{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

asst3sol - C&O 350 Linear Optimization – Winter 2009 –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
C&O 350 Linear Optimization – Winter 2009 – Solutions to Assignment 3 Question # Max. marks Part marks 1 11 2 2 2 3 2 2 6 3 3 3 8 2 2 2 2 4 7 4 3 5 6 Total 38 1. 11 marks = 2+2+2+3+2 Geometry of Basic Feasible Solutions: (a) Draw the polyhedron P , where P = { ( x 1 , x 2 ) : - 2 x 1 + 3 x 2 6 , x 1 + 2 x 2 10 , x 1 0 , x 2 0 } . (A polyhedron is the solution set of a finite system of linear inequalities.) (b) List the coordinates of all of the extreme points of P , and mark them on the picture. (c) Upon introducing slack variables, we obtain the system - 2 x 1 + 3 x 2 + x 3 = 6 x 1 + 2 x 2 + x 4 = 10 List all possible bases for this system. (d) For each basis B from (c), find the basic solution determined by B. Indicate which of these basic solutions are feasible. (e) For each basic feasible solution and for each infeasible basic solution from (d), label the corresponding point in the picture of P . Solution : (a) The polygon P is depicted in Figure 1. (b) The extreme points of P are the points: y 2 = [18 / 7 , 26 / 7] y 3 = [10 , 0] y 4 = [0 , 2] y 5 = [0 , 0] . They are marked on Figure 1. (c) Every possible set of two columns forms a basis. So the set of bases is {{ 1 , 2 } , { 1 , 3 } , { 1 , 4 } , { 2 , 3 } , { 2 , 4 } , { 3 , 4 }} . (d) The basic solutions determined by each basis are: { 1 , 2 } : [18 / 7 , 26 / 7 , 0 , 0] { 1 , 3 } : [10 , 0 , 26 , 0] { 1 , 4 } : [ - 3 , 0 , 0 , 13] { 2 , 3 } : [0 , 5 , - 9 , 0] { 2 , 4 } : [0 , 3 , 0 , 6] { 3 , 4 } : [0 , 0 , 6 , 10] . The infeasible solutions are those corresponding to the bases { 1 , 4 } and { 2 , 3 } . The rest are feasible. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Figure 1: The polygon P . (e) The points corresponding to the basic solutions are: y 2 = [18 / 7 , 26 / 7 , 0 , 0] y 3 = [10 , 0 , 26 , 0] y 6 = [ - 3 , 0 , 0 , 13] y 1 = [0 , 5 , - 9 , 0] y 4 = [0 , 3 , 0 , 6] y 5 = [0 , 0 , 6 , 10] .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}