Problem solving using GA -1

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Consider the given problem of the two bar pendulum

The total potential for the two bar pendulum is written as

Õ= - P [(L₁ Sin q₁ + L₂ Sin q₂ )]- (W₁ L₁/2)Cos q₁ – W₂ [L₂ Cos q₂ + L₁ Cos q₁]

Putting the values for P, W₁,W₂, L₁, L₂ we get:

Õ(q_1, q₂)= -4Sinq₁ - 6Cosq₁-4 Sinq₂- 2 Cos q₂ …………….. (1)

where

0<=q_1; q₂<=90

The equilibrium configuration is the one which makes Õ a minimum.

Hence equation 1 gives the objective function. Let Õ(q_1, q₂)= f

Since the objective function f is negative, instead of minimizing it ,we maximize -f=f ^’.

The max value of f=8 when q_1, q₂=0, hence the fitness function F is taken as

F=f’-7 = -f-7

Since there are 2 unknowns q_{1 ,} q₂ in the problem, we use 4 bit binary string for each unknown.

Accuracy = X^U-X^L = 90/15 = 6⁰

2⁴-1

The binary coding and corresponding angles are given as follows:

S.No.	Binary Coding	Angle
1	0000	0
2	0001	6
3	0010	12
4	0011	18
5	0100	24
6	0101	30
7	0110	36
8	0111	42
9	1000	48
10	1001	54
11	1010	60
12	1011	66
13	1100	72
14	1101	78
15	1110	84
16	1111	90

Step 1 :

Randomly generate 8 populations with 8 bit strings as follows:

Population No.	Population	q1	q2	F=-f-7
1	0000 0000	0	0	8-7=1
2	0010 0001	12	6	2.1
3	0001 0101	6	30	3.11
4	0010 1000	12	48	4.01
5	0110 1010	36	60	4.66
6	1110 1000	84	48	1.91
7	1110 1101	84	78	1.93
8	0111 1100	42	72	4.55

The average fitness from the above table

F_av = (1+2.1+3.11+4.01+4.66+1.91+1.93+4.55)/8 =2.908

Step 2:

The average fitness from the above table

F_av = (1+2.1+3.11+4.01+4.66+1.91+1.93+4.55)/8 =2.908

Now we select the mating pool depending on F/F_av (rank selection). Two Populations with highest F/F_av will have 2 copies each in the mating pool while two populations with lowest F/F_av will die out. Others will get 1 copy each in the mating pool.

Population No.	Population	F=-f-7	F/Fav	Count
1	0000 0000	8-7=1	0.38	0
2	0010 0001	2.1	0.812	1
3	0001 0101	3.11	1.203	1
4	0010 1000	4.01	1.55	1
5	0110 1010	4.66	1.802	2
6	1110 1000	1.91	0.738	0
7	1110 1101	1.93	0.746	1
8	0111 1100	4.55	1.76	2

Mating pool:

Population No.	Population	F=-f-7	F/Fav
1	0010 0001	2.1	0.812
2	0001 0101	3.11	1.203
3	0010 1000	4.01	1.55
4	0110 1010	4.66	1.802
5	0110 1010	4.66	1.802
6	1110 1101	1.93	0.746
7	0111 1100	4.55	1.76
8	0111 1100	4.55	1.76

Step 3:

Crossover

It proceeds in 3 steps:

1. Select two strings at random.
2. Select crossover sites at random
3. Swap bits between the crossover sites.

For the above mating pool:

Pop. No.	Population	Mate with	CS1	CS2	Population after crossover
1	0010 0001	5	2	6	0010 1001
2	0001 0101	6	1	5	0110 1101
3	0010 1000	7	4	8	0010 1100
4	0110 1010	8	4	6	0110 1110
5	0110 1010	1	2	6	0110 0010
6	1110 1101	2	1	5	1001 0101
7	0111 1100	3	4	8	0111 1000
8	0111 1100	4	4	6	0111 1000

Step 4: Mutation

Select bits with a small probability, here with a probability 0.031

i.e only 2 bits out of a total of 64 (8 populations X 8 bit wide) bits will be mutated.

Hence randomly select bit no.21 and bit no. 62. The bits of the crossed over populations in these locations are flipped.

Pop. No.	Population after crossover	Random bits for mutation	Population after Mutation = population for next generation
1	0010 1001		0010 1001
2	0110 1101		0110 1101
3	0010 1100	21	0010 0100
4	0110 1110		0110 1110
5	0110 0010		0110 0010
6	1001 0101		1001 0101
7	0111 1000		0111 1000
8	0111 1000	62	0111 1100

Now repeat the process on the next generation population.

Step 1:

Find q₁ and q₂ from the initial table

S.No.	Binary Coding	Angle
1	0000	0
2	0001	6
3	0010	12
4	0011	18
5	0100	24
6	0101	30
7	0110	36
8	0111	42
9	1000	48
10	1001	54
11	1010	60
12	1011	66
13	1100	72
14	1101	78
15	1110	84
16	1111	90

Population after Mutation = population for next (2nd ) generation	q1	q2	F =-f-7
0010 1001	12	54	4.11
0110 1101	36	78	4.53
0010 0100	12	24	3.15
0110 1110	36	84	4.4
0110 0010	36	12	3.00
1001 0101	54	30	3.49
0111 1000	42	48	4.44
0111 1100	42	72	4.55

F_av = (4.11+4.53+3.15+4.4+3+3.49+4.44+4.55)/8 = 3.95

The average fitness of the new set is better than the previous set(which was 2.9)

Step 2:

The average fitness from the above table

F_av = (4.11+4.53+3.15+4.4+3+3.49+4.44+4.55)/8 = 3.95

Now we select the mating pool depending on F/F_av (rank selection). Two Populations with highest F/F_av will have 2 copies each in the mating pool while two populations with lowest F/F_av will die out. Others will get 1 copy each in the mating pool.

Population (2nd ) generation	F =-f-7	F/Fav	Count
0010 1001	4.11	1.04	1
0110 1101	4.53	1.14	2
0010 0100	3.15	0.796	0
0110 1110	4.4	1.11	1
0110 0010	3.00	0.759	0
1001 0101	3.49	0.883	1
0111 1000	4.44	1.12	1
0111 1100	4.55	1.15	2

Mating pool:

Population (2nd ) generation	F =-f-7	F/Fav
0010 1001	4.11	1.04
0110 1101	4.53	1.14
0110 1101	4.53	1.14
0110 1110	4.4	1.11
0111 1100	4.55	1.15
1001 0101	3.49	0.883
0111 1000	4.44	1.12
0111 1100	4.55	1.15

Repeat Crossover and Mutation

Step 3:

Crossover

It proceeds in 3 steps:

4. Select two strings at random.
5. Select crossover sites at random
6. Swap bits between the crossover sites.

For the above mating pool:

Pop. No.	Population (2nd generation)	Mate with	CS1	CS2	Population after crossover
1	0010 1001	5	2	6	0011 1100
2	0110 1101	6	1	5	0001 0101
3	0110 1101	7	4	8	0110 1000
4	0110 1110	8	4	6	0110 1110
5	0111 1100	1	2	6	0110 1000
6	1001 0101	2	1	5	1110 1101
7	0111 1000	3	4	8	0111 1101
8	0111 1100	4	4	6	0111 1100

Step 4: Mutation

Select bits with a small probability, here with a probability 0.031

i.e only 2 bits out of a total of 64 (8 populations X 8 bit wide) bits will be mutated.

Hence randomly select bit no.21 and bit no. 62. The bits of the crossed over populations in these locations are flipped.

Pop. No.	Population after crossover	Random bits for mutation	Pop. after Mutation = population for next (3rd )generation
1	0011 1100		0011 1100
2	0001 0101		0001 0101
3	0110 1000	21	0110 0000
4	0110 1110		0110 1110
5	0110 1000		0110 1000
6	1110 1101		1110 1101
7	0111 1101		0111 1101
8	0111 1100	62	0111 1000

Again repeat the process on this new population

Finally after many iterations you will get a value that is close to (34,66) which is the optimum value of angles for stabilizing the system under consideration