ECE 307- Techniques for Engineering Decisions · ECE 307 © 2005 - 2018 George Gross, University of...
Transcript of ECE 307- Techniques for Engineering Decisions · ECE 307 © 2005 - 2018 George Gross, University of...
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ECE 307 © 2005 - 2018 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved.
George GrossDepartment of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
ECE 307- Techniques for Engineering Decisions
Lecture 4. Duality Concepts in Linear Programming
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Definition: A LP is in symmetric form if all the
variables are restricted to be nonnegative and all
the constraints are inequalities of the type:
DUALITY
objective type corresponding inequality type
max ≤
min ≥
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We first define the primal and dual problems
DUALITY DEFINITIONS
.
)
(
(
.
.
)
.
T
T
T
m
max Z c xs t P
Ax bx
in W b y
s tD
A y c
y 0
0
=
=
≥ ≥
≤ ≥
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The problems (P ) and (D) are called the symmetric
dual LP problems; we restate them as
DUALITY DEFINITIONS
( )P
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DUALITY DEFINITIONS
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EXAMPLE 1: MANUFACTURER TRANSPORTATION PROBLEM
warehousesretail stores
R 1 R 2 R 3
W 1 2 4 3
W 2 5 3 4
W 1
W 2
R 1
R 2
R 3
shipment cost coefficients
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EXAMPLE 1: MANUFACTURER TRANSPORTATION PROBLEM
We are given that the supplies stored in warehouses
satisfy
We are also given the demands needed to be met at
the retail stores :
1 2W Wand
, ,1 2 3R R Rand
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EXAMPLE 1: MANUFACTURER TRANSPORTATION PROBLEM
The problem is to determine the least-cost shipping
schedule
We define the decision variable
The shipping costs may be viewed as
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FORMULATION STATEMENT
i j i ji j
i j
min Z c x x x x x x x
s.t.x x x
x x x
x x
x x
x x
x 0 i j
2 3
11 12 13 21 22 231 1
11 12 13
21 22 23
11 21
12 22
13 23
2 4 3 5 3 4
300
600
200
300
400
1,2, 1,2,3
≤
≤
≥
≥
≥
≥
= =
= = + + + + +
+ +
+ +
+
+
+
= =
∑∑
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DUAL PROBLEM SETUP USING SYMMETRIC FORM
= =
=
− −↔
↔
+↔
+↔
+↔
= =
∑∑2 3
1 1
11 12 131
21 22 232
11 213
12 224
13 235
300600200300400
1,2 1,2,3
ij iji j
ij
min Z c x
s.t.x x xy
x x xyx xy
x xyx xy
x 0 i j
− ≥ −
− − − ≥ −
≥
≥
≥
≥
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DUAL PROBLEM SETUP
= − − + + +
− + =
− + =
− + =
− + =
− + =
− + =
=
1 2 3 4 5
1 3 11
1 4 12
1 5 13
2 3 21
2 4 22
2 5 23
300 600 200 300 400
243534
1,2, ... ,5i
maxW y y y y ys.t.
y y cy y cy y c
y y cy y cy y c
y 0 i
≤
≤
≤
≤
≤
≤
≥
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The moving company proposes to the manufac-turer to:
To convince the manufacturer to get the business, the mover ensures that the delivery fees cannot exceed the transportation costs the manufacturer would incur (the dual constraints)
THE DUAL PROBLEM INTERPRETATION
1 1
2 2
1 3
2 4
3 5
300 /600 /200 /300 /400 /
W y unitW y unitR y unitR y unitR y unit
buy all the units at atbuy all the units at atsell all the units at atsell all the units at atsell all the units at at
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THE DUAL PROBLEM INTERPRETATION
The mover wishes to maximize profits, i.e.,
− + =− + =− + =
− + =− + =− + =
1 3 11
1 4 12
1 5 13
2 3 21
2 4 22
2 5 23
243534
y y cy y cy y c
y y cy y cy y c
≤≤≤≤≤≤
− ⇒
= − − + + +1 2 3 4 5300 600 200 300 400
revenues costs dual cost objective function
maxW y y y y y
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Resource requirements
EXAMPLE 2: FURNITURE PRODUCTS
item sales price ($)
desks 60
tables 30
chairs 20
requirements
lumber board
laborcarpentry
finishing
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The Dakota Furniture Company manufacturing:
We assume that the demand for desks, tables and chairs is unlimited and the available resources are already purchased
The decision problem is to maximize total revenues
EXAMPLE 2: FURNITURE PRODUCTS
resource desk table chair available
lumber board (ft ) 8 6 1 48
finishing (h) 4 2 1.5 20
carpentry (h) 2 1.5 0.5 8
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We define decision variables
The Dakota problem is
PRIMAL AND DUAL PROBLEM FORMULATION
x number of desks producedx number of tables producedx = number of chairs produced
1
2
3
==
+ +
↔ + + ≤↔ + + ≤↔ + + ≤
≥
1 2 3
1 1 2 3
2 1 2 3
3 1 2 3
1 2 3
60 30 20
8 6 484 2 1.5 202 1.5 0.5 8
max Z = x x xs.t.y x x x lumbery x x x finishing
carpentryy x x xx , x , x 0
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PRIMAL AND DUAL PROBLEM FORMULATION
The dual problem is
+ +
≥
+ + ≥
+ + ≥
≥
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
48 20 8
8 4 2 60
6 2 1.5 30
1.5 0.5 20
min W = y y y
s.t.
y + y + y desk
y y y table
y y y chair
y , y , y 0
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PRIMAL AND DUAL PROBLEM FORMULATION
+ ++ + ≥+ + ≥+ + ≥
+ +↔ + + ≤↔ + + ≤↔ + + ≤
≥
≥
1 2 3
1 1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2
2 1 2 3
3 1 2
2 3
3
3
1
48 20 848 20 8 60
60 30 208 6 484
6 2 1.5 301.5 0.
2 1.5 202 1.5 0.5 8
5 20
max Z = x x xy x x x lumbery x x x finishingy x x x carpentry
x ,max W = y y y
y y y desky y y tabley y y chai
x , x
r, y , y
0
y 0
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An entrepreneur wishes to purchase all of Dakota’s resources
He needs, therefore, to determine the prices to pay for each unit of each resource
We solve the Dakota dual problem to determiney 1, y 2 and y 3
INTERPRETATION OF THE DUAL PROBLEM
y price paid for lumber board ft
y price paid for h of finishing labor
y price paid for h of carpentry labor
1
2
3
= 1
= 1
1
=
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To induce Dakota to sell the raw resources, the resource prices must be set sufficiently high
For example, the entrepreneur must offer Dakota at least $ 60 for a combination of resources that consists of 8 ft of lumber board, 4 h of finishing and 2 h of carpentry, since Dakota could use this combination to sell a desk for $ 60: this require-ment implies the following dual constraint:
INTERPRETATION OF THE DUAL PROBLEM
y y y1 2 38 4 2 60+ + ≥
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In the same way, we obtain the two additional
constraints for a table and for a chair
The i th primal variable is associated with the i th
constraint in the dual problem statement
The j th dual variable is associated with the j th
constraint in the primal problem statement
INTERPRETATION OF DUAL PROBLEM
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A new diet requires that all food eaten come from
one of the four “basic food groups”:
chocolate cake
ice cream
The four foods available for consumption are
brownie
chocolate ice cream
EXAMPLE 3: DIET PROBLEM
soda
cheesecake
cola
pineapple cheesecake
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EXAMPLE 3: DIET PROBLEM
Minimum requirements for each day are:
500 cal
6 oz chocolate
10 oz sugar
8 oz fat
The objective is to minimize the diet costs
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EXAMPLE 3: DIET PROBLEM
food calorieschocolate
(oz)sugar (oz) fat (oz)
costs (cents)
brownie 400 3 2 2 50
chocolate ice cream(scoop)
200 2 2 4 20
cola (bottle) 150 0 4 1 30
pineapple cheesecake
(piece)500 0 4 5 80
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Objective of the problem is to minimize the total
costs of the diet
Decision variables are defined for each day’s
purchases
PROBLEM FORMULATION
x number of brownies
x number of chocolate ice cream scoops
x number of bottles of soda
x number of pineapple cheesecake pieces
1
2
3
4
=
=
=
=
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PROBLEM FORMULATION
The problem statement is
i
min Z = x x x x
s.t.
x x x + x calx x ozx x x + x ozx x x + x oz
x 0 i =
1 2 3 4
1 2 3 4
1 2
1 2 3 4
1 2 3 4
50 20 30 80
400 200 + 150 500 5003 2 62 2 + 4 4 102 4 + 5 8
1,4
+ + +
+ ≥+ ≥+ ≥+ ≥
≥
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The dual problem is
EXAMPLE 3: DIET PROBLEM
max W = y y y y
s.t.
y y y + yy y y yy y + yy y + y
y , y , y , y 0
1 2 3 4
1 2 3 4
1 2 3 4
1 3 4
1 3 4
1 2 3 4
500 6 10 8
400 3 + 2 2 50200 2 2 + 4 20150 + 4 30500 + 4 5 80
+ + +
+ ≤+ + ≤
≤≤
≥
brownieice - cream
sodacheesecake
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We consider a salesperson of “nutrients” who is interested in assuming that each dieter meets daily requirements by purchasing calories, sugar, fat and chocolate as “goods”
The decision is to determine the prices chargedy i = price per unit of required nutrient to sell to dieters
Objective of the salesperson is to set the prices y i
so as to maximize revenues from selling to the dieter the daily ration of required nutrients
INTERPRETATION OF THE DUAL
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Now, the dieter can purchase a brownie for 50 ¢
and have 400 cal, 3 oz of chocolate, 2 oz of sugar and 2 oz of fat
The sales price y i must be set sufficiently low to entice the buyer to get the required nutrients from
the brownie:
We derive similar constraints for the ice cream, the soda and the cheesecake
INTERPRETATION OF DUAL
y y y y1 2 3 4400 3 2 2 50+ + + ≤ brownie constraint
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DUAL PROBLEMS
T
T
T
max Z = c x
s.t.
A x b
min W b y
s.t.
A y c
y 0
x 0
≤
≥
=
≥
≥
D
P( )
( )
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For any feasible for (P ) and any feasible for
( D )
Proof:
T T TT T
T T TT
A y c c y A c x y A x
c x y A x y b b y
≥ ⇒ ≤ ⇒ ≤
≤ ≤ =
WEAK DUALITY THEOREM
TTc x b y≤
x y
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COROLLARY 1 OF THE WEAK DUALITY THEOREM
( )
( )
TT
TT
x is feasible for P c x y b
for any feasible y for D
c x y b min W∗
⇒ ≤
≤ =
( )T
for any feasible x for P ,
c x min W≤
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COROLLARY 2 OF THE WEAK DUALITY THEOREM
( )
( )
TT
TT T
y is feasible for D c x y b
for every feasible x for P
max Z max c x c x y b∗
⇒ ≤
= = ≤
( )T
for any feasible y of D ,
y b max Z≥
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If ( P ) is feasible and max Z is unbounded, i.e.,
then, ( D ) has no feasible solution.
If ( D ) is feasible and min Z is unbounded, i.e.,
then, ( P ) is infeasible.
COROLLARIES 3 AND 4 OF THE WEAK DUALITY THEOREM
Z +∞→ ,
Z ∞−→ ,
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Consider the maximization problem
DUALITY THEOREM APPLICATION
P( )
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The corresponding dual is given by
With the appropriate substitutions, we obtain
DUALITY THEOREM APPLICATION
T
T
min W = b y
s.t.
A y c
y 0
≥
≥
D( )
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DUALITY THEOREM APPLICATION
min W = y y
s.t.
y y
y y
y y
y y
y 0, y 0
1 2
1 2
1 2
1 2
1 2
1 2
20 20
+ 2 1
2 + 2
2 + 3 3
3 + 2 4
+
≥
≥
≥
≥
≥ ≥
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Consider the primal decision
decision is feasible for (P) with
The dual decision
is feasible for (D) with
GENERALIZED FORM OF THE DUAL
1, 1,2,3,4 ;ix i= =
1, 1,2iy i= =
10TZ c x= =
40TW b y= =
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DUALITY THEOREM APPLICATION
Clearly,
and so clearly, the feasible decision for (P) and (D)
satisfy the Weak Duality Theorem
Moreover, we have
( ) ( )≤1 2 3 4 1 210 40 =Z x , x , x , x = W y , y
( ) Tcorollary max Z = Z x , x , x , x b y =∗ ∗ ∗ ∗⇒ ≤1 2 3 42 40
( )corollary min W W y , y∗ ∗⇒ ≤ = 1 21 10
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COROLLARIES 5 AND 6
(P) is feasible and (D) is infeasible, then,
(P) is unbounded
(D) is feasible and (P) is infeasible, then,
(D) is unbounded
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Consider the primal dual problems:
Now
EXAMPLE
( )x 0 is feasible for P=
1 2
1 2
1 2
1 2
1 2
2
. .
2 1( )
1
,
min W y y
s t
y yD
y y
y y 0
y y 0
= +
− − ≥
+ ≥ − ≥ ≥
1 2
1 2 3
1 2 3
1 2 3
. .
2 ( )
2 1
, ,
max Z x x
s t
x x x P
x x x
x x x 0
= +− + + ≤
− + − ≤ ≥
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but
is impossible for (D) since it is inconsistent with
Since (D) is infeasible, it follows from Corollary 5that
You are able to show this result by solving (P)
using the simplex scheme
EXAMPLE
Z ∞ →
1 2,y y 0≥
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We consider the primal-dual problems (P) and (D)
with
We next provide the proof:
OPTIMALITY CRITERION THEOREM
0
0
x Py D
( )( )
is feasible foris feasible for
TT 0 0Weak Duality
Theoremc x b y⇒ ≤
0
0
TT 0 0
x Py D
c x b y
( )( )
=
is feasible foris feasible for
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OPTIMALITY CRITERION THEOREM
but we are given that
and so it follows that
and so is optimal ;
similarly,
and so it follows that is optimal
TT 0 0c x b y=
TT 0 T 0c x b y c x≤ =0x
T TT 0 0b y c x b y≥ =0y
0feasible x with y feasible∀
0feasible y with x feasible∀
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(P) is feasible and (D) is feasible; then,
MAIN DUALITY THEOREM
TTc x b y∗ ∗=
x feasible for P which is optimal and( )∗∃
y feasible for D which is optimal such that( )∗∃
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COMPLEMENTARY SLACKNESS CONDITIONS
and are optimal for (P) and (D),
respectively, if and only if
We prove this equivalence result by defining the
slack variables and such that
and are feasible; at the optimum,y
x
*x *y
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COMPLEMENTARY SLACKNESS CONDITIONS
where the optimal values of the slack variables
depend on the optimal values
Now,
u v∗ ∗and
x y∗ ∗and
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COMPLEMENTARY SLACKNESS CONDITIONS
This implies that
We need to prove optimality which is true if and
only if
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However,
Also,
( ) ( )
T T T T
Optimality
Criterion Theorem
y u v x 0 b y c x
x is optimal for P and y is optimal for D
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗
+ = ⇒ =
COMPLEMENTARY SLACKNESS CONDITIONS
,Main
Duality TheoremT T T T
x y are optimal
c x b y y u v x 0
∗ ∗
∗ ∗ ∗ ∗ ∗ ∗= ⇒ + =
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COMPLEMENTARY SLACKNESS CONDITIONS
Note that
At the optimum,
and
Ti i
j j
x y u v 0 component - wise each element 0
y u v x 0 y u 0 i m
and v x 0 j n
, , ,
1, ... ,
1, ... ,
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗
> ⇒ ≥
+ = ⇒ = ∀ =
= ∀ =
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COMPLEMENTARY SLACKNESS CONDITIONS
Hence, for i = 1, 2, … , m
and
Similarly for j = 1, 2, … , n
andm
ji i j ji
a y c 0 x 01
∗ ∗
=
− > ⇒ =∑
n
i i i j jj=1
y > 0 b = a x∗ ∗⇒ ∑
m
i ji i ji=1
x > 0 a y = c∗ ∗⇒ ∑
m
i i j i ij
b a x 0 y 01
∗ ∗
=
− > ⇒ =∑
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EXAMPLE
1 2 3 4
1 2 3 4
1 2 3 4
2 3 4
. .
2 2 3 20
2 3 2 20
1, ... ,4i
max Z x x x x
s t
x x x x
x x x x
x 0 i =
= + + +
+ + + ≤
+ + + ≤
≥
( )P
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EXAMPLE
min W y y
s t
y y
y y
y y
y y
y y 0
1 2
1 2
1 2
1 2
1 2
1 2
20 20
. .
2 1
2 2
2 3 3
3 2 4
,
= +
+ ≥
+ ≥
+ ≥
+ ≥
≥
( )D
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EXAMPLE
( )
( )
x y optimal
y x x x x 0
y x x x x 0
y
min W
* *
* * * * *1 1 2 3 4
* * * * *2 1 2 3 4
*
,
20 2 2 3
20 2 3 2
1.20.2
28
⇒
− − − − =
− − − − =
=
=
is given as an optimal solution with
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EXAMPLE
so that3 4 3
3 4 4
2 3 20 4
3 2 20 4
x x x
x x x
∗ ∗ ∗
∗ ∗ ∗
+ = ⇒ =
+ = ⇒ =
x x x x
x x x x
y y x 0
y y x 0
y y
y y
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗
∗ ∗ ∗
∗ ∗
∗ ∗
⇒
⇒
1 2 3 4
1 2 3 4
1 2 1
1 2 2
1 2
1 2
+ 2 + 2 + 3 = 20
2 + + 3 + 2 = 20
+ 2 = 1.2 + 0.4 > 1 =
2 + = 2.4 + 0.2 > 2 =
2 + 3 = 2.4 + 0.6 = 3
3 + 2 = 3.6 + 0.4 = 4
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COMPLEMENTARY SLACKNESS CONDITION APPLICATIONS
Key uses of c.s. conditions are
finding optimal (P) solution given optimal (D)
solution and vice versa
verification of optimality of solution (whether a
feasible solution is optimal)
We can start with a feasible solution and attempt
to construct an optimal dual solution; if we suc-
ceed, then the feasible primal solution is optimal
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DUALITY
T
T
T
max Z = c x
s.t.
A x b
min W b y
s.t.
A y c
y 0
x 0
≤
≥
=
≥
≥
D
P( )
( )
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Suppose the primal problem is minimization, then,
DUALITY
T
T
T
min Z = c x (P)s.t. A x b max W b y (D)
s.t.
c
x
y 0
0
A
y
≥
=
≤
≥
≥
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INTERPRETATION
The economic interpretation is
Suppose, we change
In words, the optimal dual variable for each primal constraint gives the net change in the optimal value of the objective function Z for a one unit change in the constraint on resources
* T * T * *
i
i
Z max Z c x b y W minW
b constrained resource quantitiesi m
y optimal dual variables*
,1, 2, ... ,
= = = = =
− =−
*i i i i ib b b Z y b→ + ∆ ⇒ ∆ = ∆
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INTERPRETATION
Economists refer to the dual variable as the
shadow price on the constraint resource
The shadow price determines the value/worth of
having an additional quantity of a resource
In the previous example, the optimal dual
variables indicate that the worth of another unit
of resource 1 is 1.2 while that of another unit of
resource 2 is 0.2
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GENERALIZED FORM OF THE DUAL
We start out with
Tmax Z = c x
s.t.
A x b
x 0
=
≥
P( )
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GENERALIZED FORM OF THE DUAL
A bx
A b
≤ − −
To find ( D ), we first put (P ) in symmetric form
y A x b
y A x b
x 0
symmetricform
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Let
We rewrite the problem as
The c.s. conditions apply
GENERALIZED FORM OF THE DUAL
T
T
min W b y
s t
A y c
y is unsigned
. .
=
≥
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EXAMPLE 5: THE PRIMAL
max Z x x x xs ty x x x xy xy xy x Py xy xy x
x x 0x x unsigned
1 2 3 4
1 1 2 3 4
2 1
3 2
4 2
5 3
6 3
7 4
1 4
2 3
. .8844 ( )42
10,,
= − + −
↔ + + + =↔ ≤↔ ≤↔ − ≤↔ ≤↔ − ≤↔ ≤
≥
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EXAMPLE 5: THE DUAL
min W y y y y y y y
s t
x y y
x y y y D
x y y y
x y
y y 0
y unsigned
1 2 3 4 5 6 7
1 1 2
2 1 3 4
3 1 5 6
4 7
2 7
1
8 8 4 4 4 2 10
. .
1
1 ( )
1
1
, ...... ,
π= + + + + + +
↔ + ≥
↔ + − = −
↔ + − =
↔ + ≥
≥
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EXAMPLE 5: c.s. conditions We are given that
is optimal for (P)
Then the c.s. conditions obtain
x
0
844
∗
− =
( )* * *1 1 2 1x y y 0+ − =
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EXAMPLE 5: c.s. conditions
so that
The other c.s. conditions obtain
Now, implies and so
x 0 y y1 1 28 1∗ ∗ ∗= > ⇒ + =
i i j j ij
y a x b 04
1
∗ ∗
=
− =
∑
x 04∗ = x 04 10∗ − <
y 07∗ =
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EXAMPLE 5: c.s. conditions
Also, implies
Similarly, the c.s. conditions
have implications on the variable
*3 4x =
*6y 0=
7* *
1j j i i j
ix a y c 0
=
− =
∑
*iy
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EXAMPLE 5: c.s. conditions
Since , then we have
Now, with we have
Since, we have
*2 4x = −
*3y 0=
*7y 0=
*1 1y > −
= TW b y
* *2 11y y= −
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EXAMPLE 5
Suppose
and so,
Furthermore,
implies
*1 1y =
*2y 0=
* * * *1 3 4 41 1y y y y+ − = − = −
*4 2y =
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EXAMPLE 5 Also
implies
and so
*51 1y+ =
* * *1 5 6 1y y y+ − =
*5y 0=
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EXAMPLE 5 Therefore
and so
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
* 8 1 8 4 4 2
4 2 10
16
W y 0 0
0 0 0
= + + + +
+ +
=
( ) ( )* *16W Z P D= = ⇔ optimality of and
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PRIMAL – DUAL TABLE
primal (maximize) dual (minimize)
A ( coefficient matrix ) A T ( transpose of the coefficient matrix )
b ( right-hand side vector ) b ( cost vector )
c ( price vector ) c ( right hand side vector )
i th constraint is = type the dual variable y i is unrestricted in sign
i th constraint is type the dual variable y i 0
i th constraint is type the dual variable y i 0
x j is unrestricted j th dual constraint is = type
x j 0 j th dual constraint is type
x j 0 j th dual constraint is type
≥
≤
≥
≤
≥
≤
≥
≤