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COMP4121 Linear Programming
Creation date：2024-04-22 17:04:45
Belonging category：计算机computer science
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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CIVL2010 ECMT2150 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 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CSCI262 SCC.311 COMP8101 IAB320 CHC5028 TSP ICC565 CS512 320SC COMP2396 COMP207 SVM BSCH EECS 2101 COMP3230 CEG3136 INFO 3440 EE3009 MGT7174 COMP 3334 ECON20120 Econ 702 CFRM 505 CS4001 CSE 328 AMATH 482 ENG2008 MATH39522 DTS101TC CSC318 EDUC 124 NBA 5110 ECON0125 ELEC7310 EEP101 ECE 777AE MDS 6112 Bioc0036 ACCFIN5246 STA 404 CFRM505 Chinese 2241G MATH 376 AUENV 218 ELEC0006 ACCT 3230 FIN3020 ENVM3115 CMPSC 464 EE 4233 CST 8284 ENGI 2211 EEB240 COMP52715 PHYS1120 ECON0019 MAT187 FINN1037 CSci 5421 7SSMM803 STAT 400 LUBS5996M ELC5216 STATS 4A03 ENGG1003 BADM 3661 DATA423 ST332 ECO2121 6406ELE IB3K20 CSE 465 CSE115 CMPUT 206 ecs150 INU1111 QM222 DTS204TC MATH2319 CPT108 SENG6110 MATH3041 INFS3604 MA 4704 CIVIL 750 EEEE4123 CS 3800 comp3511 MEC206 CSCI-UA.202 IOM 302 NBS8295 BMS 201 ECON 3018 ESE 605 ADM3225 STAT7038 Economics 409 ECON 503 MSE800 4SSMN902 IEOR E4106 FRE6831 ISE 563 MATH 451 ECON0001 COMPSCI369 CITS4402 MA1608 MQBS1030 MUSC1982 CITS4012 ACC202 BAFI1026 MA117 ECON5074 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270M CIS 3207 Programs CSCI 2132 CS270 CSE-381 COMP3230A CIS 212 CSI 333 FINM 326 FINM-36702 Introduction to Software Analysis EEE102 GNG1106 ECS 60 CS 161 CSCI851 CSE1222 EDPS0249 TCHR5003 ANTA02 BBUS1SBY SOSC 1140 MDIA2091 FINS3641 INSY 661 ITEC3040 CPSC 481 computer DEE3519 COMP304 CSE536 Microtheory OFFICE 280 CIVE50003 COSC2673 ECS784U CMPSC/MATH 455 AMA 505 CSE 158 BUS1040 FIT3179 FIT5202 Math 104A Math 111A MBUS107 COMPSCI 761 COMP3710 COMPSCI 316 COMP559 COMP557 CP1 1902 COMP 310 ECED 3403 COMPX523 AGCM 3103 EMET 7001 ENGR3821 CISC 124 IPC comp20005 CS 323 CS105 STAT 326 MATH4007 CS4551 COMP1000 ACIT 4640 ENVX3002 CS2034 FC712 algorithms CSCI 4155 CSI4142 POL 850 MATH20811 2B03 CSCI803 Economies COMP 2016 Architect TRADE INFO1110 FIT1043 networks Robotics ICS 31 ECEC-301 FIT2099 CS 3640 CSE1010 Computational COMP1001 CSE 231 CS10 FIT5166 CSE 400 CSCI561 Math 350 CO-496 ENGG6400 CSCI 1133 COMP3055 MATH 819 CSE 6363 CIS 555 CS4420 COMP0037 ECS 140A Spanning Tree Protocol network 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4061 ECON 8820 INT104 ITE3713 CS551J ECM1410 CSMBD21 COMP 2012 CISC221 Code SCC312 COMP639 CMPSC 221 system COMP3217 COMP1721 ENG5056 Math 466 COMP0039 MFIN 5400F FINN40915 EENG31400 MODL5007M ​EEET 3050 EEET 3050 Quantitative Methods mathematics COMP1921 FIN3309 VE311 ESCI 1908 FIT5057 COSC2446 COMPUTING Math 110A EE 567 MATH10098 CISC642 EE10134 ECE 321 MTH2051 SIT 281 EEE3314 ELEC273 COMP5930M CE322 ELEC270 EE5101 MCEN90008 MTSC 887 Probability ENGN4528 MAE 424 MA2552 EBU5303 CHEM10600 EBU6018 ELEC2103 ELEE 5200 CSE 535 COMP6528 IS6153 IS3S664 BUSANA 7003 QTS0109 ECO3152B ENGR 227 LCSCI4208 MATH32051 ECS776P MGMT3004 N1551 JAVA ROB-6213 TNS3113 BEEM062 COMP47670 4SSMN903 BST 833 COEN 146 BASC0003 CAS EC501 Networks CSIT940 ELEC372 ACCFIN5034 Math 3527 DATA4000 INT102 FND109 MMM071 ROB6213 ETF5650 ECON 445 QMSS 301 Econ 320 PHY 4390 FIN 111 FINN41615 ECS 116 MTH6158 IOE 552 STA 141B Math 413 ECON3016 Physics 4303 Numerical BFF5270 EECS 111 FINANCE 251 COMP814 MGSC 410 ECON0051 EBU5376 STAT6118 ECON 20110 COMU7000 average acceleration BUSI4412 BFF3121 BIT233 ECMM462 ETC3430 DESN2348 INTE2584 ENME502 MA3XJ INTE2627 MA2815 MA3AM L0101 Math 3MB3 CPCCBC4004 EEEM061 0502E220 APPLIED 6SSMN310 ENV 3310 PM2PCOL3 CONS6201 MATH502 FIT5003 EEEN313 MATH3001 MECH3460 ECON30009 COSC1252 Assembler COMP 3023 ENGG7601 COMP20005 COMP SCI 1103 ICSI 402 INF2C CS320 COMPSCI7064 ECE 3220 PROGRAMMAMING MIPS Computer Science 2413 AM3611 MSOA TE2101 FFT CMPSC-121 CSE 109 COMP SCI 7064 HCI CMP-5013A CCN2042 COMP3400 EEEN30052 CO4215 EE101 KIT107 CSE 421 CSE521 MCD4720 EECS 280 MIT 403 GRPC VG101 APiE EECE 1080C FIT3143 CSE 325 ISOM 3029 CS3307 State SIT255 Engineering MQF Modelling ACX3150 SIA322 ACCT90004 COMU7201 MGMT90018 WRIT1000 COMP3702 BISM1201 COMP7505 IDEA9106 CSE3SMT BFC5926 BFC5280 BFC5935 QBUS3330 ECON3700 IFN521 ECON2121 COSC2626 MA413 DDES9903 ENGG1300 COMP9337 CO2201 ACCT2002 ICM289 MMM054 MSBA-204 DIGM707 CSC2250 Econ312 EVAT python CAN404 LUBS2840 CSCI-GA.2250 COMSM0093 COMP281 AFM 333 IBUS1001 CP2501 Math 105A VE320 TEC101 SOCY2166 COM240 EFIM20033 ENGI4467 EFIM20034 EC 979 BE432 7EJ502 CSSE6400 COMP90025 Econ 312 ENG2031 ECO0006 EXERSCI 207 CAN209 ECO 137W FT312 ECO 2199 MATH32052 BDW4213 APS 360 SAS 6 CHME0017 PO92Q Economics 01AV MATH39512 MATH 5486 COMP8620 ISE102 ECON235 ECON 206 FI 345 PHAY0020 MATH0099 MATH0085 EFB106 PHY 134 MAE115 circuits Chemistry CSE12 MATH-UA 233 PHAS0038 COMP2140 CONMGNT 7049 COMP151101 ECON 485 ACX5903 ENGN6536 ARCH1162 Law PHAS0042 CCF1C03 MATH 351 Phys 139 AG942 MGT001371 APH106 ECON1051 MA4AM MA3Z7 NUMBER THEORY REST0006 SESS0026 STAT0023 TABL 2741 CMP4008Y EFIM10015 Acct 560 PSYC735 GEOG 5 ACCT3104 CSEC5616 PHIL1012 ITLS6111 ECON8022 5CCM226A compX123 MA4XJ R001 EWB LLP120 COMP643 ENG 103 ECON 23950 ACCT 207 FINC 616 Bio99 Math 137A Psychology BCPM0054 MATH20212 EC3450 CENV6141 PSY 205 CE335 CS365 Calculus ACSD INFT6800 MCIT INFO6030 PHYS 111LV H63EEJ ECON2151 ENGI 2181 MTRX1701 PLIT08009 CS240 Math 270 CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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