So far we learned a lot about the structure of LPs. But how do we actually solve them? Can we do it faster than by just enumerating all basic feasible solutions (BFS)?
Simplex – 1940s.
Ellipsoid method – 1970s.
Interior point method – 1980s.
Recall that we know that for an LP (in standard form) opt always corresponds to a vertex of a polytope (a BFS).
Imagine we are given some BFS, how to check if it is optimal?
Naive approach: check all other BFS’ and compare their costs.
\(\Rightarrow\) Problem: there can be \(m \choose n\) of them!
But do we really need to look at all the other BFS’?
Without loss of generality make \(A\) have full row rank (define):
find basis in rows of \(A\), say \(a_1,\ldots,a_k\)
any other \(a_\ell\) is linear combo of those.
so \(a_\ell x = \sum \lambda_i a_i x\)
so better have \(b_l = \sum \lambda_i a_i\) if any solution.
if so, anything feasible for \(a_1,\ldots,a_\ell\) feasible for all.
\(m\) constraints \(Ax=b\) all tight/active
given this, need \(n-m\) of the \(x_i \ge 0\) constraints
also, need them to form a basis with the \(a_i\).
write matrix of tight constraints, first \(m\) rows then identity matrix adjacent to zero matrix \[\begin{array}{l} \\ \\ m\text{ constraint rows} \\ \\ \\ \\ \\ \\ n\text{ row identity} \\ \\ \\ \end{array} \left( \begin{array}{lccccccc} &\multicolumn{4}{c}{n-m\text{ columns}}&\multicolumn{3}{c}{m\text{ columns}}\\ &\multicolumn{4}{c}{\overbrace{\qquad\qquad\qquad}}&\multicolumn{3}{c}{\overbrace{\qquad\qquad}}\\ \\ &\multicolumn{3}{c}{\cdots}&A&\multicolumn{3}{c}{\cdots}\\ \\ &1&\\ &&1&&&&0\\ &&&1\\ &&&&1\\ &&&&&1\\ &&0&&&&1\\ &&&&&&&1\\ \end{array} \right)x \begin{array}{c} \\ \\ = \\ \\ \\ \\ \ge \\ \\ \\ = \\ \end{array} \left( \begin{array}{c} \\ \\ \\ b \\ \\ \\ \\ 0 \\ \\ \\ 0 \\ \\ \end{array} \right)\]
zero matrix corresponds to slack (nonzero) \(x_i\)
need linearly independent rows
equiv, need linearly independent columns
but columns are linearly independent iff \(m\) columns of \(A\) including all that correspond to nonzero \(x_i\) are linearly independent
because columns of identity matrix are clearly independent
but columns of zero matrix depend entirely on \(A\) for independence
gives other way to define a vertex: \(x\) is vertex if
\(Ax=b\)
\(m\) linearly independent columns of \(A\) include all \(x_j \ne 0\)
This set of \(m\) columns is called a basis.
\(x_j\) of columns called basic set \(B\), others nonbasic set \(N\)
given bases, can compute \(x\):
\(A_B\) is basis columns, \(m \times m\) and full rank.
solve \(A_B x_B = b\), set other \(x_N=0\).
note can have many bases (column sets) for same vertex (choice of 0 \(x_j\))
note algebra is \(m\)-dimensional, so really depends only on number of constraints not variables
Summary: \(x\) is vertex of \(P\) if for some basis \(B\),
\(x_N=0\)
\(A_B\) nonsingular
\(A_B^{-1} b \ge 0\)
The algorithm:
start with a basic feasible soluion
try to improve it
picture: move on improving edge.
math: work relative to current \(x\)
rewrite LP: \(\min c_Bx_B + c_N x_N\), \(A_Bx_B+A_N x_N=b\), \(x \ge 0\)
true for all \(x\), not just current
\(B\) is basis for bfs
since \(A_Bx_B = b-A_Nx_N\), so \(x_B = A_B^{-1}(b-A_Nx_N)\), know that \[\begin{aligned} cx &= &c_Bx_B+c_Nx_N\\ &= & c_B A_B^{-1}(b-A_Nx_N) + c_N x_N\\ &= & c_B A_B^{-1} b + (c_N-c_BA_B^{-1}A_N)x_N\\ \end{aligned}\]
reduced cost \(\tilde c_N = c_N-c_BA_B^{-1}A_N\)
if no \(\tilde c_j < 0\), then increasing any \(x_j \in N\) increases (nondecreases) cost (may violate feasiblity for \(x_B\), but who cares?), so are at optimum!
if some \(\tilde c_j < 0\), can increase \(x_j\) to decrease cost
but since \(x_B\) is func of \(x_N\), will have to stop when \(x_B\) hits a constraint.
this happens when some \(x_i\), \(i \in B\) hits 0.
we bring \(j\) into basis, take \(i\) out of basis.
we’ve moved to an adjacent basis.
called a pivot
show picture
Need initial vertex. How find? HW.
maybe some \(x_i \in B\) already 0, so can’t increase \(x_j\), just pivot to same obj value.
could lead to cycle in pivoting, infinite loop.
can prove exist noncycling pivots (eg, lexicographically first \(j\) and \(i\))
Would hope that you don’t need to visit too many of the BFS. Not true! Klee-Minty cube (a slightly twisted hypercube)
We can’t prove though that there is no pivot choice rule that leads to always fast simplex (the best ones are “only” subexponential)
Hirch conjecture (1957): diameter/number of pivots needed is at most \(m+n\) (very optimistic!) Disproved by Santos in 2010 but only barely so: diameter \(\geq (1+\eps)(m+n)\).
Still possible that, e.g., diameter \(\leq poly(m,n)\).
Kalai-Kleitmen 1992: \(m^{\log n}\) bound on path length, also \(2^n m\).
There is a linear time simplex for fixed \(n\)!
Note small diameter does not guarantee fast simplex pivot rule - as some of the pivots might require increasing the value of the objective from time to time!
In practice: works really really well (unless it doesn’t). (Needs some linear algebra tricks to make pivoting fast.)
defined reduced costs of nonbasic vars \(N\) by \[{\tilde c}_N = c_N-c_BA^{-1}_BA_N\] and argued that when all \({\tilde c}_N \ge 0\), had optimum.
Define \(y=c_BA^{-1}_B\) (so of course \(c_B=yA_B\))
nonegative reduced costs means \(c_N \ge yA_N\)
put together with \(c_B=yA_B\), see \(yA \le c\) so \(y\) is dual feasible
but, \(yb = c_B A^{-1}_B b = c_Bx_B=cx\) (since \(x_N=0\))
so \(y\) is dual optimum.
simplex finds primal and dual optima simultaneously
more generally, \(yb-cx\) measures duality gap for current solution!
another way to prove duality theorem: prove there is a terminating (non cycling) simplex algorithm.
We know a lot about structure. And we’ve seen how to verify optimality in polynomial time. Now turn to question: can we solve in polynomial time?
Yes, sort of (Khachiyan 1979):
polynomial algorithms exist
strongly polynomial unknown.
Basic idea: Reduce optimization to feasibility checking. (Remind how it is with polytope \(P\).)
Bolzano-Weierstrass Theorem—proves certain sequence has a subsequence with a limit by repeated subdividing of intervals to get a point in the subinterval.
The Bolzano-Weierstrass method: Divide the desert by a line running from north to south. The lion is then either in the eastern or in the western part. Let’s assume it is in the eastern part. Divide this part by a line running from east to west. The lion is either in the northern or in the southern part. Let’s assume it is in the northern part. We can continue this process arbitrarily and thereby constructing with each step an increasingly narrow fence around the selected area. The diameter of the chosen partitions converges to zero so that the lion is caged into a fence of arbitrarily small diameter.
Analogue of a fence: separation oracle. Given point \(x\) that is not in \(P\) come up with a separating hyperplane, i.e., \(u\) such that \(u^T x \leq 0\) but \(y^Tx' > 0\) for all \(x'\in P\).
We want to use ellipsoid instead of boxes.
Ellipsoid \(E(D,z):=\{x \mid (x-z)^TD^{-1}(x-z) \leq 1 \}\), where \(D=BB^T\) for some invertible matrix \(B\).
Note: \(E(r\cdot I,z)\) is just an radius \(r\) sphere centered at \(z\).
In general: \(E(D,z)\) is a mapping of a unit sphere via an affine change of coordinates \(x\rightarrow Bx +z\).
Goal: start with a big ellipsoid that encompasses everything (for sure).
In each step, query the center of the ellipsoid \(z\). If \(z\in P\) done. Otherwise, there is a separating hyperplane \(u\). Shift it so it passes through the center and draw a smaller ellipsoid that encompasses the half in which the feasible point might lie.
How to measure progress? Keep track of the volume of the ellipsoid. Want it to shrink significantly in each step.
Outline of algorithm:
solve feasibility, which we know is equivalent to optimizing
goal: find a feasible point for \(P=\set{Ax \le b}\)
start with ellipse containing \(P\), center \(z\)
check if \(z \in P\)
if not, use separating hyperplane to get 1/2 of ellipse containing \(P\)
find a smaller ellipse containing this 1/2 of original ellipse
until center of ellipse is in \(P\).
Consider sphere case (wlog as affine transformations change the volume by the same factor and we care about the ratio change), separating hyperplane \(x_1=0\)
try center at \((\epsilon,0,0,\ldots)\)
Draw picture to see constraints
requirements:
\(d_1^{-1}(x_1-\epsilon )^2+\sum_{i > 1}d_i^{-1}x_i^2 \le 1\)
constraint at \((1,0,0)\): \(d_1^{-1}(x-\epsilon )^2 = 1\) so \(d_1 = (1-\epsilon )^2\)
constraint at \((0,1,0)\): \(\epsilon ^2/(1-\epsilon )^2+d_2^{-1} = 1\) so \(d_2^{-1} = 1-\epsilon ^2/(1-\epsilon )^2\approx 1-\epsilon ^2\)
What is volume? about \((1-\epsilon )/(1-\epsilon ^2)^{n/2}\)
set \(\epsilon\) about \(1/n\), get \((1-1/n)\) volume ratio.
Formalize shrinking lemma:
Let \(E=(z,D)\) define an \(n\)-dimensional ellipsoid
consider separating hyperplane \(ax \le az\)
Define \(E'=(z',D')\) ellipsoid: \[\begin{aligned} z' &= &z-\frac{1}{n+1}\frac{Da^T}{\sqrt{aDa^T}}\\ D' &= & \frac{n^2}{n^2-1}(D-\frac{2}{n+1}\frac{Da^TaD}{aDa^T}) \end{aligned}\]
then \[\begin{aligned} E \cap \set{x \mid ax \le ez} &\subseteq &E'\\ {\mbox{vol}}(E') &\le &e^{1/(2n+1)}{\mbox{vol}}(E) \end{aligned}\]
for proof, first show works with \(D=I\) and \(z=0\). new ellipse: \[\begin{aligned} z' &=&-\frac1{n+1}\\ D' &= & \frac{n^2}{n^2-1}(I-\frac{2}{n+1} I_{11}) \end{aligned}\] and volume ratio easy to compute directly.
for general case, transform to coordinates where \(D=I\) (using new basis \(B\)), get new ellipse, transform back to old coordinates, get \((z',D')\) (note transformation don’t affect volume ratios.
So ellipsoid shrinks. Now prove 2 things:
needn’t start infinitely large
can’t get infinitely small
Starting size:
recall bounds on size of vertices (polynomial)
so coords of vertices are exponential but no larger
so can start with sphere with radius exceeding this exponential bound
this only uses polynomial values in \(D\) matrix.
if unbounded, no vertices of \(P\), will get vertex of box.
Ending size:
suppose polytope full dimensional
if so, it has \(n+1\) affinely indpendent vertices
all the vertices have poly size coordinates
so they contain a box whose volume is a poly-size number (computable as determinant of vertex coordinates)
Put together:
starting volume \(2^{n^{O(1)}}\)
ending volume \(2^{-n^{O(1)}}\)
each iteration reduces volume by \(e^{1/(2n+1)}\) factor
so \(2n+1\) iters reduce by \(e\)
so \(n^{O(1)}\) reduce by \(e^{n^{O(1)}}\)
at which point, ellipse doesn’t contain \(P\), contra
must have hit a point in \(P\) before.
Justifying full dimensional:
feasible points form a “lattice” of grid points with bounded number of bits
take \(\set{Ax \le b}\), replace with \(P'=\set{Ax \le b+\epsilon}\) for tiny \(\epsilon\)
any point of \(P\) is an interior of \(P'\), so \(P'\) full dimensional (only have interior for full dimensional objects)
\(P\) empty (of lattice points) iff \(P'\) is (because \(\epsilon\) so small)
can “round” a point of \(P'\) to \(P\).
Infinite precision:
built a new ellipsoid each time.
maybe its bits got big?
no.
Optimization
Could use optimization to feasibility transform
But an easier approach is binary search on value of optimum
Add constraint that optimum must exceed proposed value
Vary value
Need to find vertex?
use rounding technique
Notice in ellipsoid, were only using one constraint at a time.
didn’t matter how many there were.
didn’t need to see all of them at once.
just needed each to be represented in polynomial size.
so ellipsoid works, even if huge number of constraints, so long as have separation oracle: given point not in \(P\), find separating hyperplane.
of course, feasibility is same as optimize, so can optimize with sep oracle too.
this is on a polytope by polytope basis. If can separate a particular polytope, can optimize over that polytope.
Another good reason to optimize by binary search on objective:
just need feasibility oracle/separator
Might not have separator for combined primal-dual formulation
especially since dual can have exponentially many variables!
This is very useful in many applications. e.g. network design.
Can also show that optimization implies separation:
suppose can optimize over \(P\)
then of course can find a point in \(P\)
suppose \(0\in P\) (saves notation mess—just shift \(P\))
define \(P^* = \set{z \mid zx \le 1\ \forall x \in P}\)
can separate over \(P^*\):
given \(w\), run OPT\((p)\) with \(w\) objective
get \(x^*\) maximizing \(wx\)
if \(wx^* \le 1\) then \(w \in P^*\)
else \(wx^* > 1 \ge x^*z \ \forall z \in P^*\) so \(x^*\) is separating hyperplane
since can separate \(P^*\), can optimize it
suppose want to separate \(y\) from \(P\)
let \(z=\)OPT\((P^*,y)\).
if \(yz>1\) then (since \(z \in P^*\)) we have \(yz>1\) but \(xz \le 1 \ \forall x \in P\) (separating hyperplane)
if \(y \le 1\) then suppose \(y \notin P\).
then \(ax \le \beta\) for \(x \in P\) but \(ay > \beta\)
since \(0 \in P\), \(\beta \ge 0\)
if \(\beta > 0\) then \(\frac{a}{\beta}x \le 1\ \forall x \in P\) so its in \(P^*\) but \(\frac{a}{\beta}y > 1\) so it is a better opt for \(y\) contra
if \(\beta = 0\) then \(\lambda ax \le 0 \le 1 \forall \lambda>0\) so \(\lambda a \in P^*\) but \(\lambda a y>1\) for some \(\lambda>0\) so is better opt for \(y\) contra.
Ellipsoid has problems in practice (\(O(n^6)\) for one). So people developed a different approach that has been extremely successful.
What goes wrong with simplex?
follows edges of polytope
complex stucture there, run into walls, etc
interior point algorithms stay away from the walls, where structure simpler.
Karmarkar did the first one (1984)
huge media hype not accurate at the time
now interior point is competitive with Simplex and often better
arguably known from 1960s as “barrier methods”
but visibility from Karmarkar set off huge progress
Primal-Dual Method
Solve primal and dual together
not by combining into one LP
use current dual to tell you how to improve primal
and vice versa
(This idea has become very powerful in non-LP combinatorial optimization also)
Potential function:
Idea: use a (nonlinear) potential function that is minimized at opt but also enforces feasibility
use gradient descent to optimize the potential function.
argue make “sufficient progress per step” to finish in poly steps
Use logarithmic barrier function \[G(x) = cx-\mu(\sum\ln x_j)\] and try to minimize it
first term aims for optimal objective
second enforces positivity
note barrier prevents from ever hitting optimum since \(G \rightarrow \infty\) at boundary
but as discussed above ok to just get close.
then “round” to a better vertex, which will be opt
Early methods used gradient descent:
immediately take \(\mu\) small so objective dominates
then use gradient descent to minimize \(G_\mu(x)\)
More recent, arguably cleaner approach is central path
for each \(\mu\) there is an optimum point
varying \(\mu\) traces out the central path
Start with \(\mu\) huge, objective function term negligible
Equivalent to “start with a feasible point”
And requires just as much thinking as simplex, ellipsoid
but same kind of tricks work
shrink \(\mu\) towards 0
eventually objective dominates as \(\mu \rightarrow 0\)
so we reach opt
in the continuous domain unconcerned with runtime, this “obviously” works
we need a discrete version
Characterizing the central path
fix \(\mu\)
central path minimizes \(G_\mu(x)\)
imposes condition on gradient \(\nabla G(x) = \langle c_i-\mu/x_i \rangle\)
but gradient may not be zero
because we are constrained by \(Ax=b\)
so real condition is that gradient is perpendicular to feasible set
i.e. is a linear combination of rows of \(A\)
(because these are the constraint hyperplanes’ normal vectors)
i.e. \(yA = \langle c_i-\mu/x_i \rangle\)
write \(yA+s=c\) where \(s_i =\mu/x_i > 0\) are slack variables
so \(y,s\) is dual feasible
duality gap is \(cx-yb = (yA+s)x - yb = sx = n\mu\)
conversely, if have feasible \(x_i, s_i\) with \(x_is_i=\mu\), then we know we are on the central path at parameter \(\mu\)
this is just saying each constraint is violated “equally” under proper scaling
How discretize?
“predictor-corrector method”
use linear approximation
compute tangent to central path
move along it as far as approximation holds (“predictor”)
conclude \(\mu\) has improved on tangent and thus on corresponding central path point
“correct” back to central path
each step improves \(\mu\)
stop when \(\mu\) “close enough” to 0
I will show predictor step (moving towards better \(\mu\))
corrector step is similar but easier
just moving back to an optimum of the potential for new \(\mu\)
Termination
as with ellipsoid, analyze time needed to get from start to “sufficiently good” \(\mu\)
once duality gap smaller than bit-precision of vertices, done
as we saw before, can “round” to a vertex
since are closer to opt than second-best vertex, our starting objective is better than second best
so we can only round to opt
thus, sufficient to get \(\mu = 2^{-O(L)}\)
as with simplex, there’s the cold-start problem
using similar method (designing related LP with easy starting point and opt at real desired starting point) can achieve \(\mu=2^L\)
so solving only requires \(O(L)\) iterations of improving \(\mu\) by a constant
What is the improvement direction?
From current \(x,s, \mu\), update to \(x+dx\), \(s+ds\) that are “on central path” for new, smaller \(\mu\)
central path needs \((s_i+ds_i)(x_i+dx_i) = (\mu+d\mu)\) (same constant \(\forall i\))
i.e. \(s_i d(x_i) + d(s_i)x_i = d\mu +\) low order terms
but also need \(A(dx)=0\) (to stay primal feasible)
and \((dy)A+ds=0\) (to stay dual feasible)
solve this linear system to get directions \(ds\) and \(dx\)
these are the tangent for central path
Can we solve this?
yes: if rescale, can see this as a projection problem
rescale \(A\) so all \(x_i=1\), meaning all \(s_i = \mu\)
equations now require \[\begin{aligned} \mu dx + ds & = &\mathbf{1}d\mu\\ Adx &= &0\\ (dy)A+ds &= &0\end{aligned}\]
here \(\mathbf{1}\) is the all-ones vector
in other words, \(\mu dx\) in nullspace of \(A\) and \(ds\) in span of \(A\)
but they sum to \(\mathbf{1}d\mu\)
so \(\mu dx\) and \(ds\) are just decomposition of \(\mathbf{1}d\mu\) onto \(A\) and \(A^{\perp}\)
note this defines directions because both \(dx\) and \(ds\) (and \(d\mu\) and \(dy\)) can be multiplied by same arbitrary scalar
move in this direction until approximation breaks
How far?
Simplify
recall rescaled coordinates until all \(x_i=1\)
so all \(s_i = \mu\)
and duality gap is \(\sum s_i = n\mu\)
rescale \(dx_i\) and \(ds_i\) so “predicted” change in \(\mu\) is \(d\mu = \mu\)
i.e. linear approximation predicts each \(x_is_i\) decreases by \(\mu\)
predicts we close duality gap by moving \(dx,ds\)
how far in that direction can we move—parameter \(\theta\)—before approximation breaks?
we want to be on central path, i.e. \((x_i+\theta dx_i)(s_i+\theta ds_i) = \mu+d\mu\)
actual change is \(\theta (x_ids_i + s_idx_i) +\theta^2 ds_idx_i\)
our linear system says these quantities equal for all \(i\) to first order (ignoring \(dx_ids_i\) term)
so until our linear approximation breaks down, all \(s_ix_i\) are changing by about same amount
i.e., we are near central path
approximation breaks when second order term is comparable to first order
so want \(\theta^2dx_ids_i \ll \theta (x_ids_i + s_idx_i)\).
i.e. \(\theta \ll x_i/dx_i + s_i/ds_i\)
we rescaled \(x_i=1\) and \(s_i=\mu\), so need \(\theta \ll \mu/ds_i + 1/dx_i\).
so how big can \(dx_i\) be?
remember we solved \(\mu dx + ds =\) orthogonal decomposition of \(d\mu\)
and rescaled so length of \(d\mu\) is \(\mu\)
i.e. \(\mu dx\) is projection of \(\mathbf{1}\mu\), so \(dx\) is projection of \(\mathbf{1}\)
all one’s vector has length \(\sqrt{n}\)
so any \(dx_i \le \sqrt{n}\)
similarly, any \(ds_i \le \mu\sqrt{n}\)
so requirement that \(\theta \ll \mu/ds_i+1/dx_i\) met by \(\theta \ll \mu/(\mu \sqrt{n}) + 1/\sqrt{n} = O(1/\sqrt{n})\)
so we can set \(\theta \approx 1/\sqrt{n}\) and have second-order term negligible
How much improvement?
we said if \(\theta=1\) we close duality gap to zero (in linear approximation)
since we only move \(1/\sqrt{n}\) we shave a \(1/\sqrt{n}\) factor.
but that means we reduce \(\mu\) by a constant in \(\sqrt{n}\) steps
which means we finish in poly\((n)\) steps!
Interior point has recently been revolutionizing maxflow.