3.9 BJS3 p.136
2x1 + 2x2 + 4x3 + 0x4 + 5x5 + 3x6
3x1 + 6x2 + 3x3 + 3x4 + 3x5 + 4x6 ≤ 60
x1,...,x6 ≥ 0
Maximize
subject to
transform to:
2x1 + 2x2 + 4x3 + 0x4 + 5x5 + 3x6 + 0x7
3x1 + 6x2 + 3x3 + 3x4 + 3x5 + 4x6 + x7 = 60
x1,...,x6,x7 ≥ 0
Maximize
subject to
(a)
Find all basic feasible solutions and solve.
Basic feasible solution
i Ratio: 60/A1i ci(60/A1i)
1: 60/3 = 20 2(20) = 40
(20,0,0,0,0,0,0)
2: 60/6 = 10 2(10) = 20
(0,10,0,0,0,0,0)
3: 60/3 = 20 4(20) = 80
(0,0,20,0,0,0,0)
4: 60/3 = 20
0(20) = 0
(0,0,0,20,0,0,0)
5: 60/3 = 20 5(20) = 100
(0,0,0,0,20,0,0)
6: 60/4 = 15 3(15) = 45
(0,0,0,0,0,15,0)
7: 60/1 = 60
0(60) = 0
(0,0,0,0,0,0,60)
Optimal solution:
(b)
(0,0,0,0,20,0,0)
See (b).
Objective value
40
20
80
0
100
45
0
with objective value 100.
Prescribe a general rule for solving a problem of the form:
Maximize
subject to
Σj=1,...,n cj xj
Σj=1,...,n aj xj ≤ b , x1,...,xn ≥ 0 , where a1,...,an > 0 , b > 0 , and c 1,...,cn ≥ 0.
We can add a slack variable x n+1 with an+1 = 1 and c n+1 = 0 to put the problem in the form:
Maximize
subject to
Σj=1,...,n+1 cj xj
Σj=1,...,n+1 aj xj
= b , x1,...,xn+1 ≥ 0 , where a1,...,an+1 > 0 , b > 0 , and c 1,...,cn+1 ≥ 0.
The structural constraint can be written in the form: Ax = b , where A = [ a 1 a2 ... an an+1 ] is 1 by n+1 .
Since a1,...,an+1 ≥ 0 , Ax = b and x ≥ 0 imply that the feasible region is bounded; thus there is an exteme
point optimum at a basic feasible solution.
All bases are 1x1 matrices.
Each basis inverse has as its only entry the reciprocal of the only entry of the basis matrix.
Thus for each j , the corresponding basic feasible solution is x j = b/aj with objective value c j xj = b( cj/aj ) .
Since a1,...,an+1 ≥ 0 , b > 0 , and c 1,...,cn+1 ≥ 0 , maxj=1,...,n+1 { b( cj/aj ) } = b
will be the optimal objective value.
( maxj=1,...,n+1 { cj / aj } )
Let M = maxj=1,...,n+1 { cj / aj } .
j
Each xj such that c j/aj = M will define an optimal extreme point solution (b/a j) e with objective value bM .
(c)
Comment on how you would treat the case c j < 0 for any j.
Such a modification has no effect on the
boundedness of the feasible region. Thus there is an extreme point optimum at a basic feasible solution.
Having added the slack variable x n+1 with an+1 = 1 and c n+1 = 0 , so that the problem is in the form:
Maximize
subject to
Σj=1,...,n+1 cj xj
Σj=1,...,n+1 aj xj
= b , x1,...,xn+1 ≥ 0 , where a1,...,an+1 ≥ 0 , b > 0 , and c 1,...,cn+1 ≥ 0 ,
One basic feasible solution is x n+1 = b with objective value 0 .
For each j such tha c j < 0 the corresponding basic feasible solution is x j = b/aj with objective value c j xj < 0.
Thus we can ignore the ratios c j/aj with cj < 0 when looking for the optimal solution.
In this case we let M = max j=1,...,n+1 { cj / aj : cj ≥ 0 } and identify optimal basic feasible solutions as before.
(d)
Comment on how you would treat the case c j > 0 and aj ≤ 0 for any j.
(i) Subcase: assume c j > 0 and a j = 0 for some j.
When a j = 0, the only constraint on x j is xj ≥ 0 .
(Note that A has a "zero" column.)
Thus we can make xj as large as we like and increase the objective value as much as we like since c j > 0 .
j
The objective value is unbounded along the ray {x + λe : λ ≥ 0} for any feasible solution x.
Example:
Maximize
2x1 + 2x2 + 4x3 + 0x4 + 5x5 + 3x6 + 1x7 + 0x8
subject to
3x1 + 6x2 + 3x3 + 3x4 + 3x5 + 4x6 +0x7 + x8 = 60
x1,...,x7,x8 ≥ 0
x = (0,0,0,0,0,0,λ,60) is a feasible solution with objective value λ for any λ ≥ 0.
(ii) Subcase: assume c j > 0 and a j < 0 for some j.
When a j < 0 the corresponding basic solution is x j = b/aj < 0 and is not feasible.
However, for any basic feasible solution x k = b/ak with ak > 0 an improving extreme direction is possible.
Assume, without loss of generality, that k = 1 with a1 > 0 and N .1 = A. j = [ aj ] .
Then B = A.1 = [ a1 ] and B
1
-1
1
-1
= [1/a1] .
1
Consider p = [-B N.1 | e ] = [ -aj / a1 | e ] .
1
Since a1 > 0 and a j < 0 , -aj/a1 > 0 so that p is an extreme direction for the feasible region.
B
N
1
And [ c | c ] · p = c1(-aj/a1) + cj > 0 since c 1 > 0 , (-aj/a1) > 0 and cj > 0 .
1
j
Thus the optimal objective is unbounded along the ray { (-a j/a1)e + λ e : λ ≥ 0 }
Example:
Maximize
2x1 + 2x2 + 4x3 + 0x4 + 5x5 + 3x6 + 1x7 + 0x8
3x1 + 6x2 + 3x3 + 3x4 + 3x5 + 4x6 -6x7 + x8 = 60
x1,...,x7,x8 ≥ 0
x = (20+2λ,0,0,0,0,0,λ,0) is a feasible solution with objective value 40+5λ for any λ ≥ 0.
subject to
STD
H6.1 Prove that if d is a direction for X
then d is an extreme direction for X
Proof
with exactly two positive components,
STD
.
Assume that d is a direction for X
Thus Ad = 0 , d ≥ 0, and d ≠ 0.
STD
with exactly two positive components.
Let A. j and A.k (with j ≠ k) be the two columns of A corresponding to the positive
components of d. Thus d j > 0 and dk > 0 with all other components of d zero.
d
Now Ad = dj A. j + dk A.k = 0 with dj > 0 and dk > 0 which implies Q = { A. j , A.k }
is linearly dependent.
d
Consider Q / { A. j } = { A.k } and
d
Q / { A. k } = { A. j } .
Since we have assumed the structural matrix A has no zero columns,
both { A. j } and { A.k } are linearly independent.
d
ED
So Q is nearly linearly independent and by the Independence Characterization of X
Theorem (see lecture 17), d is an extreme direction for X
STD
.
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