Download MATH3033 Information Sheet 1. The Standard Maximum Problem

MATH3033 Information Sheet
1
Linear Programming
1. The Standard Maximum Problem: Find a vector x that maximises
the quantity c1 x1 + . . . + cn xn subject to the constraints
a11 x1 + . . . + a1n xn ≤ b1
..
.
am1 x1 + . . . + amn xn ≤ bm
where x1 , . . . , xn ≥ 0. We often write c = (c1 , . . . , cn )T , b = (b1 , . . . , bm )T


a11 · · · a1n
.. .
..
and A =  ...
.
.
am1 · · · amn
For the rest of the statements in this section, this standard maximum
problem will be referred to as (1).
2. The Standard Minimum Problem: Find a vector y that minimises
the quantity b1 y1 + . . . + bm ym subject to the constraints
a11 y1 + . . . + am1 ym ≥ c1
..
.
a1n y1 + . . . + amn ym ≥ cn
where y1 , . . . , ym ≥ 0.
For the rest of the statements in this section, this standard minimum
problem will be referred to as (2).
3. Theorem 1: If x is feasible for the standard maximum problem (1) and
y is feasible for its dual (2) then
cT x ≤ yT b.
4. Corollary 1: If a standard problem and its dual are both feasible, then
both are bounded feasible.
5. Corollary 2: If there are points x0 and y0 feasible for a standard maximum problem (1) and its dual (2) respectively, such that cT x0 = y0T b
then x0 is optimal for (1) and y0 is optimal for (2).
1
6. The Duality Theorem: If a standard maximum or minimum problem
is bounded feasible, then so is its dual, optimal vectors for both problems
exist, and their objective functions are equal.
7. The Equilibrium Theorem: Let x and y be feasible vectors for a
standard maximum problem (1) and its dual (2) respectively. Then x and
y are optimal if and only if
yi = 0 for every i such that
n
X
aij xj < bi
j=1
and
xj = 0 for every j such that
m
X
yi aij > cj .
i=1
8. The Simplex Method: The simplex tableau for a standard maximum
problem and its dual looks like this:
y1
y2
..
.
x1
a11
a21
..
.
x2
a12
a22
..
.
···
···
···
...
ym am1 am2 · · ·
−c1 −c2 · · ·
xn
a1n
a2n
..
.
b1
b2
..
.
amn bm
−cn 0
If ai0 j0 6= 0 then yi0 and xj0 can exchange places pivoting on the entry ai0 j0 ,
in which case each aij is replaced with âij according to these formulae:
1
ai0 j0
a
âi0 j = aii0jj for j 6= j0 . (Entries in the same row)
0 0
a
âij0 = − aiijj0 for i 6= i0 . (Entries in the same column)
0 0
a a
âij = aij − ijai0 jio j for i 6= i0 and j 6= j0 .
0 0
(i) âi0 j0 =
(ii)
(iii)
(iv)
The simplex method is to follow these rules:
Rule 1: If b ≥ 0, choose any column j0 with the last entry negative. Of the
b
positive aij0 choose i0 such that ai i0j ≤ abiji for all i, and pivot on
0 0
0
ai0 j0 . If there is a tie, any of the tied entries may be used.
Rule 2: If some of the bi are negative, pick the first bk such that bk < 0,
b1 , . . . , bk−1 ≥ 0. In row k, pick any negative entry akj0 < 0. Calculate
bk
and abiji for all i with bi ≥ 0 and aij0 > 0. Pick i0 for which this
akj0
0
ratio is smallest (in the case of a tie, pick any one of the tied entries)
and pivot on ai0 j0 .
2
Alternatively, the simplex method can be applied to the dual problem, in
which case the rules are:
Rule 1*: If −c ≥ 0, choose any row i0 with the last entry negative. Of the
−c
−c
negative ai0 j choose j0 such that ai jj0 ≥ aij j for all j, and pivot on
0 0
0
ai0 j0 . If there is a tie, any of the tied entries may be used.
Rule 2*: If some of the −cj are negative, pick the first −ck such that −ck < 0,
−c1 , . . . , −ck−1 ≥ 0. In column k, pick any positive entry ai0 k > 0.
−c
k
Calculate a−c
and ai jj for all j with −cj ≤ 0 and ai0 j > 0. Pick j0 for
i0 k
0
which this ratio is closest to zero (in the case of a tie, pick any one of
the tied entries) and pivot on ai0 j0 .
9. Sensitivity Analysis
Suppose the standard maximum problem has been solved with dual solution y = (y1 , . . . , ym ). To calculate the new CPF solution under changes
in parameter(s) A → Â, b → b̂ and c → ĉ, follow the following steps:
(a) Construct the matrix [Â : I] and remove the columns corresponding
to nonbasic variables. What remains is B.
(b) Calculate B −1 . (Alternatively, and probably more quickly, find B −1
from the coefficients of the slack variables in the extended solved
tableau.)
(c) Calculate:
b∗ = B −1 b̂,
A∗ = B −1 Â,
z ∗ − ĉ = y − ĉ and
Z ∗ = yb̂.
(d) Construct the simplex matrix
B −1 b∗
A∗
z ∗ − ĉ y Z ∗
(e) Apply Gauss-Jordan elimination so that the columns of this matrix
corresponding to basic variables have zeroes everywhere, except for
the entry in the row of that basic variable, which must be 1.
(f) Removing the columns of the basic variables leaves the new simplex
tableau.
10. Integer Programming
We use the branch and bound procedure. Set Z ∗ = −∞ (or to +∞ for
a standard minimum IP problem).
(a) Divide the feasible region according to values of one decision variable.
3
(b) Solve the LP relaxations of the two new sub-problems, to get bounds
on Z in each sub-region.
(c) A sub-problem is fathomed if any of the following conditions is true:
(i) The solution to the LP relaxation is integer. If the Z-value here
is better than Z ∗ then update the value of Z ∗ to this new value.
(ii) The sub-problem is infeasible.
(iii) The Z-value of the sub-problem is worse than Z ∗ .
A fathomed sub-problem needs no further consideration.
If all sub-problems have been fathomed, stop; the IP solution is Z ∗ .
If there are unfathomed sub-problems remaining, re-start the procedure
on the most recent one. In the case of ties, pick the one with the best
Z-value.
2
Nonlinear Programming
1. A function f (x) is convex (resp. concave) if the line segment joining any
two points of the graph of f lies on or above (resp. below) the graph of
f.
If f is a function of one variable, then f is convex (resp. concave) iff
2
d2 f
≥ 0 (resp. dd xf2 ≤ 0).
d x2
If f is a function of two variables, then f is convex (resp. concave) iff
∆ = fxx fyy − (fxy )2 ≥ 0 and fxx , fyy ≥ 0 (resp. fxx , fyy ≤ 0).
2. A set S is convex if the line segment joining any two elements of S lies
entirely in S.
3. Newton’s Method for functions of one variable: Provided that xi
is sufficiently close to x∗ , where f (x∗ ) = 0, an approximation xi+1 to x∗
can be found using the formula
xi+1 = xi −
f 0 (xi )
.
f 00 (xi )
4. Karush-Kuhn-Tucker conditions: If x∗ minimises f (x) subject to the
conditions that for i = 1, . . . , m and j = 1, . . . , l, gi (x) ≤ 0 and hj (x) = 0,
4
then there exist scalars µi and λj (called KKT multipliers) such that
∇f (x∗ ) +
m
X
µi ∇gi (x∗ ) +
i=1
l
X
λj ∇hj (x∗ ) = 0
j=1
gi (x∗ )
hj (x∗ )
µi gi (x∗ )
µi
≤
=
=
≥
0
0
0
0,
for i = 1, . . . , m and j = 1, . . . , l.
Corollary: If f is convex and the feasible region is also convex, any point
x∗ that satisfies the KKT conditions must be a global minimum.
5. Quadratic Programming: We assume that f is of the form f (x) =
cT x− 21 xT Qx, and that all constraints are linear. To minimise f (x) subject
to the linear constraints Ax ≤ b and x ≥ 0, we do the following:
(i) Write down the KKT conditions for the problem and re-arrange them
in the form
Qx + (−IAT )µ = c
Ax + s = b,
along with a complementarity condition.
(ii) For any ci > 0 and bj < 0 introduce artificial variables z1 , z2 , etc.
(iii) Solve the LP problem of minimising Z = z1 + z2 + · · · (i.e. maximise
−Z = −z1 − z2 − · · · ), with the constraints you found in step (i).
Use the simplex method to do this, rewriting the objective function
in terms of non-z variables and avoiding any pivots that violate the
complementarity conditions.
3
Game Theory
1. Dominance: Player i’s strategy ŝi is strictly dominated by player i’s
strategy si if
ui (si , s−i ) > ui (ŝi , s−i )
for all possible s−i .
2. Weak Dominance: Player i’s strategy ŝi is weakly dominated by player
i’s strategy si if
ui (si , s−i ) ≥ ui (ŝi , s−i )
5
for all possible s−i , and
ui (si , s−i ) > ui (ŝi , s−i )
for some s−i .
3. Best Response: Player i’s strategy ŝi is a best response to other players’
strategies s−i if
ui (ŝi , s−i ) ≥ ui (si , s−i )
for all si .
Equivalently, ŝi solves the problem: choose si to maximise ui (si , s−i ).
4. Nash Equilibrium: A strategy profile (s∗1 , s∗2 , . . . , s∗N ) is a Nash Equilibrium if s∗i is a best response to s∗−i for each i ∈ {1, 2, . . . , N }.
4
Markov Chains
We assume that time increases in discrete units, t = 0, 1, 2, . . . and that the
system in question has a finite number of states, numbered 0 to N .
(n)
1. We define the n-step transition probability pij as the probability that in
n units of time from now, the system will be in state j, given that it is
in state i now. We usually write the one-step transition probabilities in
a transition matrix P = (pij ). The n-step probabilities can be found by
computing P n .
2. For any irreducible ergodic Markov chain, lim pnij = πj exists and is
n→∞
independent of i. The row vector (π0 , . . . , πN ) can be found by solving
(π0 , . . . , πN ) = (π0 , . . . , πN )P.
(n)
3. fij is the probability that the system will transition from state i to state j
N
∞
X
X
(1)
(n)
(n−1)
(n)
in n steps. We have fij = pij and fij =
pik fkj . If
fij = 1
k=0,k6=j
n=1
then the expected first-passage-times µij (the expected time to transition
from state i to state j) satisfy
µij = 1 +
N
X
k=0,k6=j
6
pik µkj .
4. If state k is an absorbing state, fik is the probability that the system will
ever enter state k given that it is in state i now. then the column vector
(f0k , f1k , . . . , fN k )T can be computed from




f0k
f0k
 f 
 f1k 

 .  = P  1k
 ...  ,
 .. 
fN k
fN k
along with the conditions that fkk = 1, and if state l is any other absorbing
state, flk = 0.
7