The Primal Problem: Maximize C1*X1 + C2*X2 + .... + Cn*Xn subject to A11* X1 + A12*X2 + ... + A1n*Xn <= B1 A21* X1 + A22*X2 + ... + A2n*Xn <= B2 Am1* X1 + Am2*X2 + ... + Amn*Xn <= Bm X1, X2, ... , Xn >= 0 The Dual Problem Minimize B1*Y1 + B2*Y2 + .... + Bm*Ym subject to A11* Y1 + A21*Y2 + ... + Am1*Ym >= C1 A12* Y1 + A22*Y2 + ... + Am2*Ym >= C2 A1n* Y1 + A2n*Y2 + ... + Amn*Ym >= Cn Y1, Y2, ... , Ym >= 0
Theorem
For every primal feasible solution
x= (X1, X2, ... , Xn)
and
for every dual feasible solution
y= (Y1, Y2, ... , Ym)
C1*X1 + C2*X2 + .... + Cn*Xn <=
B1*Y1 + B2*Y2 + .... + Bm*Ym
Proof
Because the constraints of the dual are that A11* Y1 + A21*Y2 + ... + Am1*Ym >= C1 A12* Y1 + A22*Y2 + ... + Am2*Ym >= C2 A1n* Y1 + A2n*Y2 + ... + Amn*Ym >= Cn we have that S= (A11* Y1 + A21*Y2 + ... + Am1*Ym)*X1 + (A12* Y1 + A22*Y2 + ... + Am2*Ym)*X2 + (A1n* Y1 + A2n*Y2 + ... + Amn*Ym)*Xn >= C1*X1 + C2*X2 + .... + Cn*Xn .Regroup the terms on the left hand side so that they are grouped according the Yi instead of Xi:
S= (A11* X1 + A12*X2 + ... + A1n*Xn)*Y1 + (A21* X1 + A22*X2 + ... + A2n*Xn)*Y2 + (Am1* X1 + Am2*X2 + ... + Amn*Xn)*Ym From the primal problem, A11* X1 + A12*X2 + ... + A1n*Xn <= B1 A21* X1 + A22*X2 + ... + A2n*Xn <= B2 Am1* X1 + Am2*X2 + ... + Amn*Xn <= Bm So, S <= B1*Y1 + B2*Y2 + .... + Bm*Ym So in conclusion: C1*X1 + C2*X2 + .... + Cn*Xn <= S <= B1*Y1 + B2*Y2 + .... + Bm*Ym as required.
Why?
Every primal feasible solution satisfies
C1*X1 + C2*X2 + .... + Cn*Xn <= S
= B1*Y1' + B2*Y2' + .... + Bm*Ym'
so the primal solution must be optimal.
Every dual feasible solution satisfies
C1*X1' + C2*X2' + .... + Cn*Xn' = S
<= B1*Y1 + B2*Y2 + .... + Bm*Ym
so the dual solution must be optimal.