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Chap 3. The simplex method

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Title: Chap 3. The simplex method


1
Chap 3. The simplex method
  • Standard form problem
  • minimize cx
  • subject to Ax b
  • x ? 0
  • A mxn, full row rank
  • From earlier results, we know that there exists
    an extreme point (b.f.s.) optimal solution if LP
    has finite optimal value.
  • Simplex method searches b.f.s to find optimal
    one.
  • If LP unbounded, there exists an extreme ray di
    in the recession cone K ( K x Ax 0, x ? 0
    ) such that cdi lt 0. Simplex finds the
    direction di if LP unbounded, hence providing the
    proof of unboundedness.

2
3.1 Optimality conditions
  • A strategy for algorithm Given a feasible
    solution x, look at the neighborhood of x for a
    feasible point that gives improved objective
    value. If no such point exists, we are at a
    local optimal point. In general such local
    optimal point is not global optimal. However, if
    we minimize a convex function over a convex set
    (convex program), local min is a global min
    point, which is the case for linear programming
    problem. (HW)
  • Def P is polyhedron. x ? P, d ?Rn is a feasible
    direction at x if ? ? gt 0 such that x ?d ? P
  • Given a b.f.s. x ? P, ( xB B-1b, xN 0 for
    N nonbasic)
  • ( B AB(1), AB(2), , AB(m) )
  • want to find a new point x ?d such that it
    satisfies Ax b and x ? 0 and the new point
    gives an improved objective value.

3
  • Consider moving to x ?d
  • where dj 1 for some nonbasic variable xj
  • 0 for other nonbasic variables
    except xj
  • and xB ? xB ?dB
  • We require
  • A(x ?d) b for ? gt 0
  • ? need Ad 0 (iff condition to satisfy A(x
    ?d) b, for ? gt 0)
  • ? 0 Ad ?i1n Aidi ?i 1m AB(i)dB(i)
    Aj BdB Aj
  • ? dB -B-1Aj
  • Assuming that columns of A are permuted so that
    A B, N and x (xB, xN), d (dB, dN),
  • d is called j-th basic
    direction.

4
  • Note that ? (n m) basic directions when B is
    given.
  • Recall d Ad 0 is the null space of A and
    its basis is given by columns of
    , where P is a permutation matrix for
  • AP B, N

Each column here gives a basic direction.
  • Those n m basic directions constitute a basis
    for null space of A (from earlier). Hence we can
    move along any direction d which is a linear
    combination of these basis vectors while
    satisfying A(x ?d) b, ? gt 0.

5
  • However, we also need to satisfy the
    nonnegativity constraints in addition to Ax b
    to remain feasible.
  • Since dj 1 gt 0 for nonbasic variable index j
    and xj 0 at the current solution, moving along
    (- dj ) will make xj ? 0 violated immediately.
    Hence we do not consider moving along -dj
    direction.
  • Therefore, the direction we can move is the
    nonnegative linear combination of the basic
    directions, which is the cone generated by the n
    m basic directions.
  • Note that if a basic direction satisfies dB
    -B-1Aj ? 0, it is an extreme ray of recession
    cone of P (recall HW).
  • In simplex method, we choose one basic direction
    as the direction of movement.

6
  • Two cases
  • (a) current solution x is nondegenerate xB gt 0
    guarantees that xB ?dB gt 0 for some ? gt 0
  • (b) x is degenerate some basic variable xB(i)
    0. It may happen that i-th component of dB
    -B-1Aj is negative
  • ? Then xB(i) becomes negative if we move
    along d. So we cannot make ? gt 0. Details
    later.

7
x3 0
F
E
x5 0
x1 0
x4 0
G
x2 0
  • Figure 3.2 n 5, n m 2
  • x1, x3 nonbasic at E. x3, x5 nonbasic at F (x4
    basic at 0).

8
  • Now consider the cost function
  • Want to choose the direction that improves
    objective value
  • ( c(x ?dj) cx ?cdj lt 0 )
  • cdj (cB, cN) cj
    cBB-1Aj ? (reduced cost)
  • If lt 0, then objective value improves if
    we move to x ?dj
  • Note) For i-th basic variable, may be
    computed using above formula.

9
  • Thm 3.1 (optimality condition)
  • Consider b.f.s. x with basis matrix B. be
    the reduced cost vector.
  • (a) If ? 0 ? x is optimal
    (sufficient condition for opt)
  • (b) x is optimal and nondegenerate ?
    ? 0

Pf) (a) assume ? 0. y is an
arbitrary point in P. Let d y x ? Ad
0 ? BdB ?i?N Aidi 0 ? dB -
?i?N B-1Aidi cd cy cx cBdB ?i?Ncidi
?i?N (ci cBB-1Ai)di (
, di ? 0 since yi ? 0, xi 0 for i
?N and d y x )
10
  • (Pf-continued)
  • ( Also may prove by contraposition
  • If x not optimal, ? y x d s.t. cy lt cx
    (note P is convex)
  • Feasible direction d at x is in the cone
    generated by basic directions. Hence cy cx
    cd c(?i ?idi) lt 0 , ?i ? 0
  • ? for some basic direction dj , cdj (cB,
    cN)(-B-1Aj, ej)
  • cj cBB-1Aj lt 0 )
  • (b) Suppose x is nondegenerate b.f.s. and
    for some j.
  • xj must be a nonbasic variable and we can
    obtain improved solution by moving to x ?dj , ?
    gt 0 and small.
  • Hence x is not optimal. ?

11
  • Note that the condition c ? 0 is a sufficient
    condition for optimality of a b.f.s. x, but it is
    not necessary. The necessity holds only when x
    is nondegenerate.
  • Def 3.3 Basis matrix B is said to be optimal if

12
3.2 Development of the simplex method
  • (Assume nondegenerate b.f.s. for the time being)
  • Suppose we are at a b.f.s. x and computed
    , j?N
  • If ? 0, ? j ?N, current solution is
    optimal.
  • Otherwise, choose j ? N such that lt 0
    and find d vector
  • ( dj 1, di 0 for i ? B(1), , B(m), j,
    and dB -B-1Aj )
  • Want to find ? max ? ? 0 x ?d ?P .
  • Cost change is ?cd ?
  • The vector d satisfies A(x ?d) b, also want
    to satisfy
  • (x ?d) ? 0

13
  • (a) If d ? 0, then (x ?d) ? 0 for all ? ?
    0 . Hence ? ?
  • (b) If di lt 0 for some i, (xi ?di ) ? 0 ?
    ? ? - xi / di
  • For nonbasic variables, di ? 0. Hence only
    consider
  • basic variables

Let y x ?d. Have yj ? gt 0 for
entering nonbasic variable xj. (we assumed
nondegeneracy, hence xB(i) gt 0 for all basic
variables) Let l be the index of the basic
variable selected in the minimum ratio test,
i.e.
14
  • Replace xB(l) in the basis with the entering
    variable xj.
  • New basis matrix is

Also replace the set B(1), , B(m) of basic
indices by B(1), , B(m) given by B(i)
B(i), i ? l, j, i l
15
  • Thm 3.2
  • (a) AB(i) , i ? l, and Aj are linearly
    independent. Hence is a basis matrix.
  • (b) y x ?d is a b.f.s. with basis

Pf)
16
  • Pf-continue)
  • (b) Have y ? 0, Ay b, yi 0,
  • Columns of B linearly independent.
    Hence b.f.s. ?
  • See the text for a complete description of an
    iteration of the simplex method.
  • Thm 3.3 Assume standard polyhedron P ? ? and
    every b.f.s. is nondegenerate. Then simplex
    method terminates after a finite number of
    iterations. At termination, two possibilities
  • (a) optimal basis B and optimal b.f.s
  • (b) Have found a vector d satisfying Ad 0, d
    ? 0, and cd lt 0, and the optimal cost is - ?.

17
Remarks
  • 1) Suppose x nondegenerate b.f.s. and we move
    to x ?d, ? gt 0
  • Consider the point y x ?d, ? gt 0 and y
    feasible
  • (nondegeneracy of x guarantees the existence
    of ? gt 0 and y feasible)
  • Then A( x ?d ) b
  • y ( yB , yN ) ? yB xB ?dB gt 0
    for sufficiently small ? gt 0
  • yN xN ?dN 0 ?ej (0, , ?,
    0, , 0)
  • Since ( n m 1 ) of constraints xj ? 0 are
    active and m constraints Ax b active, we have
    ( n 1 ) constraints are active at ( x ?d )
    (also the active constraints are lin. Ind.) and
    no more inequalities are active.

18
  • (continued)
  • Hence y is in the face defined by the active
    constraints, which is one-dimensional since the
    equality set of the face is ( n 1
    )-dimensional. So y is in one-dimensional face
    of P (edge) and no other proper face of it.
  • When ? is such that at least one of the basic
    variable becomes 0 (say xl ), then entering
    nonbasic variable replaces xl in the basis and
    the new basis matrix is nonsingular and the
    leaving basic variable xl 0 ? xl ? 0
    becomes active.
  • Hence we get a new b.f.s., which is a
    0-dimensional face of P.
  • For nondegenerate simplex iteration, we start
    from a b.f.s. ( 0-dim face), then follow an edge
    ( 1-dim face ) of P until we reach another b.f.s.
    ( 0-dim face)

19
  • (continued)

The recession cone of P is K y Ay 0, y
? 0 ( P K Q). Since d ? K and ( n 1 )
independent rows active at d, d is an extreme ray
of K (recall HW) and cd cj cBB-1Aj lt
0 ? LP unbounded. Hence, given a basis
(b.f.s.), finding an extreme ray d (basic
direction) in the recession cone with cd lt 0
provides a proof of unboundedness of LP.
20
Simplex method for degenerate problems
  • If degeneracy allowed, two possibilities
  • (a) current b.f.s. degenerate ? ? may be
    0 (if, for some l,
  • xB(l) 0 and dB(l) lt 0 )
  • Perform the iteration as usual with ? 0
  • New basis B is still nonsingular ( solution not
    changed,
  • only basis changes), hence the current solution
    is b.f.s
  • with different basis B.
  • ( Note that we may have nondegenerate iteration
    although
  • we have a degenerate solution.)
  • (b) although ? may be positive, new point may
    have more than one of the original basic
    variables become 0 at the new point. Only one of
    them exits the basis and the resulting solution
    is degenerate. (It happens when we have ties in
    the minimum ratio test.)

21
-g
x
g
f
x5 0
x4 0
h
x3 0
x6 0
y
x2 0
x1 0
  • Figure 3.3 n m 2. x4, x5 nonbasic. (g, f
    are basic dir.)
  • Then pivot with x4 entering, x6 exiting basis.
    (h, -g are basic dir)
  • Now if x5 enters basis, we follow the direction
    h until
  • x1 ? 0 becomes active, in which case x1 leaves
    basis.

22
  • Cycling a sequence of basis changes that leads
    back to the initial basis. ( only basis changes,
    no solution change)
  • Cycling may occur if there exists degeneracy.
    Finite termination of the simplex method is not
    guaranteed. Need special rules for entering
    and/or leaving variable selection to avoid
    cycling (later).
  • Although cycling hardly occurs in practice,
    prolonged degenerate iterations might happen
    frequently, especially in well-structured
    problems. Hence how to get out of degenerate
    iterations as early as possible is of practical
    concern.

23
Pivot Selection
  • (a) Smallest (largest) coefficient rule choose
    xj with
  • argminj?N cj cj lt 0
  • (b) largest increase rule xj with cj lt 0
    and ? cj is max.
  • (c) steepest edge rule
  • (d) maintain candidate list
  • (e) smallest subscript rule ( avoid cycling).

24
Review of calculus
  • Purpose Interpret the value cd in a different
    way and derive the logic for the steepest edge
    rule
  • Def p gt 0 integer. h Rn ? R, then h(x) ?
    o( xp) if and only if limxk ? 0 h(xk)/ xk
    p 0 for all sequences xk with xk ? 0
    for all k, that converges to 0.
  • Def f Rn ? R is called differentiable at x
    iff there exists a vector ?f(x) (called
    gradient) such that
  • f(z) f(x) ?f(x)( z x ) o( z x )
    or in other words,
  • lim z ? x f(z) f(x) - ?f(x)( z x ) /
    z x 0
  • (Frechet differentiability)

25
  • Def f Rn ? R. One sided directional
    derivative of f at x with respect to a vector y
    is defined as
  • f( x y ) ? lim?? 0 f(x ?y) f(x) / ?
    if it exists.
  • Note that - f( x -y) lim?? 0 f(x ?y)
    f(x) / ? .
  • Hence the one-sided directional derivative f(
    x y) is two-sided iff f( x -y) exists and
    f( x -y) -f( x y)
  • Def i-th partial derivative of f at x
  • ?f(x)/?xi lim ?? 0 f( x?ei ) f(x) /?
    if it exists (two sided)
  • ( Gateuax differentiability )
  • ( f is called Gateaux differentiable at x if all
    directional derivatives of f at a vector x exist
    and f( x y) is a linear function of y. F
    differentiability implies G differentiability,
    but not conversely. We do not need to
    distinguish F and G differentiability for our
    purposes here.)

26
  • Suppose f is F differentiable at x, then for any
    y ? 0

Hence f( x y) exists and f( x y) ?f(x)y
(linear function of y) If f is F differentiable,
then it implies f( x -y) -f( x y) from
above, hence f( x y) is two-sided. In
particular,
27
  • In simplex algorithm, moving direction d
  • for xj entering. Then
  • Hence the rate of change cd in the objective
    function when move in the direction d from x is
    the directional derivative.
  • So cj cBB-1Aj is the rate of change of f
    when we move in the direction d.
  • But, f(x d) is sensitive to the size (norm)
    of d.
  • ( f( x kd) ?f(x)(kd) k f( x d) )
  • To make fair comparison among basic directions,
    use normalized vector d / d to compute f(
    x d)

28
  • Hence, among basic directions with cj lt 0, choose
    the one with smallest normalized directional
    derivative. (steepest edge rule)
  • Problem here is that we need to compute dj
    (additional efforts needed). But, once dj
    is computed, it can be updated efficiently in
    subsequent iterations. Competitive (especially
    the dual form) against other rules in real
    implementation.
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