The document discusses complementary slackness and Farkas' lemma within the context of linear programming, addressing strong duality theorem and optimal solutions for primal and dual problems. It outlines the implications of these theories, providing examples and proving key relationships between primal and dual variables. Additionally, it explores various formulations of Farkas' lemma and their applications in assessing the feasibility of systems of linear inequalities.

1. Complementary slackness & Farkas’ lemmaSHAMEER P HDEPT OF FUTURES STUDIES

2. STRONG DUALITY THEOREMIf ZLP or WLP is finite, then both P and D have finite optimal value and ZLP= WLPCOROLLARY:There are only four possibilities for a dual pair of problems P and DZLP or WLP are finite and equalZLP= ∞ and D is infeasibleWLP = -∞ and P is infeasibleboth P and D are infeasibleDEPT OF FUTURES STUDIES

3. Complementary Slackness PRIMAL: u = max{cx : Ax≤ b, x≥ 0}DUAL: w = min{by: Ay≥c, y ≥0}let‘s’ be the vector of slackvariables of theprimal&‘t’ isthe vector of surplus variables of duals= b-Ax≥0 and t= yA-c≥0DEPT OF FUTURES STUDIES

4. PrincipleIf x* isanoptimalsolution of Primal (P) and y *isanoptimalsolution of Dual (D), thenxj* tj * =0for all j &yi* si * =0 foralli(The theorem identifies a relationship between variables in one problem and associated constraints in the other problem.)DEPT OF FUTURES STUDIES

5. proofUsing 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)DEPT OF FUTURES STUDIES

6. In other wordsGiven 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 kthcomponent of the optimal solution of its dual is zeroORIf a variable is positive, then its dual constraint is tight.

7. If a constraint is loose, then its dual variable is zero.DEPT OF FUTURES STUDIES

8. examplePRIMAL:max z=6x1 +5x2such that: x1+x2≤5 3x1+2x2≤12 x1,x2 ≥0OR x1+x2+S1=5 3x1+2x2+S2=12 x1,x2,s1,s2 ≥0Solution: x1 =2, x2 =3, z=27 S1=0, s2=0DUAL:Min. W=5y1+12y2Such that y1+3y2≥6y1+2y2≥5y1,y2≥0OR y1+3y2-t1 =6 y1+2y2-t2=5 y1,y2,t1,t2≥0Solution: y1 =3, y2 =1, w=27t1 =0, t2=0DEPT OF FUTURES STUDIES

9. applicationsUsed in finding an optimal primal solution from the given optimal dual solution and vice versa

10. Used in finding a feasible solution is optimal for the primal problem DEPT OF FUTURES STUDIES

11. FARKAS’ LEMMAProposed 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 ∈ Rn+. Then exact one of the below is true:There exists an x∈ Rn+suchthat Ax≤ b; orThere exists a y∈ Rm+such that y ≥ 0, yA = 0 and yb < 0. ORAx≤ b, x ≥ 0 is infeasibleiff yA ≥0, y b < 0 is feasibleDEPT OF FUTURES STUDIES

12. 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 . ZLP =WLP =0. hence {xԐRn+ :Ax ≤b}≠ф and Yb≥0 for all yԐRm+ with yA ≥0WLP = -∞. hence {xԐRn+ :Ax ≤b}=ф and there exists yԐRm+ with yA ≥0 and yb˂0.DEPT OF FUTURES STUDIES