Transcript x - Dalton State College

INTRODUCTION TO OPERATIONS
RESEARCH
Duality Theory
AUGMENTING THE DUAL PROBLEM


Since dual problems are also a linear
programming problem, it also has CP
solutions.
Since dual problems have function constraints
in ≥ form, must augment by subtracting
surplus variables from functional constraints.
Use surplus variable: zj – cj for each constraint j
 After augmenting dual problem:
(y1, y2, ..., ym, z1-c1, z2-c2, ..., zn-cn)

PRIMAL/DUAL PROBLEMS
Primal Variable
Decision Variables: xj
Slack Variables: xn+i


Associated Dual Variable
Surplus Variables: zj-cj
Decision Variables: yi
Complementary Basic Solution Property – Each basic solution
in the primal problem has a complementary basic solution in
the dual problem s.t. Z = W
Complementary Slackness Property – Given the association
between variables, the variables in the primal basic solution
and the complementary dual basic solution satisfy the
complementary slackness relationship:
Primal
Dual
Basic
Non-Basic
Non-Basic
Basic
WYNDOR GLASS CO. EXAMPLE
Maximize Z = 3x1 + 5x2
Subject to:
x1
+ x3
2x2
+ x4
3x1 + 2x2
+ x5
=4
= 12
= 18
Minimize W = 4y1 + 12y2 + 18y3
Subject to:
y1
+ 3y3 – (z1 – c1)
2y2
+ 2y3
– (z2 – c2)
=3
=5
WYNDOR GLASS CO. EXAMPLE
Optimal Solution
Basic
Variable
Z
x3
x2
x1
Z
1
0
0
0
x1
0
0
0
1
Dual Solution
Coefficient of:
x2
x3
x4
x5
0
0
3/2
1
0
1
1/3 -1/3
1
0
1/2
0
0
0
-1/3 1/3
Right
Side
36
2
6
2
Primal Solution
WYNDOR GLASS CO. EXAMPLE
Iteration
1
2
3
4
5
6
7
8
Basic
Feasible?
Solution
(0,0,4,12,18)
Yes
(4,0,0,12,6)
Yes
(6,0,-2,12,0)
No
(4,3,0,6,0)
Yes
(0,6,4,0,6)
Yes
(2,6,2,0,0)
Yes
(4,6,0,0,-6)
No
(0,9,4,-6,0)
No
Z=W
0
12
18
27
30
36
42
45
Basic
Feasible?
Solution
(0,0,0,-3,-5)
No
(3,0,0,0,-5)
No
(0,0,1,0,-3)
No
(-9/2,0,5/2,0,0)
No
(0,5/2,0,-3,0)
No
(0,3/2,1,0,0)
Yes
(3,5/2,0,0,0)
Yes
(0,0,5/2,9/2,0)
Yes
COMPLEMENTARY BASIC SOLUTIONS

Complementary Optimal Basic Solutions
Property – Each optimal basic solution in the
primal problem has a complementary
optimal basic solution in the dual problem
such that Z = W.
Satisfies Condition for Optimality?
Yes
No
Feasible?
Yes
No
Optimal
Super-Optimal
Sub-Optimal
Neither Feasible nor Optimal
NONSTANDARD LP
The approach here is similar to when we dealt
with non-standard formulations in the context of
the simplex method.
 There is one exception: we do not add artificial
variables. We handle “=“ constraints by writing
them as “<=“ constraints.

 This
is possible here because we do not require
here that the RHS is non-negative.
NONSTANDARD LP

Greater than or equal to constraints:
 Multiply
through the inequality constraint by -1 to
make it a less than constraint:
x1 + 2x2 ≥ 5
(x1 + 2x2 ≥ 5) (-1)
-x1 – 2x2
≤ –5
NONSTANDARD LP

Equality constraints:
 Convert
the equality constraint to a pair of
inequality constraints:
x1 + 2x2 = 5
and
x1 + 2x2
x1 + 2x2
≤5
≥5
x1 + 2x2 ≤ 5
-x1 – 2x2 ≤ –5
NONSTANDARD LP

Unrestricted variables:
 Replace
the variable unrestricted in sign, by the
difference of two nonnegative variables:
x1 + 2x2
≤5
x1 ≥ 0,
x2 unrestricted
x1 + 2(x2 – x2) ≤ 5
EXAMPLE
Maximize Z = x1 + x2 + x3
Subject to:
2x2 – x3
≥4
x1 – 3x2 + 4x3
=5
x1 – 2x2
≤ 18
x1, x2 ≥ 0,
x3 unrestricted
Maximize Z = x1 + x2 + (x3 – x3)
Subject to:
2x2 –
(x3 – x3)
≥4
x1 – 3x2 + 4(x3 – x3)
=5
x1 – 2x2
≤ 18
– 2x2 + x3 – x3
x1 – 3x2 + 4x3 – 4x3
–x1 + 3x2 – 4x3 + 4x3
x1 – 2x2
≤ –4
≤5
≤ –5
≤ 18
EXAMPLE
Primal Problem
Maximize Z = x1 + x2 + x3 – x3
Subject to:
– 2x2 + x3 – x3
≤ –4
x1 – 3x2 + 4x3 – 4x3
≤5
–x1 + 3x2 – 4x3 + 4x3
≤ –5
x1 – 2x2
≤ 18
Dual Problem
Minimize W = –4y1+5y2–5y3+3y4
Subject to:
y2 – y3 + y4 ≥ 1
–2y1 – 3y2 + 3y3 – 2y4 ≥ 1
–y1 – 4y2 + 4y3
≥ –1
y1 + 4y2 – 4y3
≥1
EXAMPLE
Dual Problem
Dual Problem
Minimize W = –4y1+5y2–5y3+3y4
Subject to:
y2 – y3 + y4 ≥ 1
–2y1 – 3y2 + 3y3 – 2y4 ≥ 1
–y1 – 4y2 + 4y3
≥ –1
y1 + 4y2 – 4y3
≥1
Equality constraint
(y2 – y3)
–3(y2 – y3)
–4(y2 – y3)
4(y2 – y3)
Unrestricted
variable (y2 – y3)
Minimize W = –4y1+5y2+3y3
Subject to:
y2 + y3 ≥ 1
–2y1 – 3y2 – 2y3 ≥ 1
y1 + 4y2
=1
y1, y3 ≥ 0,
y2 unrestricted
STREAMLINING THE CONVERSION


An equality constraint in the primal generates a dual
variable that is unrestricted in sign.
An unrestricted in sign variable in the primal generates an
equality constraint in the dual.
Primal Problem
Dual Problem
opt=max
opt=min
Constraint i :
<= form
= form
Variable i :
yi >= 0
yi urs
Variable j:
xj >= 0
xj urs
Constraint j:
>= form
= form