optimization part a solution
DESCRIPTION
optimisation solutionsTRANSCRIPT
BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANIFIRST SEMESTER 2012-13AAOC C222: Optimization
Comprehensive Examination (05/12/2012)
1 Asssignment
-19 -16 -14 -17-11 -13 -10 -11-16 -15 -9 -13-12 -13 -12 -15
⇔
0 3 5 22 0 3 20 1 7 33 2 3 0
⇔
0 3 2 22 0 0 20 1 4 33 2 0 0
⇔
0 2 1 13 0 0 20 0 3 24 2 0 0
Arjun-Facebook, Bhaskar-Intel, Chaitanya-Google and Dravid-IBM. Total earnings $5900
2 Transportation Problem and Dual
min c11x11 + c12x12 + c13x13 + c21x21 + c22x22 + c23x23
such that
x11 + x12 + x13 = a1
x21 + x22 + x23 = a2
x11 + x21 = b1
x12 + x22 = b2
x13 + x23 = b3
All x ≥ 0. Number of basic variables=2+3-1=4.
The dual problem is max w = u1a1 + u2a2 + v1b1 + v2b2 + v3b3
such that
u1 + v1 ≤ c11
u1 + v2 ≤ c12
u1 + v3 ≤ c13
u2 + v1 ≤ c21
u2 + v2 ≤ c22
u2 + v3 ≤ c23
all u and v unrestricted
3 Transportation Problem with NortWestCorner
Optimal allocation is as follows
155 25
5 80
a cij = ui + vj for basic xij. Therefore c11 = 0, c21 = 5, c22 = 8, c32 = 10, c33 = 15 andoptimal cost= $1475
b ui + vj − cij ≤ 0 for nonbasic xij. Therefore c12 ≥ 3, c13 ≥ 8, c23 ≥ 13, c31 ≥ 7
4. Branch and Bound Tree
5. Convexity and Concavity
No conclusion can be drawn from Hessian matrix Method. (Although, function is convex using
other methods).
IPP-0:
X1=15, x2=2.5, z=51.25
IPP-2:
X1=44/3, x2=3, z=51.5
IPP-1: X1=16, x2=2, z=53
Fathomed
IPP-4: X1=15, x2=3, z=52.5
Fathomed
IPP-3: x1=14, x2=4, z=52
Fathomed
X2 ≥ 3 X2 ≤ 2
X1 ≥ 15 X1 ≤ 14
Optimal solution
;0)()()()(
8300
3200
0040
0000
)(
;8;3;0;0
;3;2;0;0;0;0
;4;0;0;0;0;0
4321
2
4
2
34
2
24
2
14
2
43
2
2
3
2
23
2
13
2
42
2
32
2
2
2
2
12
2
41
2
31
2
21
2
2
1
2
xHxHxHxH
xH
x
f
xx
f
xx
f
xx
f
xx
f
x
f
xx
f
xx
f
xx
f
xx
f
x
f
xx
f
xx
f
xx
f
xx
f
x
f
6. QPP Maximize z = 2x1+3x2-2x12 , Subject to x1+4x2 ≤ 4; x1+x2 ≤ 2, and x1, x2≥ 0.
)()(
)2()44()232(),,,,,,,(
0
0
2
4
11
41
00
04)32(
222111
22121211
2
12121212121
22
11
2
1
2
1
2
1
21
2
1
xx
sxxsxxxxxssxxL
x
x
s
s
x
x
toSubject
x
xxx
x
xz
21
21
221
121
2221
11211
22112211
mod
var&,
24
44
34
24
0,0,0,0
RRrMinimize
simplexifiedAppplying
oiableartificialareRRWhere
sxx
sxx
R
Rx
basistheinaresIf
xsxx
QPPforConditionKKT
i
Initial table: Iteration-0,
From below table: x1, 1, 2 possible to enter. But x1 is entering into basis, R1 leaves
BV X1 X2 1 2 1 2 S1 S2 R1 R2 Solution
r 0 0 0 0 0 0 0 0 -1 -1 0
r 4 0 5 2 -1 -1 0 0 0 0 5
S1 1 4 0 0 0 0 1 0 0 0 4
S2 1 1 0 0 0 0 0 1 0 0 2
R1 4 0 1 1 -1 0 0 0 1 0 2
R2 0 0 4 1 0 -1 0 0 0 1 3
Iteration-1,
From below table: 1, 2,x2 possible to enter. But x2 is entering into basis, S1 leaves
BV X1 X2 1 2 1 2 S1 S2 R1 R2 Solution
r 0 0 4 1 0 -1 0 0 0 0 3
S1 0 4 -1/4 -1/4 1/4 0 1 0 -1/4 0 7/2
S2 0 1 -1/4 -1/4 1/4 0 0 1 -1/4 0 3/2
x1 1 0 1/4 1/4 -1/4 0 0 0 1/4 0 1/2
R2 0 0 4 1 0 -1 0 0 0 1 3
Iteration-2,
From below table: 1, 2 possible to enter. But 1 is entering into basis, R2 leaves
BV X1 X2 1 2 1 2 S1 S2 R1 R2 Solution
r 0 0 4 1 0 -1 0 0 0 0 3
x2 0 1 -1/16 -1/16 1/16 0 1/4 0 -1/16 0 7/8
S2 0 0 -3/16 -3/16 3/16 0 -1/4 1 -3/16 0 5/8
x1 1 0 1/4 1/4 -1/4 0 0 0 1/4 0 1/2
R2 0 0 4 1 0 -1 0 0 0 1 3
Iteration-3,
From below table: R1 and R2 are not in the basis, Hence solution is Optimal
BV X1 X2 1 2 1 2 S1 S2 R1 R2 Solution
r 0 0 0 0 0 0 0 0 0 0 0
x2 0 1 0 -3/64 1/16 -1/64 1/4 0 -1/16 0 59/64
S2 0 0 0 -9/64 3/16 -3/64 -1/4 1 -3/16 0 49/64
x1 1 0 0 3/16 -1/4 1/16 0 0 1/4 0 5/16
1 0 0 1 1/4 0 -1/4 0 0 0 1 3/4
The solution is l
x1 = 5/16, x2= 59/64, s2 = 49/64, 1=3/4, 2=0, s1=0, 1=0, 2=0.
z = 3.19.
7. Genetic Algorithm
Tabulate the solution as given below (Show all calculation up to three decimal places).
B.V D.V x f(x) F(x) Pi
=Fi/Fj
Cum Pi
Rand No
N MP H After PC
Mutation Pm
Next iteration
100 4 1.795 0.031 0.969 0.166 0.166 0.815 5 101 0 001 001 000
010 2 0.897 0.015 0.985 0.168 0.334 0.466 3 001 101 101 101
001 1 0.448 0.008 0.992 0.17 0.504 0.94 6 011 2 010 010 010
110 6 2.692 0.047 0.955 0.163 0.667 0.235 2 010 011 011 111
101 5 2.244 0.039 0.962 0.165 0.832 0.784 5 101 1 110 110 110
011 3 1.346 0.023 0.977 0.167 0.999 0.214 2 010 001 001 001
Sum =5.84
3 M 1 M 5M 2M 2M 1M
Or
B.V D.V x f(x) F(x) Pi
=Fi/Fj
Cum Pi
Rand No
N MP H After PC
Mutation Pm
Next iteration
100 4 1.795 1.776 0.360 0.139 0.139 0.815 5 101 0 001 001 000
010 2 0.897 0.859 0.537 0.207 0.346 0.466 3 001 101 101 101
001 1 0.448 0.458 0.685 0.264 0.61 0.94 6 011 2 010 010 010
110 6 2.692 2.692 0.270 0.104 0.714 0.235 2 010 011 011 111
101 5 2.244 2.234 0.309 0.119 0.833 0.784 5 101 1 110 110 110
011 3 1.346 1.317 0.431 0.166 0.999 0.214 2 010 001 001 001
Sum 2.592
3 M 1 M 5M 2M 2M 1M
Note: Marks Distribution
All the set of populations are considered then 3 Marks awarded up to F(x). Rest of the procedure is not
considered, therefore no marks will be awarded.
Feasible Binary variable sets (100, 010, 001, 110, 101, 011) are 6 listed in the above table and rest (000,
111) are violating the constraint x (0, : 3 Marks
Pi- 1 Marks, MP- 5Marks, Pc -2Marks, Pm- 2Marks, Population for the next iteration- 1Marks