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COMP4121 Linear Programming

创建日期：2024-04-22 17:04:45

所属分类：计算机computer science

标签： COMP4121

Linear Programming

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COMP4121 Lecture Notes

Linear Programming

We now move to one of the most important cases of convex programming,
called Linear Programming, (LP), in which the objective is a linear function
and the convex domain over which the extremum is sought is defined by con-
straints which are also linear functions. We follow closely our old COMP3121
textbook by Cormen, Leiserson, Rivest and Stein, introduction to Algorithms.
We start by defining a common representations of Linear Programming opti-
mization problems.
In the standard form the objective to be maximized is given by
f(x) =
n∑
j=1
cj xj ,
and the constraints are of the form
n∑
j=1
aijxj ≤ bi, 1 ≤ i ≤ m; (1)
xj ≥ 0, 1 ≤ j ≤ n, (2)
Note that x represents a vector, x> = 〈x1, . . . , xn〉; we assume that vectors are
columns; thus, to write them as rows we have to transpose them. The numbers
aij are assumed to be real.
Clearly, a minimization problem of the form:
minimize f(x) =
n∑
j=1
cj xj ,
subject to the constraints of the form
n∑
j=1
aijxj ≥ bi, 1 ≤ i ≤ m; (3)
xj ≥ 0, 1 ≤ j ≤ n, (4)
can easily be reduced to a corresponding maximization problem in the standard
form
maximize f∗(x) =
n∑
j=1
c∗j xj ,
with constraints of the form
n∑
j=1
a∗ijxj ≤ b∗i , 1 ≤ i ≤ m;
xj ≥ 0, 1 ≤ j ≤ n,
by taking c∗ = −c, a∗ij = −aij and b∗ = −b.
1
Example: Humans require 13 vitamins V1, . . . , V13 in total; the daily re-
quired amounts d1, . . . , d13 for each of them are known. Your local grocery store
caries 130 types of food staples, and for each food staple Sk and each of the 13
vitamins Vi the content of Vi per gram of Sk is also a known value, denoted by
C(k, i). For each food staple Sk you are also given its price per gram, denoted
by Pk. Your goal is to determine the quantities xk of each food staple Sk which
you should buy, such that your daily requirements of all vitamins are met, but
the total price you pay for your daily food provision is as small as possible.
Solution: You have to:
minimize the objective
130∑
k=1
Pk xk
subject to 13 constraints
130∑
k=1
C(k, i)xk > di, 1 ≤ i ≤ 13,
x1 ≥ 0, . . . , x130 ≥ 0.
Somewhat messy formulation of an LP program we have given can be refor-
mulated in a much more compact matrix form. To get a very compact represen-
tation of linear programs, let us introduce a partial ordering on vectors x ∈ Rn
by x ≤ y if and only if such inequalities hold coordinate-wise, i.e., if and only if
xj ≤ yj for all 1 ≤ i ≤ n. Since vectors are written as a single column matrices,
letting c> = 〈c1, . . . , cn〉 ∈ Rn and b> = 〈b1, . . . bm〉 ∈ Rm, and letting A be
the matrix A = (aij)i,j of size m × n, we get that the above problem can be
formulated simply as:
maximize c>x
subject to the following two (matrix-vector) constraints:1 Ax ≤ b and x ≥ 0.
Thus, to specify a Linear Programming optimisation problem we just have
to provide a triplet (A,b, c), and the information that the triplet represents an
LP problem in the standard form.
While in general the “natural formulation” of a problem into a Linear Pro-
gram does not necessarily produce the non-negativity constrains (4) for all of
the variables, in the standard form such constraints are indeed required for all
of the variables. However, this poses no problem, because each occurrence of
an unconstrained variable xj can be replaced by the expression x
′
j − x∗j where
x′j , x
∗
j are new variables satisfying the constraints x
′
j ≥ 0, x∗j ≥ 0.
Any vector x which satisfies the two constraints is called a feasible solution,
regardless of what the corresponding objective value cTx might be. Note that
the set of feasible solutions, i.e., the domain over which we seek to maximize the
1Note that the inequality involved in the constraints is the partial ordering on vectors
which we have just introduced above. Also, in the non-negativity constraint, 0 represents a
vector 0 ∈ Rn with all coordinates equal to 0.
2
objective, is a convex set because it is an intersection of the half spaces defined
by each of the constraints.
As an example, let us consider the following optimization problem:
maximize 3x1 + x2 + 2x3 (5)
subject to the constraints
x1 + x2 + 3x3 ≤ 30 (6)
2x1 + 2x2 + 5x3 ≤ 24 (7)
4x1 + x2 + 2x3 ≤ 36 (8)
x1, x2, x3 ≥ 0 (9)
How large can the value of the objective z(x1, x2, x3) = 3x1 + x2 + 2x3 be,
without violating the constraints? If we add inequalities (6) and (7), we get
3x1 + 3x2 + 8x3 ≤ 54
Since all variables are constrained to be non-negative, we are assured that
3x1 + x2 + 2x3 ≤ 3x1 + 3x2 + 8x3 ≤ 54
The far left hand side of this equation is just the objective (5); thus z(x1, x2, x3)
is bounded above by 54, i.e., z(x1, x2, x3) ≤ 54.
Can we obtain a tighter bound? We could try to look for coefficients
y1, y2, y3 ≥ 0 to be used to for a linear combination of the constraints:
y1(x1 + x2 + 3x3) ≤ 30y1
y2(2x1 + 2x2 + 5x3) ≤ 24y2
y3(4x1 + x2 + 2x3) ≤ 36y3
Then, summing up all these inequalities and factoring, we get
x1(y1 + 2y2 + 4y3) + x2(y1 + 2y2 + y3) + x3(3y1 + 5y2 + 2y3)
≤ 30y1 + 24y2 + 36y3 (10)
If we compare this with our objective (5) we see that if we choose y1, y2 and y3
so that:
y1 + 2y2 + 4y3 ≥ 3
y1 + 2y2 + y3 ≥ 1
3y1 + 5y2 + 2y3 ≥ 2
then
x1(y1 + 2y2 + 4y3) + x2(y1 + 2y2 + y3) + x3(3y1 + 5y2 + 2y3) ≥ 3x3 + x2 + 2x3
3
Combining this with (10) we get:
30y1 + 24y2 + 36y3 ≥ 3x1 + x2 + 2x3 = z(x1, x2, x3)
Consequently, in order to find an as tight as possible upper bound for our
objective z(x1, x2, x3), we have to look for y1, y2, y3 which produce the smallest
possible value of z∗(y1, y2, y3) = 30y1 + 24y2 + 36y3, but which do not violate
the constraints
y1 + 2y2 + 4y3 ≥ 3 (11)
y1 + 2y2 + y3 ≥ 1 (12)
3y1 + 5y2 + 2y3 ≥ 2 (13)
y1, y2, y3 ≥ 0 (14)
Thus, trying to find the best upper bound for our objective z(x1, x2, x3) obtained
by forming a linear combination of the constraints only reduces the original
maximization problem to a minimization problem:
minimize 30y1 + 24y2 + 36y3
subject to the constraints (11-14)
Such minimization problem P ∗ is called the dual problem of the initial problem
P .
Let us now repeat the whole procedure with P ∗ in place of P , i.e., let us find
the dual program (P ∗)∗ of P ∗. We are now looking for non negative multipliers
z1, z2, z3 ≥ 0 to multiply inequalities (11-13) and obtain
z1(y1 + 2y2 + 4y3) ≥ 3z1
z2(y1 + 2y2 + y3) ≥ z2
z3(3y1 + 5y2 + 2y3) ≥ 2z3
Summing these up and factoring produces
y1(z1 + z2 + 3z3) + y2(2z1 + 2z2 + 5z3) + y3(4z1 + z2 + 2z3) ≥ 3z1 + z2 + 2z3
(15)
If we choose these multiples so that
z1 + z2 + 3z3 ≤ 30 (16)
2z1 + 2z2 + 5z3 ≤ 24 (17)
4z1 + z2 + 2z3 ≤ 36 (18)
we will have:
y1(z1 + z2 + 3z3) + y2(2z1 + 2z2 + 5z3) + y3(4z1 + z1 + 2z3) ≤ 30y1 + 24y2 + 36y3
Combining this with (15) we get
3z1 + z2 + 2z3 ≤ 30y1 + 24y2 + 36y3
Consequently, finding the dual program (P ∗)∗ of P ∗ amounts to maximizing
the objective 3z1 + z2 + 2z3 subject to the constraints (16-18). But notice that,
4
except for having different variables, (P ∗)∗ is exactly our starting program P .
Thus, the dual program (P ∗)∗ for program P ∗ is just P itself, i.e., (P ∗)∗ = P .
So, at the first sight, looking for the multipliers y1, y2, y3 did not help much,
because it only reduced a maximization problem to an equally hard minimization
problem. Well, perhaps it still maybe easier to somehow solve both problems
at the same, and this is precisely what the SIMPLEX algorithms does.2
To find the maximal value of 3x1 + x2 + 2x3 subject to the constraints
x1 + x2 + 3x3 ≤ 30
2x1 + 2x2 + 5x3 ≤ 24
4x1 + x2 + 2x3 ≤ 36
let us start with x1 = x2 = x3 = 0 and ask ourselves: how much can we increase
x1 without violating the constraints? Since x1 +x2 +3x3 ≤ 30 we can introduce
a new variable x4 such that x4 ≥ 0 and
x4 = 30− (x1 + x2 + 3x3) (19)
Since such variable measures how much “slack” we got between the actual value
of x1+x2+3x3 and its upper limit 30, such x4 is called a slack variable. Similarly
we introduce new variables x5, x6 requiring them to satisfy x5, x6 ≥ 0 and let
x5 = 24− 2x1 − 2x2 − 5x3 (20)
x6 = 36− 4x1 − x2 − 2x3 (21)
Since we started with the values x1 = x2 = x3 = 0, this implies that these new
slack variables must have values x4 = 30, x5 = 24, x6 = 36, or as a single
vector, the initial basic feasible solution is (
x1
0 ,
x2
0 ,
x3
0 ,
x4
30,
x5
24,
x6
36). Note that “fea-
sible” refers merely to the fact that all of the constraints are satisfied.3
Now we see that (19) implies that x1 cannot exceed 30, and (20) implies
that 2x1 ≤ 24, i.e., x1 ≤ 12, while (21) implies that 4x1 ≤ 36, i.e., x1 ≤ 9. Since
all of these conditions must be satisfied we conclude that x1 cannot exceed 9,
which is the upper limit coming from the constraint (21).
If we set x1 = 9, this forces x6 = 0. We now swap the roles of x1 and x6:
since we cannot increase x1 any more, we eliminate x1 from the right hand sides
of the equations (19-21) and from the objective, introducing instead variable x6
to the right hand side of the constraints and into the objective. To do that, we
solve equation (21) for x1:
x1 = 9− x2
4
− x3
2
− x6
4
2It is now useful to remember how we proved that the Ford - Fulkerson Max Flow algorithm
in fact produces a maximal flow, by showing that it terminates only when the flow reaches
the capacity of a (minimal) cut!
3Clearly, setting all variables to 0 does not always produce a basic feasible solution because
this might violate some of the constraints; this would happen, for example, if we had a
constraint of the form −x1 +x2 +x3 ≤ −3; choosing an initial basic feasible solution requires
a separate algorithm to “bootstrap” the SIMPLEX algorithm - see for the details our (C-L-
R-S) textbook.
5
and eliminate x1 from the right hand side of the remaining constraints and the
objective to get:
z = 3
(
9− x2
4
− x3
2
− x6
4
)
+ x2 + 2x3
= 27− 3
4
x2 − 3
2
x3 − 3
4
x6 + x2 + 2x3
= 27 +
1
4
x2 +
1
2
x3 − 3
4
x6
x5 = 24− 2
(
9− x2
4
− x3
2
− x6
4
)
− 2x2 − 5x3
= 6 +
x2
2
+ x3 +
x6
2
− 2x2 − 5x3
= 6− 3
2
x2 − 4x3 + x6
2
x4 = 30−
(
9− x2
4
− x3
2
− x6
4
)
− x2 − 3x3
= 21 +
x2
4
+
x3
2
+
x6
4
− x2 − 3x3
= 21− 3
4
x2 +
5
2
x3 +
x6
4
To summarise: the “new” objective is
z = 27 +
1
4
x2 +
1
2
x3 − 3
4
x6
and the “new constraints” are
x1 = 9− x2
4
− x3
2
− x6
5
(22)
x4 = 21− 3
4
x2 − 5
2
x3 +
x6
4
(23)
x5 = 6− 3
2
x2 − 4x3 + x6
2
(24)
Our new basic feasible solution replacing (
x1
0 ,
x2
0 ,
x3
0 ,
x4
30,
x5
24,
x6
36) is obtained by set-
ting all the variables on the right to zero, thus obtaining (
x1
9 ,
x2
0 ,
x3
0 ,
x4
21,
x5
6 ,
x6
0 ).
NOTE: These are EQUIVALENT constraints and objectives; the old ones were
only transformed to an equivalent form. Any values of the variables will produce
exactly the same value in both forms of the objective and they will satisfy the
first set of constraints if and only if they satisfy the second set.
So x1 and x6 have switched their roles; x1 acts as a new basic variable,
and the new basic feasible solution is: (
x1
9 ,
x2
0 ,
x3
0 ,
x4
21,
x5
6 ,
x6
0 ); the new value of the
objective is z = 27 + 140 +
1
20− 340 = 27. We will continue this process of find-
ing basic feasible solutions which increase the value of the objective, switching
which variables are used to measure the slack and which are on the right hand
side of the constraint equations and in the objective. The variables on the left
are called the basic variables and the variables on the right are the non basic
variables.
6
We now choose another variable with a positive coefficient in the objective,
say x3 (we could also have chosen x2). How much can we increase it?
From (22) we see that x32 must not exceed 9, otherwise x1 will become
negative. Thus x3 cannot be larger than 18. Similarly,
5
2x3 cannot exceed 21,
otherwise x4 will become negative, and so x3 ≤ 425 ; similarly, 4x3 cannot exceed
6, ie, x3 ≤ 32 . Thus, in order for all constraints to remain valid, x3 cannot exceed
3
2 . Thus, we increase x3 to
3
2 ; equation (24) now forces x5 to zero.

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CS 333 CP1406 KIT308 Economics 701 ECON3740 ECE 5759 BFC3340 FINC2012 COMP4500 MATH3871 FIT3080 18-819C Economic ECO202Y5Y PHY100 FINC-780 Econ 100B EC602 IE582 EE450 COMP9517 WRDS 150B ECON 570 LECTURE 2 ECA5101 ECE 4420 ADAD9311 FIT3031 CS 544 MF 703 CS 6140 KIT501 ENGSCI 344 CSC171 IE 332 FINM8007 21S2 COSC 1114 FE545 E471 SIT102 CSC108 MAT301 IAE 101 COP3503 STAT2004 EC 303 CS1026 STAT1203 MATH3171 ISOM5160 IFN657 FIT2014 ACCT2102 DSCI550 STATS 330 ELEC421 COMP3670 QBUS6840 MEES:698B MAST30027 MIET1077 Economics IFT 6758 FIT5122 BUSN5101 CptS 111 HRMT5502 Math2018 PP 312 INFS3200 CZ3005 CSSE1001 SENG 2011 MKTG 553Q BUFN 640 COMP507 ETF5952 DATA 311 ECE4043 BANK3011 finance M3H623544 HW1 INFS 7410 AI6103 STAT3888 INFS3208 ACCT7103 ECOS3013 PSYCH 20B F79MA STAT3006 ENGG952 2X MGSC 7000 ECE4132 MSF 526 MATH 3033 CDS 511 EEEE4120 EEET2465 ECA5103 MATH2070 ELEC3104 BU51019 EEE8147 ENGR20003 COSC 1107 ETF3200 FNCE 435 MSFI 435 ISyE6420 GENG4405 HW ETC5523 IE 7315 MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A PHIL2617 PHYS3116 MIE250 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