MIE376 Lec9 DWD Incomplete.page3

MIE376 Lec9 DWD Incomplete.page3 - practical algorithm....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
MIE376 Mathematical Programming Lecture Notes Daniel Frances 2011 3 The coordinator would then either decline, partially accept or fully accept the offers, and run some other LP to update the objective function prices to be used by the individual LPs in the next iteration Reformulating the 2-Plant LP Suppose we accept that each of the plants is going to derive their offers based on their own local LP. While the local objective may change based on the signal from the coordinator, the constraint set remains unchanged. Therefore the set of corner point of each of the local constraint set represents the set of possible offers that could be received from each plant: For Plant 1 the set of possible offers is {(P 1 0 , P 2 0 ), (P 1 1 , P 2 1 )…, (P 1 i , P 2 i )…} For Plant 2 the set of possible offers is {(Q 1 0 , Q 2 0 ), (Q 1 1 , Q 2 1 )…, (Q 1 i , Q 2 i )…} Clearly, for any LPs of realistic size it would be computationally infeasible to list all the basic feasible solutions (corner points), and so these sets are only of theoretical importance on the way to deriving a
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: practical algorithm. Lets ASSUME for a moment that we have all the possible (P 1 i ,P 2 i ) and (Q 1 i , Q 2 i ). Then it follows from convexity of each of the constraint sets that any linear combinations of the (P 1 i ,P 2 i ) offers is also feasible o i.e. for any set of i such that i = 1 an offer i (P 1 i ,P 2 i ) is also feasible any linear combinations of the (Q 1 i ,Q 2 i ) offers is also feasible o i.e. for any set of i such that i = 1 an offer i (Q 1 i ,Q 2 i ) is also feasible Thus any solution (P 1 , P 2 ) can be rewritten as i (P 1 i ,P 2 i ), or P 1 = i P 1 i and P 2 = i P 2 i Similarly we can rewrite Q 1 = i Q 1 i and Q 2 = i Q 2 i Substituting in the 2-Plant model we obtain Max 90 i P 1 i + 80 i P 2 i + 70 i Q 1 i + 60 i Q 2 i subject to 8 i P 1 i + 6 i P 2 i + 7 i Q 1 i + 5 i Q 2 i 80; i =1 i = 1 i , i 0...
View Full Document

Ask a homework question - tutors are online