) Y5 Y6 Y 510 7/5 9/10 0 -f X1 90 -2/5 -9/10 0 Y X7 360 7/5 -3/5 2/3 Y
X8 150 0 3/2 1/3 Y X4 50 0 -1/2 1/3 Y X5 50 0 0 2/3 Y X6 50 0 0 1/3 Y X2 X3 X Primal solution (360 150 0 90 0 0 50 50 50) Dual solution (0 7/5 9/10 0 0 0 0 0 0)
3 d)
Y1 Y2 Y 0 2 2 10 -f -13 -2 -4 -2 Y
-10 -2 -1 -3 Y -2 0 0 -1* Y 5/2 1 0 0 Y
Y1 Y2 Y -20 2 2 10 -f -9 -2 -4 -2 Y -4 -2* -1 -3 Y 2 0 0 -1 Y 5/2 1 0 0 Y
Y5 Y2 Y -24 1 1 7 -f -5 -1 -3* 1 Y 2 -1/2 1/2 3/2 Y 2 0 0 -1 Y 1/2 1/2 -1/2 -3/2 Y
X1 X2 X3 X4 X5 X6 X 24 -5 0 0 1/2 2 0 2 X2 1 1 1 0 -1/2 1/2 0 0 X6 1 3* 0 0 1/2 -1/2 1 0 X3 7 -1 0 1 3/2 -3/2 0 1
X1 X2 X3 X4 X5 X6 X 77/3 0 0 0 4/3 7/6 5/3 2 X2 2/3 0 1 0 -2/3 2/3 -1/3 0 X1 1/3 1 0 0 1/6 -1/6 1/3 0 X3 22/3 0 0 1 5/3 -5/3 1/3 1
You can find the relationships from this tableau with the previous solutions.
P20 ex1 First, 0 is a solution. Then for two
solutions (^) ( s 1 (^) x 1 (^) f 1 η 1 ) and (^) ( s 2 (^) x 2 (^) f 2 η 2 ) it is easy to show that
( s 1^ x 1^ f 1^ η 1 ) +^ ( s 2^ x 2^ f 2 η 2 )also satisfies^ cx^ =^ ds^ +^ f and^ bs^ =^ Ax +^ η^.
Next, for a solution (^) ( s 1 (^) x 1 (^) f 1 η 1 ), α( s 1 (^) x 1 (^) f 1 η 1 )satisfies cx = ds + f and
bs = Ax + η.
Notice that if x , , s η is determined, then − f is uniquely determined by cx = ds + f.
Since ( ) 0
T x
A I b
s
, where the dimension of the vector
T x
s
is m+q+1, and
the number of independent equations is q, so the dimension for the solution
x^ T
s
is given
by m+q+1-q=m+1.
Same argument can be given for the part (^) ( − g ξ t y ).
2 From cx = ds + f we get t ( − f )= dst − tcx
From bs = Ax + η we get y η = ybs − yAx
From − td = yb − g we get − gs = − dst − ybs
From tc = − yA + ξwe get ξ x = tcx + yAx
Sum the above up.
3 Use the result of problem 2, ( )( ) 0
T
− g ξ t y s x − f η =
4 Max f = cx − d
St Ax + η= b
x , η ≥ 0
The same thing for its dual.
5 From 2, cx − d + yb + d = cx + yb = f + g = ξ x + y η, (t=s=1), where x , η , ξ , y ≥ 0 ,
so the above equation must be greater than or equal to zero, and it equals zero only when
ξ x = 0 and y η = 0.
Exercises for Part VIII (Px means exercise on page x)
P6 ex Problem P is consistent if and only if b is in
( ,^ )|^ 0,^0
u r u r i r i
A xeu A xer A xiu A xir η x η
+^ +^ +^ ≥^ ≥
its dual Q is consistent if and only if c is in
( ,^ )|^ 0,^0
c i c i r i r
y Aeu y Aiu y Aer y Air ξ y ξ
−^ −^ −^ −^ +^ ≥^ ≥
The consistency of P and Q is independent. P Q are consistent if and only if P is bounded in which case
i) Q is bounded, and 0 = f ∗^ + g ∗
ii) The optimal solution sets for P and Q are not empty. iii) A feasible solution for P and a feasible solution for Q are optimal if and only if their
corresponding ,
i r
η ξ satisfy 0, 0
i i r r
y η = ξ x =.
P11 ex1 For dual problem Q(the column problem), we consider the labels of the Dual
problem. The rows with the artificial labels can be deleted along with the artificial variables(non-basic and hence always be zero). Check the elements in the distinguished row corresponding to the columns label artificial. If they all equals zero, then the columns labeled artificial can be eliminated. Otherwise, the dual is not consistent. Check the elements in the distinguished columns corresponding to the rows labeled unrestricted. If some of them are not zero, then the dual is unbounded if consistent. Now we end up with the lower-right sub-tableau.
P12 ex2 Consider the main duality proposition constructed for canonical form. If we extend
the problem a little and allow unrestricted variables, then the corresponding dual slack variables (the slack variables of the corresponding dual constraints which now become the dual artificial variables) should be zero. If we extend this problem a little bit and allow
P14 ex 3 The statements are quite similar to ex1 above, except that for property iii), at the optimal
we have η i yi = 0, xj ξ j = 0 for yi , x j restricted variables. (Since for unrestricted yi , x j ,
η i yi = 0, xj ξ j = 0 is always true because η i = 0, ξ j = 0 ).
Exercises for Part IX (Px means exercise on page x)
P5 ex1 Let ai be the row vector of A, then for the given u
∗ , we have a column vector
1
2 .
.
p
a u
a u
a u
∗
∗
∗
Suppose the maximum element is the a ui ∗
∗ , then v
∗
must be ( 0, 0...0,1, 0..0 ) where 1 is
the i
∗
th element (since E ( u , v ) E u ( , v )
∗ ∗ ∗
≤ ). Now consider ( v A 1 ,... v A m )
∗ ∗ where Aj is the
column vector of A. Suppose the minimum one is v A^ ∗^ j (^) ∗ Then the optimal u ∗^ must be
( 0, 0...0,1, 0..0)^
′ (^) with 1 in the j ∗^ entry (since
E u ( , v ) E u v ( , )
∗ ∗ ∗ ≤ ). Now ai j (^) ∗ ∗ is the
saddle point.
P6 ex2 The reverse argument of ex 1.
P12 ex1 The primal and dual are always feasible. The optimal value is given by
Min s (or the dual problem) s.t. Au ≤ s
1
m j j
u
u
=
The complementary slackness is: at the optimal solution
vi ( a ui − s ) = 0, u j ( t + vAj )= 0
P12 ex2 Similar to ex 1
P12 ex3 The primal is given in ex1. The dual is
Min t s.t. − vA ≤ t
1
m i i
v
v
=
It is equivalent to Min t
s.t. − A ′ ′ v ≤ t
1
m i i
v
v
=
∑
or Min t s.t. Av ′ ≤ t
1
m i i
v
v
=
∑
which is identical to the primal.
P21 ex1 Example 1:
Add 3 and get
.
Min − x 1 (^) − x 2
s.t. x 1 (^) + 5 x 2 ≤ 1
2 x 1 (^) + 6 x 2 ≤ 1
4 x 1 (^) + 5 x 2 ≤ 1
xi ≥ 0
The solution is (1/4,0), the optimal value is -1/4. So the optimal value for φ g is 4.
Remember we add 3 to A, so the value of the original problem is 4-3=1.
Example 2:
. Add 3 and get
Min − x 1 (^) − x 2
s.t. 5 x 1 (^) + x 2 ≤ 1
2 x 1 (^) + 4 x 2 ≤ 1
xi ≥ 0
The optimal value is (1/6,1/6) with the optimal value -1/3. So the optimal value for
φ g is 3. Remember we add 3 to A, so the value of the original problem is 3-3=0.
P21 ex2 From the matrix of example 2, we see that the expected pay for V is 0 no matter which
