Final project ECE 505-Applied Optimization for Engineers

Qiaoyu Zhang,A20331991

Weixiong Wang,A20332258

Background:

In the power market, both generators and loads will bid price, and different

generators and loads might bid for different price. According to the price that they bid,

we need to decide how many electricity can generator produce and how many loads

can be supply in order to maximize the societal surplus. In the meantime, we should

consider the limits in real power system. These limits include: limit of transmission

line, limit of generator and limit of load.

Statement of Objectives:

The optimization problem going to solve is about power market. As other kinds of

market, seller will bid a price for selling its produce. At the same time buyer will bid a

price that they can accept for buying the produce. Only when both the seller and buyer

agree the price, the deal can be done. And this price is called market clearing price. If

seller and buyer can’t reach the agreement, the seller have to keep his produces.

Design Procedure:

Now suppose that there is a three buses system including bus 1, bus 2 and bus 3.

Three buses are connected with each other with three transmission lines. Assume that

line1 (l1) is between bus 1 and bus 2. Line2 (l2) is connecting bus 2 and bus 3 and

line 3(l3) is between bus 3 and bus 1. Each of them has a maximize power flow limit

l1max, l2max and l3max. If power exceeds these limit, power system will become

unstable.

Figure 1

As showed in figure 1. There are two generator in this power system G1 and G2.

G1 is located at bus 1, and G2 is located at bus 2. The power generated by G1 is

called P1, and the power generated by G2 is called P2. What’s more, G1 and G2

has lower and upper limit because of the properties of generator. Generators can’t

generator too small power and can’t generate power that exceed its limit. For G1,

the limit is P1min and P1max and for G2, the limit is P2min and P2max.

Also, there are two load in the power system: L1 and L2. L1 is located at bus 3

and L2 is located at bus 2. In the realization, not all the load need to be supplied.

Some of the load can cut off for saving money. On the other hand, most of the

load have to be supplied. Then the lower limit of L1 is L1min and the upper limit

is L1max. The same as L2, we have L2min and L2max

Bus1 Bus2 Bus3

Generator G1 G2

Load L2 L1

When the power market is open, both generators and loads will bid prices.

Assume that G1 bid for price C1, G2 bid for price C2. And L1 bid for price B1,

and L2 bid for price B2. After biding price, we need to decide how much power

should be produce base on the objective function. In this case, we want to

maximize the societal surplus.

𝑠𝑜𝑐𝑖𝑒𝑡𝑎𝑙 𝑠𝑢𝑟𝑝𝑙𝑢𝑠 = (𝐿1𝐵1 + 𝐿2𝐵2) − (𝑃1𝐶1 + 𝑃2𝐶2)

What’s more, according to Kirchhoff’s Current Law (KCL), the total injection

should be equal to the total withdrawal. First, I assume that the power flow from

bus1 to bus 2 is f12, the power flow from bus2 to bus3 is f23, and the power flow

from bus3 to bus1 is f31. And then, we can get:

𝑏𝑢𝑠1: 𝑃1 + 𝑓31 = 𝑓12

𝑏𝑢𝑠2: 𝑃2 + 𝑓12 = 𝑓23 + 𝐿2

𝑏𝑢𝑠3: 𝑓23 = 𝐿1 + 𝑓31

Formulation:

As the definition of linear programming, the standard format is:

min𝑥∈𝑅𝑛

{𝑐𝑇𝑥|𝐴𝑥 = 𝑏, 𝐶𝑥 ≤ 𝑑}

We separate the situation into two different cases.

Case1:

In this case, we consider that every load is important, none of these load can be cut

off. It probably means that all the load is residential load or important industry load.

In this case, only generator can bid for the price.

(i)Variables:

First, we need to decide the variables for this problem. There are three kinds of

variables in this question: power of generators, loads and power flow. In this case, as

the load can’t be cut off, we can consider the load to be constant value. Therefore, the

variable should be:

𝑥 =

[ 𝑃1

𝑃2

𝑓12

𝑓23

𝑓31]

(ii). Objective function:

The objective of this problem is to maximize the societal surplus: (𝐿1𝐵1 + 𝐿2𝐵2) −

(𝑃1𝐶1 + 𝑃2𝐶2).

As the loads are constant in this case, maximizing the societal surplus equation should

become minimizing (𝑃1𝐶1 + 𝑃2𝐶2). And then, we can get the objective function:

𝑓(𝑥) = [𝐶1 𝐶2 0 0 0]

[ 𝑃1

𝑃2

𝑓12

𝑓23

𝑓31]

(iii). Equality constraints:

The Equality constraints for this problem is the KCL equation for every bus:

𝑏𝑢𝑠1: 𝑃1 + 𝑓31 = 𝑓12

𝑏𝑢𝑠2: 𝑃2 + 𝑓12 = 𝑓23 + 𝐿2

𝑏𝑢𝑠3: 𝑓23 = 𝐿1 + 𝑓31

[1 0 −1 0 100

1 1 −1 00 0 1 −1

]

[ 𝑃1

𝑃2

𝑓12

𝑓23

𝑓31]

= [0𝐿2

𝐿1

]

Therefore, we can get:

𝐴 = [1 0 −1 0 100

1 1 −1 00 0 1 −1

]

𝑏 = [0𝐿2

𝐿1

]

(iv). Inequality constraints:

As the loads are constant in this case, we only have generator and power flow limit.

And these limits are:

𝑃1𝑚𝑖𝑛 ≤ 𝑃1 ≤ 𝑃1𝑚𝑎𝑥

𝑃2𝑚𝑖𝑛 ≤ 𝑃2 ≤ 𝑃2𝑚𝑎𝑥

|𝑓12| ≤ 𝑙1𝑚𝑎𝑥

|𝑓23| ≤ 𝑙2𝑚𝑎𝑥

|𝑓31| ≤ 𝑙3𝑚𝑎𝑥

According this, we can get:

𝑃1 ≤ 𝑃1𝑚𝑎𝑥

𝑃2 ≤ 𝑃2𝑚𝑎𝑥

−𝑃1 ≤ −𝑃1𝑚𝑖𝑛

−𝑃2 ≤ −𝑃2𝑚𝑖𝑛

𝑓12 ≤ 𝑙1𝑚𝑎𝑥

𝑓23 ≤ 𝑙2𝑚𝑎𝑥

𝑓31 ≤ 𝑙3𝑚𝑎𝑥

−𝑓12 ≤ 𝑙1𝑚𝑎𝑥

−𝑓23 ≤ 𝑙2𝑚𝑎𝑥

−𝑓31 ≤ 𝑙3𝑚𝑎𝑥

Therefore, in the format of 𝐶𝑥 ≤ 𝑑, we can get:

C=

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

-1 0 0 0 0

0 -1 0 0 0

0 0 -1 0 0

0 0 0 -1 0

0 0 0 0 -1

And

𝑑 =

[ 𝑃1𝑚𝑎𝑥

𝑃2𝑚𝑎𝑥

𝑙1𝑚𝑎𝑥

𝑙2𝑚𝑎𝑥

𝑙3𝑚𝑎𝑥

−𝑃1𝑚𝑖𝑛

−𝑃2𝑚𝑖𝑛

𝑙1𝑚𝑎𝑥

𝑙2𝑚𝑎𝑥

𝑙3𝑚𝑎𝑥 ]

Case 2:

In this case, some of the load can be cut off. It means that both loads and generator

can bid for price, and loads can determine not to accept the electricity for the purpose

of saving money. In this case, some load might be less important than others and they

can using electricity at other time when the electricity price is lower. Also, people can

use computer program to bid the price automatically and let computer to decide

whether to use electricity or not. This is a new trend of power system called smart

grid.

(i)Variables:

We need to consider load as variables too, as load can change and not be decided yet.

Therefore the variables for this case is:

𝑥 =

[ 𝑃1

𝑃2

𝐿1

𝐿2

𝑓12

𝑓23

𝑓31]

(ii). Objective function:

The objective of this problem is to maximize the societal surplus: (𝐿1𝐵1 + 𝐿2𝐵2) −

(𝑃1𝐶1 + 𝑃2𝐶2).

As the loads in this case is not a constant but a variables in this case, we should

consider loads to be included in the objective function. And maximizing the (𝐿1𝐵1 +

𝐿2𝐵2) − (𝑃1𝐶1 + 𝑃2𝐶2) means minimizing (𝑃1𝐶1 + 𝑃2𝐶2) − (𝐿1𝐵1 + 𝐿2𝐵2). We can

get the objective function:

𝑓(𝑥) = [𝐶1 𝐶2 −𝐵1 −𝐵2 0 0 0]

[ 𝑃1

𝑃2

𝐿1

𝐿2

𝑓12

𝑓23

𝑓31]

(iii). Equality constraints:

According to the KCL, the equation of equality constraints should be same as before:

𝑏𝑢𝑠1: 𝑃1 + 𝑓31 = 𝑓12

𝑏𝑢𝑠2: 𝑃2 + 𝑓12 = 𝑓23 + 𝐿2

𝑏𝑢𝑠3: 𝑓23 = 𝐿1 + 𝑓31

However, the loads in this case are no longer constants. Therefore, we need to adjust

the equation when writing in the standard format.

[1 0 0 0 −1 0 10 1 0 −1 1 −1 00 0 −1 0 0 1 −1

]

[ 𝑃1

𝑃2

𝐿1

𝐿2

𝑓12

𝑓23

𝑓31]

= 0

In this case,

𝐴 = [1 0 0 0 −1 0 10 1 0 −1 1 −1 00 0 −1 0 0 1 −1

]

𝑏 = 0

(iv). Inequality constraints:

As loads are not constant in this case, we need to consider the limit of loads too.

Thought some of the loads can be cut off to save money, not all the loads can be cut

off. There are still large amount of loads need to have power supply all the time. Also,

not all the residential user of industry are willing to join the market. Therefore, the

loads should only change between the lower limit Lmin and upper limit Lmax.

Then we can get the inequality constraints:

𝑃1𝑚𝑖𝑛 ≤ 𝑃1 ≤ 𝑃1𝑚𝑎𝑥

𝑃2𝑚𝑖𝑛 ≤ 𝑃2 ≤ 𝑃2𝑚𝑎𝑥

𝐿1𝑚𝑖𝑛 ≤ 𝐿1 ≤ 𝐿1𝑚𝑎𝑥

𝐿2𝑚𝑖𝑛 ≤ 𝐿2 ≤ 𝐿2𝑚𝑎𝑥

|𝑓12| ≤ 𝑙1𝑚𝑎𝑥

|𝑓23| ≤ 𝑙2𝑚𝑎𝑥

|𝑓31| ≤ 𝑙3𝑚𝑎𝑥

According this, we can get:

𝑃1 ≤ 𝑃1𝑚𝑎𝑥

𝑃2 ≤ 𝑃2𝑚𝑎𝑥

−𝑃1 ≤ −𝑃1𝑚𝑖𝑛

−𝑃2 ≤ −𝑃2𝑚𝑖𝑛

𝐿1 ≤ 𝐿1𝑚𝑎𝑥

𝐿2 ≤ 𝐿2𝑚𝑎𝑥

−𝐿1 ≤ −𝐿1𝑚𝑖𝑛

−𝐿2 ≤ −𝐿2𝑚𝑖𝑛

𝑓12 ≤ 𝑙1𝑚𝑎𝑥

𝑓23 ≤ 𝑙2𝑚𝑎𝑥

𝑓31 ≤ 𝑙3𝑚𝑎𝑥

−𝑓12 ≤ 𝑙1𝑚𝑎𝑥

−𝑓23 ≤ 𝑙2𝑚𝑎𝑥

−𝑓31 ≤ 𝑙3𝑚𝑎𝑥

Therefore, in the format of 𝐶𝑥 ≤ 𝑑, we can get:

C=

1 0 0 0 0 0 0

0 1 0 0 0 0 0

0 0 1 0 0 0 0

0 0 0 1 0 0 0

0 0 0 0 1 0 0

0 0 0 0 0 1 0

0 0 0 0 0 0 1

-1 0 0 0 0 0 0

0 -1 0 0 0 0 0

0 0 -1 0 0 0 0

0 0 0 -1 0 0 0

0 0 0 0 -1 0 0

0 0 0 0 0 -1 0

0 0 0 0 0 0 -1

And

𝑑 =

[ 𝑃1𝑚𝑎𝑥

𝑃2𝑚𝑎𝑥

𝐿1𝑚𝑎𝑥

𝐿2𝑚𝑎𝑥

𝑙1𝑚𝑎𝑥

𝑙2𝑚𝑎𝑥

𝑙3𝑚𝑎𝑥

−𝑃1𝑚𝑖𝑛

−𝑃2𝑚𝑖𝑛

−𝐿1𝑚𝑖𝑛

−𝐿2𝑚𝑖𝑛

𝑙1𝑚𝑎𝑥

𝑙2𝑚𝑎𝑥

𝑙3𝑚𝑎𝑥 ]

Design Results:

Case 1

For case1, I set up a small instant:

The bidding price of generator is:

MWhC

MWhC

/$14

/$10

2

1

The limits of power generated by generators are:

MWPMW

MWPMW

10040

20030

2

1

The loads need to be supplied are:

𝐿1 = 150𝑀𝑊

𝐿2 = 60𝑀𝑊

The power flow limits are:

MWf

MWf

MWf

140

150

40

31

23

12

Then we can get:

31

23

12

2

1

]0001410[)(

f

f

f

P

P

xf

And

140

150

40

40-

30-

140

150

40

100

200

3

2

1

2

1

3

2

1

2

1

max

max

max

min

min

max

max

max

max

max

l

l

l

P

P

l

l

l

P

P

d

Substitute c, A, b, C and d matrix into the linprog function in matlab. We can get:

Optimization terminated.

X =

170.0000

40.0000

34.6446

14.6446

-135.3554

FVAL =

2.2600e+03

EXITFLAG =

1

OUTPUT =

iterations: 5

algorithm: 'interior-point'

cgiterations: 0

message: 'Optimization terminated.'

constrviolation: 2.6409e-10

firstorderopt: 4.0029e-07This result means:

MWf

MWf

MWf

MWP

MWP

3554.135

6446.14

6446.34

0000.40

0000.170

31

23

12

2

1

Form this result, G1 is the power provider,we can see that G1 generate power much

more than lower limit. The reason is the biding price C1 of G1 is lower than the price

C2 for G2. We want the G1 generate as much power as possible. And f31 is negative

value so the the power flow should from bus1 to bus3.

The minimum of generator cost is 2260$/h. In order to compare with case 2, I set the

price of loads the same as case 2:

𝐵1 = 9$/𝑀𝑊ℎ

𝐵2 = 17$/𝑀𝑊ℎ

Then we can get the maximum 𝑠𝑜𝑐𝑖𝑒𝑡𝑎𝑙 𝑠𝑢𝑟𝑝𝑙𝑢𝑠 = (9 × 150 + 17 × 60) − 2260 =

110$/ℎ

Case 2

For case2, I set up a small instant:

The bidding price of generator is:

𝐶1 = 10$/𝑀𝑊ℎ

𝐶2 = 14$/𝑀𝑊ℎ

The bidding price of loads are:

𝐵1 = 9$/𝑀𝑊ℎ

𝐵2 = 17$/𝑀𝑊ℎ

The limits of power generated by generators are:

30𝑀𝑊 ≤ 𝑃1 ≤ 200𝑀𝑊

40𝑀𝑊 ≤ 𝑃2 ≤ 100𝑀𝑊

The limits of load are:

130𝑀𝑊 ≤ 𝐿1 ≤ 150𝑀𝑊

40𝑀𝑊 ≤ 𝐿2 ≤ 60𝑀𝑊

The power flow limits are:

|𝑓12| ≤ 40𝑀𝑊

|𝑓23| ≤ 150𝑀𝑊

|𝑓31| ≤ 140 𝑀𝑊

Then we can get:

𝑓(𝑥) = [10 14 −9 −17 0 0 0]

[ 𝑃1

𝑃2

𝐿1

𝐿2

𝑓12

𝑓23

𝑓31]

And

𝑑 =

[

2001001506040150140−30−40−130−4040150140 ]

Substitute c, A, b, C and d matrix into the linprog function in matlab. We can get:

Optimization terminated.

X =

150.0000

40.0000

130.0000

60.0000

27.7910

7.7910

-122.2090

FVAL =

-130.0000

EXITFLAG =

1

OUTPUT =

iterations: 6

algorithm: 'interior-point'

cgiterations: 0

message: [1x24 char]

constrviolation: 8.6686e-12

firstorderopt: 1.1841e-08

This result means:

𝑃1 = 150.0000MW

𝑃2 = 40.0000MW

𝐿1 = 130.0000MW

𝐿2 = 60.0000MW

𝑓12 = 27.7910MW

𝑓23 = 7.7910MW

𝑓31 = −122.2090MW

Form this result, we can see that G1 generate power much more than lower limit. The

reason is the biding price C1 of G1 is lower than the price C2 for G2. We want the G1

generate as much power as possible. However, we can see that the lower limit of P2

force G1 to operate under 200MW. As G2 is bidding for a higher price, it can only

operate at it lower limit. And we can see that L2 is bidding for high price, so all of its

load can be supplied. For L1, it operate at the lower limit because it is bidding for a

lower price. We can see that all the variable is within the limit, and the limit of P2 and

L2 is activate.

The maximum of societal surplus is 140$/h.

Conclusion:

Form these two cases, we can see that the results are different. The reason is we can

cut of some low price load to reduce the total power demand. In this way we can

reduce the usage of generator to save more money. Therefore the societal surplus in

case 2 have 20$/h more than case1. Form this two case, we can see that changing the

limit and variables can change the result of optimization problem.

Appendix

Matalb code for case 1:

A = [1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

-1 0 0 0 0

0 -1 0 0 0

0 0 -1 0 0

0 0 0 -1 0

0 0 0 0 -1];

b = [200 100 40 150 140 -30 -40 40 150 140];

Aeq = [1 0 -1 0 1

0 1 1 -1 0

0 0 0 1 -1];

beq = [0 60 150];

f = [10

14

0

0

0];

[X,FVAL,EXITFLAG,OUTPUT] = linprog(f,A,b,Aeq,beq)

Matlab code for case 2:

f=[10;14;-9;-17;0;0;0];

Aeq=[1 0 0 0 -1 0 1;

0 1 0 -1 1 -1 0;

0 0 -1 0 0 1 -1];

beq=[0,0,0];

A=[1 0 0 0 0 0 0;

0 1 0 0 0 0 0;

0 0 1 0 0 0 0;

0 0 0 1 0 0 0;

0 0 0 0 1 0 0;

0 0 0 0 0 1 0;

0 0 0 0 0 0 1;

-1 0 0 0 0 0 0;

0 -1 0 0 0 0 0;

0 0 -1 0 0 0 0;

0 0 0 -1 0 0 0;

0 0 0 0 -1 0 0;

0 0 0 0 0 -1 0;

0 0 0 0 0 0 -1];

b=[200;100;150;60;40;150;140;-30;-40;-130;-40;40;150;140];

[X,FVAL,EXITFLAG,OUTPUT] =linprog(f,A,b,Aeq,beq)