1.
Consider the LP problem: minimize
subject to
-2x1 - x2
x1 + x2 ≤ 5
2x1 + 3x2 ≤ 12
x1, x2 ≥ 0
The LP problem may be solved by the simplex method using the tableau format.
(a)
(b)
Complete the starting tableau for simplex solution of this problem (after adding slack variables).
x1
x2
x3
x4
row
basis
z
RHS
ratio
0
z
1
2
1
0
0
0
test
1
x3
0
1
1
1
0
5
5/1=5
2
x4
0
2
3
0
1
12
12/2=6
Perform one simplex iteration in the tableau format giving the result below.
Choose x 1 to enter using Dantzig's rule.
The minimum for the ratio test occurs in row 1. Pivot in row 1 of the x 1 column.
row
basis
z
x1
x2
x3
x4
RHS
0
z
1
0
-1
-2
0
-10
1
x1
0
1
1
1
0
5
2
x4
0
0
1
-2
1
2
(c)
Is the tableau in (b) optimal? Yes (all d j s are nonpositive)
(d)
Obtain the following quantities using information from the tableau for the basic
feasible solution in (b). All components must be in consistent order.
B = [ A.1 A.4 ]
(must be in this order since the x 1 value is in row 1)
N = [ A.2 A.3 ]
-1
B =
(the other ordering for these columns is possible)
(appears in the same position as the identity matrix in the initial tableau)
-1
B b = (appears in fhe RHS structural rows)
B
-1
c B N= z
N
N
= d +c
N
=
(d
N
=
1
0
2
1
1
1
3
0
=
1
-2
0
1
=
5
2
=
appears in row 0 above the nonbasic columns)
= [ -1 -2 ] + [ -1 0 ] = [ -2 -2 ]
B
-1
c B b = (the objective value appears in row 0 of the RHS)
B
-1
= -10
c B = (appears in row 0 above the position of the identity matrix in the initial tableau)
=
[ -2 0 ]
2.
Consider the following Knapsack Problems:
In problem 3.9 (homework#6) we transformed the problem:
Maximize
subject to
Σ j=1,...,n
Σ j=1,...,n
cj xj
Maximize
a j x j ≤ b , x 1 ,...,x n ≥ 0 ,
where a 1 ,...,a n > 0 , b > 0 , and c 1 ,...,c n ≥ 0.
to
subject to
Σ j=1,...,n+1
Σ j=1,...,n+1
cj xj
a j x j = b , x 1 ,...,x n+1 ≥ 0 ,
by defining a n+1 = 1 and c n+1 = 0 .
j
We discovered that a bfs with the largest objective value would be x j e where x j = M with
M = max
j=1,...,n+1
{ cj / aj } .
As long as a 1 ,...,a n+1 > 0 , b > 0 , and c 1 ,...,c n+1 ≥ 0,
any such largest objective value bfs will be an optimal solution.
(a) Find all optimal solutions to:
maximize 1x1 + 2x2 + 3x3
subject to 4x1 + 3x2 + 2x3 ≤ 12
In this problem instance M = max{1/4,2/3,3/2}=3/2,
so (0,0,12/2)=(0,0,6) is the only optimal solution and the
optimal objective is (3)(6) = 18.
x1, x2, x3 ≥ 0
(b) Find all optimal solutions to:
maximize 1x1 + 2x2 + 4x3
subject to 4x1 + 3x2 + 6x3 ≤ 12
x1, x2, x3 ≥ 0
In this problem instance M = max{1/4,2/3,4/6}=2/3 (tie),
so (0,12/3,,0)=(0,4,0) and (0,0,12/6)=(0,0,2) are two optimal
solutions with optimal objective (2)(4) = (4)(2) = 8. Thus the
set of optimal solutions is { (0,4λ,2(1-λ)) : 0 ≤ λ ≤ 1 } which
has optimal objective (2)(4λ) + (4)(2(1-λ)) = 8λ+8-8λ = 8.
(c) Find all optimal solutions to:
maximize 1x1 + 2x2 + 3x3
Approach 1: make the variable substitution xj = x'j + 1; the problem
becomes:
subject to 4x1 + 3x2 + 2x3 ≤ 12
maximize 1x'1 + 2x'2 + 3x'3 + 6
subject to 4x'1 + 3x'2 + 2x'3 ≤ 12 - 9 = 3
x1, x2, x3 ≥ 1
x'1, x'2, x'3 ≥ 0
In this problem instance M = max{1/4,2/3,3/2}=3/2,
so x' = (0,0,3/2) is the only optimal solution and the
optimal objective is (3/2)(3) + 6 = 21/2. Transforming,
x = (0,0,3/2) + (1,1,1) = (1,1,5/2) with optimal objective 21/2 .
Approach 2: since all variables must be at least at level 1, note that
x1=x2=x3=1 uses up 4(1)+3(1)+2(1)=9 units of the scarce
resourse (12 units) of the constraint,while yielding
1(1)+2(1)+3(1)=6 units of objective value (payoff). Removing
the 9 units of available scarce resourse and defining x'j to be
the additional units of decision variable xj to employ, we can
determine the optimal additional units by solving the related
problem:
maximize 1x'1 + 2x'2 + 3x'3 + 6
subject to 4x'1 + 3x'2 + 2x'3 ≤ 12 - 9 = 3
x'1, x'2, x'3 ≥ 0 ,
which leads to the same solution as in Approach 1.
In problem 3.9 (homework#6) we also discovered that when we have a column with a j < 0 we need to
allow for the possibility of directions for the feasible region being present.
(d) Find all optimal solutions to:
In this problem instance x3 with a3 < 0 cannot yield a bfs
maximize -1x1 + 2x2 - 2x3
and M = max{-1/3,2/3} = 2/3, so (0,4,0) is the best bfs.
subject to 3x1 + 3x2 - 3x3 ≤ 12
Two extreme directions d = (1,0,1) and d = (0,1,1) need to
x1, x2, x3 ≥ 0
1
be examined.
2
1
(-1,2,-2)·d = -3 so moving in this direction
2
decreases the objective value. But (-1,2,-2)·d = 0, so moving
in this direction maintains the level of the objective value.
thus the set of optimal solutions is { (0,4+λ,λ) : λ ≥ 0 } with
objective value (2)(4) = (2)(4+λ)+(-2)(λ) = 8.
(e) Find all optimal solutions to:
In this problem instance x3 with a3 < 0 cannot yield a bfs
maximize 1x1 + 3x2 - 2x3
and M = max{1/3,3/3} = 3/3 = 1, so (0,4,0) is the best bfs.
subject to 3x1 + 3x2 - 3x3 ≤ 12
Two extreme directions d = (1,0,1) and d = (0,1,1) need to
x1, x2, x3 ≥ 0
1
be examined.
2
1
(1,3,-2)·d = -1 so moving in this direction
2
decreases the objective value. But (1,3,-2)·d = 1, so moving
in this direction increases the objective value. Thus the
objective value is unbounded along the ray
{ (0,4,0) + λ(0,1,1) : λ ≥ 0 } and there is no optimal solution.
3.
Extreme Directions
The structural matrix for the constraint Ax = b of a linear programming problem in
standard form is:
A=
1
d =
(a)
1
-1
0
-1
1
0
1
1
0
0
-1
1
1
1
0
0
1
Is direction d an extreme direction for the feasible region X
Explain why or why not.
-1
0
-1
-1
1
0
0
0
STD
?
We proved in problem H6.1 (homework#6) that a direction for X
STD
with
1
exactly two positive components is an extreme direction. Since d has
1
exactly two positive components, d is an extreme direction.
2
d =
(b)
2
1
0
0
0
1
2
Write direction d as a nonnegative linear combination of two extreme directions
for the feasible region X
0
Let d =
0
1
STD.
0
0
0
0
0
0
1
0
0
Ad = 0, d ≥ 0, and d ≠ 0, so d is a direction and as was argued in (a), d is
2
0
1
also an extreme direction. Now d = d + d , a nonnegative linear
combination of two extreme directions.
(c)
3
d =
1
1
1
1
1
3
Is direction d an extreme direction for the feasible region X
Explain why or why not.
d3
Let Q
0
STD
?
= {A.1 , A.2 , A.3 , A.4 , A.5 } be the set of columns of A corresponding
d3
to positive components of A. Now Q /{A.1} = { A.2 , A.3 , A.4 , A.5 } is a set
3
of four (column) vectors from R and thus cannot be linearly independent.
3
By the extreme direction independence characterization theorem, d is not
an extreme direction.
3
You could also argue that d is not an extreme direction since it is a positive
3
1
4
linear combination of two distinct directions: d = (1)d + (1)d .
1
4
(Directions d and d are obviously distinct because they have different
patterns of zeros.)
4
d =
(d)
0
0
1
4
1
1
Is direction d an extreme direction for the feasible region X
Explain why or why not.
d4
Let Q
0
STD
?
= { A.3 , A.4 , A.5 } be the set of columns of A corresponding
4
to positive components of A. Now 0 = Ad = (1)A.3 + (1)A.4 + (1)A.5
so that Q
d4
is linearly dependent.
d4
Consider Q /{A.3} = { A.4 , A.5 } =
0
-1
-1
1
0
-1
-β
Now 0 = αA.4 + βA.5=
-α
α-β
implies α =β = 0, so
d4
Q /{A.3} is linearly independent.
d4
Consider Q /{A.4} = { A.3 , A.5 } =
1
-1
1
0
0
1
α-β
Now 0 = αA.3 + βA.5=
α
-β
implies α =β = 0, so
d4
Q /{A.4} is linearly independent.
d4
Consider Q /{A.5} = { A.3 , A.4 } =
1
0
1
0
-1
1
α
Now 0 = αA.3 + βA.4=
α-β
β
implies α =β = 0, so
d4
Q /{A.5} is linearly independent.
4
By the extreme direction independence characterization theorem, d is an extreme
direction.
4.
The simplex associated with an iteration of the Simplex Algorithm
The following tableau represents a specific simplex iteration.
row
0
basis
z
z
1
x1
x2
x3
x4
x5
x6
0
-5
0
4
-10
0
RHS
0
1
x6
0
0
3
0
-2
-1
1
12
2
x3
0
0
2
1
3
0
0
6
3
x1
0
1
-1
0
0
-4
0
0
(a)
Is the tableau optimal?
No.
(b)
Obtain the vertices of the simplex (of adjacent extreme points) associated with the basic
feasible solution given by the tableau. All components must be in the same order as the
variables in the tableau.
(The quantity d 4 in row 0 of the x 4 column is positive.)
B
N
The ordering of the basic and nonbasic variables will be: x =[x6 x3 x1] and X =[x2 x4 x5] .
(Other orderings of the nonbasic variables could be used; this seems most natural.)
The current bfs will be a vertex of its associated simplex so we let
vertex0 = (0 0 6 0 0 12) as obtained from the basis list and the RHS.
Approach 1: all adjacent extreme points can be obtained by pivoting in a nonbasic variable.
We examine the nonbasic variables.
x2: since there are positive values in rows 1 and 2 of the column associated with x2,
a blocking variable is present; thus the ratio test will tell where to pivot.
basis
basis
z
x2
RHS
ratio test
z
x2
RHS
x6
3
12
12/3 = 4
x6
0
12-(3)(3)=3
x2
1
3
x1
0
0-(-1)(3)=3
x3
2
6
6/2 = 3
x1
-1
0
-
→pivot→
vertex1 = (3 3 0 0 0 3).
x4: since there is a positive value in row 2 of the column associated with x4,
a blocking variable is present; thus the ratio test will tell where to pivot.
basis
basis
z
x4
RHS
ratio test
z
x2
RHS
x6
-2
12
-
x6
0
12-(-2)(2)=16
x3
3
6
6/3 = 2
x4
1
2
x1
0
0
-
x1
0
0-(-0)(2)=0
→pivot→
vertex2 = (0 0 0 2 0 16).
x5: since there is no positive value the column associated with x5,
STD
no blocking variable is present; thus we have a direction (4 0 0 0 1 1) for X
and will
not be able to obtain an adjacent extreme point by increasing this nonbasic variable.
This simplex of adjacent extreme points has only three vertices.
B
N
The ordering of the basic and nonbasic variables will be: x =[x 6 x 3 x 1 ] and X =[x 2 x 4 x 5 ] .
vertex0 = (0 0 6 0 0 12) as obtained from the basis list and the RHS.
j
Approach 2: generate the p vectors for the nonbasic variables and examine the
result of moving away from vertex0 using those vectors.
1
x2: vertex0 + β p = (0 0 6 0 0 12) + β ( -(-1) 1 -(2) 0 0 -(3) ) .
To maintain nonnegativity β must satisfy:
0 + β(1) ≥ 0
6 + β(-2) ≥ 0
12 + β(-3) ≥ 0
which implies β = 3 .
vertex1 = (0 0 6 0 0 12) + 3 ( -(-1) 1 -(2) 0 0 -(3) ) = (3 3 0 0 0 3) .
2
x4: vertex0 + β p = (0 0 6 0 0 12) + β ( -(0) 0 -(3) 1 0 -(-2) ) .
To maintain nonnegativity β must satisfy:
0 + β(0) ≥ 0
6 + β(-3) ≥ 0
12 + β(2) ≥ 0
which implies β = 2 .
vertex2 = (0 0 6 0 0 12) + 2 ( -(0) 0 -(3) 1 0 -(-2) ) = (0 0 0 2 0 16) .
3
x5: vertex0 + β p = (0 0 6 0 0 12) + β ( -(-4) 0 -(0) 0 1 -(-1) ) .
To maintain nonnegativity β must satisfy:
3
STD
Thus p is a direction for X
3
0 + β(4) ≥ 0
6 + β(0) ≥ 0
12 + β(1) ≥ 0
which implies β is
not bounded above.
and we will not be able to move to an adjacent
extreme point using p .
This simplex of adjacent extreme points has only three vertices.