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Complementary slackness

& Farkas’ lemma

SHAMEER P H

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STRONG DUALITY THEOREM

• If ZLP or WLP is finite, then both P and D have finite optimal value and ZLP = WLP

• COROLLARY:• There are only four possibilities for a dual pair of problems P

and Di. ZLP or WLP are finite and equal

ii. ZLP = ∞ and D is infeasible

iii. WLP = -∞ and P is infeasible

iv. both P and D are infeasible

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Complementary Slackness

• PRIMAL: u = max{cx : Ax ≤ b, x≥ 0}• DUAL: w = min{by : Ay ≥ c, y ≥ 0}

let ‘s’ be the vector of slack variables of the primal& ‘t’ is the vector of surplus variables of dual• s= b-Ax≥0 and t= yA-c≥0

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Principle

• If x* is an optimal solution of Primal (P) and y * is an optimal solution of Dual (D), then

• xj * tj

* =0 for all j &

• yi * si

* =0 for all i • (The theorem identifies a relationship between variables

in one problem and associated constraints in the other problem.)

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proof

Using the definitions of s and t cx* = (y*A-t*)x*

= y*Ax*- t*x*= y*(b-s*)-t*x* (since Ax*+s=b)= y*b-y*s*-t*x*

but by strong duality theorem, cx*=y*bie; cx* = cx*-(y*s*+t*x*)0= y*s*+t*x*y*s*=0 and t*x*=0 (Since, y*,x*,s*, t*≥0)

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In other words

• Given a dual pair of LPPs have optimal solutions, then if the kth constraint of one system holds inequality (i.e, associated slack or surplus variable is positive) then the kth component of the optimal solution of its dual is zero

OR If a variable is positive, then its dual constraint is

tight. If a constraint is loose, then its dual variable is

zero.

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example• PRIMAL:max z=6x1 +5x2

such that:x1+x2≤5

3x1+2x2≤12

x1, x2 ≥0

ORx1+x2+S1=5

3x1+2x2+S2=12

x1,x2,s1,s2 ≥0

Solution:x1 =2, x2 =3, z=27

S1=0, s2=0

DUAL:Min. W=5y1+12y2

Such that y1+3y2≥6y1+2y2≥5y1,y2 ≥0

ORy1+3y2-t1 =6y1+2y2-t2=5y1,y2,t1, t2 ≥0

Solution:y1 =3, y2 =1, w=27 t1 =0, t2 =0

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applications Used in finding an optimal primal solution from the given

optimal dual solution and vice versa Used in finding a feasible solution is optimal for the primal

problem

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FARKAS’ LEMMA• Proposed by Farkas in 1902.• Used to check whether a system of linear inequalities is

feasible or not • Let A be an m × n matrix, b R∈ n

+. Then exact one of the below is true:1) There exists an x R∈ n

+such that Ax ≤ b; or

2) There exists a y R∈ m+ such that y ≥ 0, y A = 0 and yb < 0.

ORAx ≤ b, x ≥ 0 is infeasible iff yA ≥ 0, y b < 0 is feasible

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Proof:

• Consider the LPP, ZLP =max{0x:Ax≤b, xԐRn+}

and its dual WLP =min{yb:yA≥0,yԐRm+} .

• As y=0 is a feasible solution to the dual problem, the only possibilities to occur are i & iii of corollary of strong dual theorem .I. ZLP = WLP =0. hence {xԐRn

+ :Ax ≤b}≠ф and Yb ≥0 for all yԐRm+

with yA ≥0II. WLP = -∞. hence {xԐRn

+ :Ax ≤b}=ф and there exists yԐRm+ with

yA ≥0 and yb˂0.

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Variants of Farkas’ Lemma• There are many variants of Farkas’ Lemma.• Some are:a) Either {xԐRn

+ :Ax =b}≠ф, or {yԐRm : yA ≥0 , yb˂0} ≠ф

b) Either {xԐRn :Ax ≤b}≠ф, or {yԐRm+ : yA =0 , yb˂0} ≠ф

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THANK YOU….