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Lecture 12 Chapter 6 Interior Point Algorithm

Lecture 12 Chapter 6 Interior Point Algorithm. Example Max 90x 1 + 150x 2 Subject To 0.5x 1 + x 2 < 3 x 1 , x 2 > 0. optimum. Gradient Direction.

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Lecture 12 Chapter 6 Interior Point Algorithm

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  1. Lecture 12Chapter 6 Interior Point Algorithm • Example • Max 90x1 + 150x2 • Subject To 0.5x1 + x2< 3 • x1, x2> 0 optimum

  2. Gradient Direction • 6.1 Page 253 Best direction (for a max problem) is gradient of objective function. Objective function is cx; therefore, gradient at every point is c. • 6.6 Page 254, Interior point algorithm only touches the boundary at optimality. • Simplex stays on the boundary at every step (iteration).

  3. Interior Points • Standard Form • max cx • s.t. Ax = b • x > 0 • 6.3 Page 256 What is the interior of the set • {x: Ax = b, x > 0}? • Interior points have every component > 0 !!!!!!!!!!!!

  4. Interior Points & Feasible Directions • Let y be a point with y > 0 and Ay = b, then y is an interior point. • 6.4 Page 257 Given an interior point z > 0, a direction d is feasible if and only if Ad = 0. • That is, the new point y = z + d satisfies Ay = b if and only if Ad = 0. • Ay = A(z+d) = Az + Ad = b + 0 = b

  5. Projection • If the gradient direction, c (the best direction of movement) does not satisfy Ac = 0, then we project c onto {x:Ax = 0}. • 6.6 Page 258 How do you project d onto Ax = 0? • đ = Pd where • P = (I-A’(AA’)-1A) • this comes from linear algebra

  6. Example x3 0.5x1 + x2 + x3 = 3 A = x1 AA’ = = 9/4 x2

  7. Example Continued • (AA’)-1 = (9/4)-1 = (4/9) • A’(4/9)A = (4/9)A’A = (4/9) = (4/9) = (1/9)

  8. Example Continued • I – A’(AA’)-1A = = - (1/9) = (1/9)

  9. Example Continued • Gradient direction for max 90x1 + 150x2 + 0x3 is • d = (1/9) =(1/9)

  10. Example Continued =(1/9) =(1/3) =(1/30) This direction is the same as that given on page 258. It differs by a scale factor.

  11. Example Continued • Check this direction to make sure that Ađ=0. = 0 Is this direction improving? cđ > 0? = 4110

  12. Scaling • To keep from getting too close to the boundary, we scale the region. Consider the point [3, 0.5, 1] and create • X = Let Xy = x or y = X-1 x

  13. Scaled Problem • Then {max cx: Ax=b, x > 0} • becomes {max cXy: AXy=b, Xy > 0} • Or {max ĉy: Ây=b, y > 0} where ĉ = cX and Â=AX. • Try it • Max 90x1+150x2 • s.t. 0.5x1+x2+x3 = 3 • x1, x2, x3> 0 with interior point [3, 0.5, 1]

  14. Scaled Problem • Ĉ = =

  15. Scaled Problem • Â = = Scaled problem is Max 270y1 + 75y2 s. t. 1.5y1 + 0.5y2 + y3 = 3, all var > 0 Is [1,1,1] feasible and interior?

  16. Step Size • 6.15 A LP is unbounded if the search direction đ > 0. • 6.16 What remains is to determine the step size? •  = 1/|| đ|| Move to edge of sphere with radius 1. Try It!

  17. One Step • Begin at [1, 0.5, 2] an interior point. • đ = [14, 19, -26] • || đ || = [142 + 192 + (-26)2]0.5 = 35.1141 • 1/(|| đ || ) = 0.0285 • [1,0.5,2] + 0.0285[14,19,-26] = [1.40, 1.04, 1.26]. • Old Obj = 165, New Obj = 282 Improvement!

  18. Alg 6A Affine Scaling P 274 • Step 0. Select a starting feasible solution. x0 and set t to 0 • Step 1. Check For Optimality (This is a bit tricky- see text) • Step 2. Direction. Scale ct+1 = Xtct and At+1 = AtXt • dt+1 = Pt+1ct+1. If dt+1 > 0, then problem is unbounded. • Step 3. Step Size. = 1/[||dt+1(Xt)-1||] • Step 4. Move To New Point. xt+1 = xt + dt+1 • Increment t and return to step 1.

  19. Where is all the work? • Recall P = [I-A’(AA’)-1A] page 260 • Each time we get to step 3, the A Matrix has changed. • CPLEX has an interior point algorithm implemented. • The one implemented is more sophisticated than affine scaling.

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