ece 760 assignmen 2 german escobar pdf

7/28/2019 ECE 760 Assignmen 2 German Escobar PDF

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March 6, 2009

ECE760 Electricity MarketsAssignment #2

German Escobar Zuluaga

ID 20284730

Problem1

The Ybus for the system is given by

Ybus=[10 j 0 10 j

0 10 j 10 j10 j 10 j 20 j]

Bus 1 is the slack bus. Bus 2 is a PV bus. Bus 3 is a PQ bus. We will use the Newton-Raphson method

in the solution, noting that V2 and V3 are given, and therefore we can simplify the computation. Theunknown variables are 3 and V3. Thus, the Jacobian will be a 3X3 matrix.

Defining the unknown vector

x i=[ iVi] the three unknown variables to be solved are x i =[ 3V3]Since there are N = 3 buses for problem 1, powerflow equations constitute 2(N>)However, there is one voltage-controlled bus, bus 2. Therefore V2and the equation forQ2(x) andP2(x)could be eliminated,since is 2 ,constant.

P3=V3V1Y31 sin 31V3V2Y32 sin32

Q3=[V3V1 Y31 cos 31V3 V2 Y32 cos 32Y33V32]

Putting in the known values ofP3 = PD3 = 0.9 and Q3 = QD3 = 0.4, 2= 3 = 0; V2 = 1.05 and V1= 1.0, result:

0.9=10V3

sin310.5V

3sin

3

0.4=[10V3 cos 310.5V3cos 320 V32]

In using Newton-Raphson the next step is to put the equations in the form f(x = 0. The simplest wayis to subtract the left sides from the right sides, or viceversa. It does not matter which we do: thesolution algorithm will be identical in either case, we get:

f1x =20.5V

3sin

30.9=0

f2x =20.5 V3 cos 320V320.4=0


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The first component ofx is 3,the second is V3; with that understanding we will not always use thex1,x2 notation.

Next, we find the Jacobian matrix,

Jx =

[ f1

x1

f1

x2 f2 x1

f2 x2 ]

=[20.5V3cos3 20.5sin320.5 V3sin3 20.5cos340 V3]

In the 2X2 case it is so easy to find the inverse that we will solve using:

xv1=xv[Jxv ]1 fxv (1.1

Starting with a flat profile (i.e 3,= 0 and V3 = 1.0, we get

J0=[20.5 0

0 19.5] J01=[

0.04878 00 0.05128] fx

0=[0.90.1]

Then, using (1.1 yields

x1=[01][0.04878 00 0.05128][ 0.90.1]=[0.043900.005128 ]

Continuing to the next iteration gives us

J1

=[20.58531 0.899660.90428 19.72495 ]

J1

1

=[0.04868 0.00222

0.00012 0.05080]fx1=[0.004280.02041 ]

Note that x1 is really a very good estimate. We want f(x = 0 and already f(x1 is almost zero.Continuing the next iteration we have

x2=[0.043901.005128 ][ 0.04868 0.002220.00012 0.05080][0.004280.02041 ]=[0.043741.00409 ]

Noting that the change between x1 and x2 is less than 0.00005, we stop the iteration. Alternatively,we can compute f(x2. We get

fx2=[0.000050.00022]

Which is substantially small.


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Thus, the objective of finding the x for which f(x = 0 seems very well met. Finally, to complete theproblem, we findP1 andP2:

PG1=P1=V1V2Y12 sin1 2V1V3Y13 sin13

=010.0409sin 0.04374=0.439

QG1=Q1=[V1V2Y12cos 12V1V3Y13cos 13V12Y11]

=[ 010.0409 cos0.0437410]=0.0313

PG2=P2=V1V2Y21 sin 21 V2V3Y23 sin 23

=010.5429sin 0.04374=0.461

QG2=Q2=[V1 V2 Y21cos 21V2V3Y23cos 23V22Y22]

=[ 010.5429 cos0.0437411.025]=0.4922

This completes the problem

Problem 2


Ybus=[20 j 10 j 10 j10 j 10 j 010 j 0 10 j]

Bus 1 is the slack bus. Bus 3 is a PV bus. Bus 2 is a PQ bus. We will use the Newton-Raphson methodin the solution, noting that V3 is given, and therefore we can simplify the computation. The unknownvariables are 2, 3 and V2. Thus, the Jacobian will be a 3X3 matrix.

Defining the unknown vector

x i=[ iVi]

the three unknown variables to be solved are x i =

[

2

3V

2]Since there are N = 3 buses for problem 6.25, (6.6.2 and (6.6.3 constitute 2(N>)However, there is one voltage-controlled bus, bus 3. Therefore V3and the equation forQ3(x) could beeliminated.


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yk=Pk=Pkx =Vkn=1

N

Ykn Vncos knkn (6.6.2

ykN=Q k=Q kx=Vkn=1

N

Ykn Vn sin knkn (6.6.3

k= 2, 3, ... ,N

Using (6.6.2 and (6.6.3 to compute

y i =[PiQi]=[PP[x i]QQ [x i]]P20=P2P2 x=P2V20 {Y21V1 cos[2 01021]

Y22V

2cos[

22]Y

23V

3cos[

20

30

23]

(1.1

P30=P3P3x =P3V30{Y31V1 cos [301031 ]

(1.2Y

32V

2cos [

30

20

32]Y

33V

3cos[

33]

Because we know V3, we can eliminate the equation involving Q3(x). We need the equation involvingQ2(x). to solve forV2, :Q20 =Q2Q2x =Q2V20 {Y21V1 sin[201021]

(1.3Y

22V

2sin [

22]Y

23V

3sin [

20

30

23]

Step 1 compute y 0

P20=11.0 [10 1.0cos 90101.0cos 9001.0cos 90] P22=1

P30=0.41.0 [10 1.0cos 900 1.0 cos9010 1.0cos 90] P

30=0.4

Q20 =0.41.0 [101.0 sin9010 1.0 sin9001.0sin 90] Q20 =0.4

Step 2 compute the 3X3 Jacobian matrix

J122=P22

=V2[Y21V1sin 2121Y23V3 sin 2323]

J123

=P

2

3=V

2Y

23V

3sin

2

3

23


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J132=P32

=V3Y32V2 sin 3232

J133=P33

=V3[Y31 V1sin 3131Y32V2 sin3232]

J222=P2V2

=V2Y22 cos22Y21 V1 cos 2121Y22V2 cos 2222Y23V3 cos 2323

J232=P3V2

=V3Y32cos 3232

J322=Q22

=V3[Y31V1 cos 3131Y32V2cos 3232]

J323=

Q2

3 =V2 Y23 V3 cos2323

J422=Q2V2

=V2Y22sin22Y21V1 sin2121Y22 V2 sin 2222Y23 V3sin 2323

Therefore,

J122=1.0[10 1.0sin/20 1.0 sin/2]=10

J123

=1.00 1.0sin /2=0

J132=1.00 1.0sin /2=0

J133

=1.0[ 101.0 sin/ 201.0 sin/2]=10

J222=1.010cos /2101.0cos /210 1.0cos /201.0cos/2=0

J232

=1.0 01.0 cos/ 2=0

J322

=1.0[ 101.0cos/201.0 cos / 2]=0

J323

=1.00 1.0cos /2= 0

J422=1.00sin /2101.0sin /210 1.0sin / 2101.0sin /2=10

The Jacobian matrix J(0:

J0=[10 0 0

0 10 0

0 0 10] (1.4 [10 0 0

0 10 0

0 0 10][20

30

V20]=[

10.40.4]


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Solving

20=110

=0.1 20=

0.410

=0.04 V2

0=0.4

10=0.04

Step 4

x 1=x 0x 0 =[0

0

1][

0.10.040.04]=[

0.10.04

0.96]

We proceed to the next iteration using the new values 2>3>2 = 0.96.Substituting in (1.1, we get P2(x

1)>12>>>)>Similarly, using (1.2 and (1.3, we get an updated mismatch vector:

[P20P30

Q20]1

=[0.04160.00011

0.06396]Note: In one iteration the mismatch vector has been reduced by a factor of about 10. Calculating J1, wefind that

J1=[9.55204 0 0

0 9.992 0.399890 0.39989 10.008 ] (1.5

The matrix should be compared with J0 in (1.4. It has not changed much. The off-diagonal matricesare not longer zero, but their elements are small compared to the terms in the matrix diagonal. Thematrix diagonal itself has not changed much. The same observation is true about the inverses. Theupdated inverse is

J11=[

0.10470 0 0

0 0.1002 0.004

0 0.004 0.1001] (1.6

Comparing (1.6 with inverse ofJ

0

, we do not see much change. For the second iteration we find that

x 2=x 1x 1 =[0.10.04

0.96][

0.004360.000010.00639]=[

0.104360.03973

0.9536]

To continue to the third iteration we need to find the current mismatch


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[P20

P30

Q20]

2

=[0.006630.002800.00941]

x 3=x 2x 2 =

[0.104360.03973

0.9536 ]

[0.00070.000280.00094 ]

=

[0.105060.040050.95265 ]

To continue to the fourth iteration we need to find the current mismatch

[P20

P30

Q20]

2

=[0.000990.00039

0.00145]The mismatch has been reduced by at least a factor of 200 and is small enough. On that basis we couldstop here. So we stop with the values 2>3>2 = 0.95265.

It remains to calculate the generator outputs SG1 and QG2 using the calculated values of 2, 3, and V2.Ge get:

PG1=P1=V1V2Y12 sin1 2V1V3Y13 sin13

=9.5265sin 0.105069.5265sin 0.04005=1.38045

QG1=Q1=[V1V2Y12cos 12V1V3Y13cos 13V12

Y11]

=[ 9.5265 cos0.1050610 cos0.0400520 ]=0.534

QG2=Q2=[V1V2Y21 cos 21V2V3Y23 cos 2 3V22Y22 ]

=[9.5265cos0.105060cos0.040059.0754]=0.3986

This completes the problem.Problem 3

PG2 = 0.3V2 = 0.95P2 = 0.3P3 = >0.5Q3 = >0.2


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Defining

x i=[ iVi] the three unknown variables to be solved are x i =[

2

3

V3

]Since there are N = 3 buses for problem 1, (1.1 and (1.2 constitute 2(N>)there is one voltage-controlled bus, bus 2. Therefore V2and the equation forQ2(x) could be eliminated.

yk=Pk=Pkx =Vkn=1

N

Ykn Vncos knkn (1.1

ykN=Q k=Q kx=Vkn=1

N

Ykn Vn sin knkn (1.2

k= 2, 3, ... ,N

Using (1.1 and (1.2 to compute

y i =[PiQi]=[PP[x i]QQ [x i]]P20=P2P2 x=P2V20 {Y21V1 cos[2 01021]

Y22V

2cos[

22]Y

23V

3cos[

20

30

23]

P30=P

3P

3x =P

3V

30Y

31V

1cos [

3 0

10

31]

Y32V2cos [302032]Y33 V3 cos[33]

Q30=Q

3Q

3x =Q

3V

30[Y

31V

1sin [

30

10

31] ]

Y33V

3sin [

33]Y

32V

2sin[

30

20

32]

Step 1 compute y 0

P20=0.30.95[ 25.39681.0 cos 90 24.24570.95cos 90 21.07111.0cos 90]

P22=0.3

P30=0.51.0 [11.11111.0 cos 9010 1.0cos 9021.07111.0 cos 90]

P30=0.5


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Q30=0.21.0[11.11111.0sin 9021.07111.0sin 9021.0711 1.0sin 90]Q

20 =0.2


Ybus=

[25.37683 j 14.2857 j 11.1111 j

14.2857 j 24.26571 j 1011.1111 j 10 21.09111 j]

Bus 1 is the slack bus. Bus 2 is a PV bus. Bus 3 is a PQ bus. We will use the FDLF method in thesolution, noting that V2 is given, and therefore we can simplify the computation. The unknownvariables are 2, 3 and V3.

B=[24.26571 j 1010 21.09111 j]Aplying Bv= Pxv

[24.26571 j 1010 21.09111 j][23]0

=[P20.95

P3V3

]0

and aplying BVv= Qxv , with the V2 equation stripped away, becomes

21.09111

V

3

v

=

Q3

V3

v

In this simple case we can use matrix inversion to get the explicit iteration formulas,

[23]0

=[0.05122 0.024280.02428 0.05893][P20.95

P3V3

]0

V3v=0.04741Q 3V3

v

The results of the iterations are tabulated in Table 1 starting with initial values of2= 3 = 0 and V3 =1.0. The tabulated values of 2and 3 are in radians.


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Iterationnumber

2 3 V3 P2 P3 Q3

0 0 0 1 0.32 -0.5 -0.681 0 -0.02 0.97 0.01 -0.03 -0.01

2 0 -0.02 0.97 0 0 0

Problem 4First, we will assume that each utility operates independently; that is, each will supply its own loadfrom its own generation. The results of an independent economic dispatch are given here.

Utility 1 P1 =400 MW = 90 $/MWh

Operating cost, utility 1 = 21,605.1 $/h

Utility 2 P2 = 250 MW = 42.5 $/MWh


Utility 3 P3 = 350 MW = 37.1 $/MWh


Total operating cost for all utilities = 21,605.1 + 6,877.9 + 8,086.6 $/h = 36,569.6

Now suppose the three utilities are interconnected by several transmission circuits such that the three

utilities may be thought of, and operated, as one system. If we now dispatch them as one system,considering the loads in each utility to be the same as just shown, we get a different dispatch for theunits. (using results from GAMS

LOWER LEVEL UPPER MARGINAL

p1 50.000 177.456 250.000 -9.63E-10p2 60.000 312.018 400.000 EPSp3 100.000 510.526 600.000 .

LOWER LEVEL UPPER MARGINAL

---- VAR cost -INF 30850.762 +INF .

P1=177.456MW

P2=312.018MW

P3=510.526MW

Total generation for the entire system = 1000.0 MW

= 49.942 $/MWh


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Operating cost, utility 1 = 6,033.46 $/hOperating cost, utility 2 = 9,744,44 $/hOperating cost, utility 3 = 15,072.86 $/h

Total operating cost = 30,850.76 $/h

b. Interchange power = 62.018 MW from utility 2 to utility 1

Interchange power = 160.526 MW from utility 3 to utility 1

Note that the utility 1 is now generating less, than when it was isolated, and utilities 2 and 3 aregenerating more. If we ignore losses, we can see that the change in generation in each utilitycorresponds to the net powerflow over the interconnecting circuits. This is called the interchangepower. Note also that the overall cost of operating both systems is now less than the sum of the costs tooperate the utilities when each supplied its own load.

Let's price the sale at the cost of generation plus one-half the savings in operating costs of thepurchaser. This splits the savings equally between the three operating utilities.

c. First, lets see what the cost would be under a split-savings pricing policy if the interchangeagreements were made with transaccion 2-1 first, then transaccion 3-1.

Utility 1Gen.(MW)

Utility 1Cost($/h)

Utility 2Gen.(MW)

Utility 2Cost($/h)

Utility 3Gen.(MW)

Utility 3Cost($/h)

Start 400 21605.1 250 6877.9 350 8086.6After tran 2-1 337.98 16369.64 312.02 9744.44 350 8086.6After tran 3-1 177.46 6033.46 312.02 9744.44 510.53 15072.86

Transaction 2-1: Saves utility 1 5,235.46 $Costs utility 2 2,866.54 $

After splitting savings, utility 1 pays utility 2 2,866.54 + 1,184.46 = 4,051 $


After splitting savings, utility 1 pays utility 3 6,986.26 + 1,674.96 = 8,661.22 $

Summary of payments:

Utility 1 pays a net 12,712.22 $Utility 2 receives 4,051 $Utility 3 receives 8,661.22 $

Now let the transactions be costed assuming the same split savings pricing policy but with theinterchange agreements made with transaction 3-1 first, then transaction 2-1


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Utility 1Gen.(MW)

Utility 1Cost($/h)

Utility 2Gen.(MW)

Utility 2Cost($/h)

Utility 3Gen.(MW)

Utility 3Cost($/h)

Start 400 21605.1 250 6877.9 350 8086.6After tran 3-1 239.47 9476.93 250 6877.9 510.53 15072.86After tran 2-1 177.46 6033.46 312.02 9744.44 510.53 15072.86


After splitting savings, utility 1 pays utility 2 6,986.26 + 2570.96 = 9,557.21 $


After splitting savings, utility 1 pays utility 3 2,866.54 + 288.47 = 3,155 $

Summary of payments:

Utility 1 pays a net 12,712.22 $Utility 2 receives 3,155 $Utility 3 receives 9,557.21 $

Except for utility 1, the payments for the interchanged power are different, depending on the orderin which the transactions were carried out. If transaction 2-1 were carried out first, utility 2 would beselling power to utility 1 at a lower incremental cost than if transaction 3-1 were carried out first.Obviously, it would be to a seller's (utility 2 in this case advantage to sell when the buyer's (utility 1incremental cost is high.

When several two-party interchange transactions are made, the pricing must follow the propersequence. In this problem, the utility supplying the energy receives more than its incrementalproduction costs no matter which transaction is costed initially. The rate that the other two utilities payper MWh are different and depend on the order of evaluation. These differences may be summarizedas follows in terms of $/MWh.

Cost Rates ($/MWh)2-1 costed first 3-1 costed first

1 pays 12712.22 12712.222 receives 4051 3155

3 receives 8661.22 9557.21

d. The optimal allocations savings based on the Shapley Value are:


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Coalition utilities

Coalition Savings

Independent 01-2 5235.462-3 0

1-3 10336.181-2-3 15571.64

Shapley value C= s1!ns !n !

Utility 1

Coalition Marginal Contribution Weight 1 0 0 = 0 11! 31 !

3 !=

1

3

1-2 5235.46 0 = 5235.46 21! 32!

3 !=

1

6

1-3 10336.18 0 = 10336.18 1

6

1-2-3 15571.64 0 =15571.64 1

3

Shapley value = 1=1

30

1

65235.46

1

610336.18

1

315571.64

1=$9508.52/h

Utility 2

Coalition Marginal Contribution Weight 2 0 0 = 0 11! 31!

3 !=

1

3

2-1 5235.46 0 = 5235.46 21! 32!

3 !=

1

6

2-3 0 1

6

2-1-3 15571.64 5235.46 =10336.18 1

3

Shapley value = 2=1

30

1

65235.46

1

60

1

310.336.18

2=$ 4317.97/h


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Utility 3

Coalition Marginal Contribution Weight

3 0 0 = 0 11! 31!

3 !=

1

3

3-1 10336.18 0 = 10336.18 21! 3 2!

3 ! =

1

6

3-2 0 0 = 0 1

6

3-1-2 15571.64 10336.18 =5235.46 1

3

Shapley value = 3=1

30

1

610336.18

1

60

1

35235.46

1=$9508.52/h

2=$ 4317.97/h

3=$3467.85/h

Problem 5

In this problem, three power systems have sent, their buy/sell offers to the broker. In the table thatfollows, these are tabulated and the maximum pool savings possible is calculated

UtilitiesSellingEnergy

IncrementalCost ($/MWh)

MWhfor Sale

Seller's TotalIncrease in Cost ($)

3 38.7 20 7743 40.3 20 806

3 41.9 20 8383 43.5 20 8702 44.9 20 8983 45.1 20 9023 46.7 20 9342 47.3 20 9463 48.3 20 9662 49.7 20 9943 49.9 20 998

UtilitiesBuying

Energy

Decremental

Cost ($/MWh)

MWh

for Purchase

Buyer's Total

Decrease in Cost ($)1 86.4 20 17281 82.8 20 16561 79.2 20 15841 75.6 20 15121 72 20 14401 68.4 20 13681 64.8 20 12961 61.2 20 1224


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1 57.6 20 11521 54 20 10801 50.4 20 1008

Net pool savings = $ 15048 $ 9926 = $ 5122


16/21

The broker sets up transactions as shown in the following table

Transaction Savings Computation

TotalTransactionSavings ($)

3 sells 20 MWh to 1 20 MWh (86.4 38.7 $/MWh = 9543 sells 20 MWh to 1 20 MWh (82.8 40.3 $/MWh = 850

3 sells 20 MWh to 1 20 MWh (79.2 41.9 $/MWh = 746

3 sells 20 MWh to 1 20 MWh (75.6 43.5 $/MWh = 642

2 sells 20 MWh to 1 20 MWh (72 44.9 $/MWh = 5423 sells 20 MWh to 1 20 MWh (68.4 45.1 $/MWh = 4663 sells 20 MWh to 1 20 MWh (64.8 46.7 $/MWh = 3622 sells 20 MWh to 1 20 MWh (61.2 47.3 $/MWh = 2783 sells 20 MWh to 1 20 MWh (57.6 48.3 $/MWh = 1862 sells 20 MWh to 1 20 MWh (54 49.7 $/MWh = 863 sells 20 MWh to 1 20 MWh (50.4 49.9 $/MWh = 10

Total 5122

The rates and total payments are easily computed under the split-savings arrangement. These areshown in the following table

TransactionPrice

($/MWh)Total

Costs ($)

3 sells 20 MWh to 1 62.55 = 1251

3 sells 20 MWh to 1 61.55 = 1231

3 sells 20 MWh to 1 60.55 = 12113 sells 20 MWh to 1 59.55 = 1191

2 sells 20 MWh to 1 58.45 = 1169

3 sells 20 MWh to 1 56.75 = 1135

3 sells 20 MWh to 1 55.75 = 1115

2 sells 20 MWh to 1 54.25 = 1085

3 sells 20 MWh to 1 52.95 = 1059

2 sells 20 MWh to 1 51.85 = 1037

3 sells 20 MWh to 1 50.15 = 1003

Total 12487

Utility 3 receives $9196 from Utility 1 and Utility 2 receives $3291. Note that each participantbenefits: Utility 3 receives $2108 above its costs; Utility 2 receives $453 above its costs; and, Utility 1saves $2561.

b. Considering a transmission constraint on line 1-3 of 60 MW: In this problem, three power systemshave sent, their buy/sell offers to the broker. In the table that follows, these are tabulated and themaximum pool savings possible is calculated


17/21

UtilitiesSellingEnergy

IncrementalCost ($/MWh)

MWhfor Sale

Seller's TotalIncrease in Cost ($)

3 38.7 20 7743 40.3 20 806

3 41.9 20 8382 44.9 20 8982 47.3 20 9462 49.7 20 9942 52.1 20 10422 54.5 20 10902 56.9 20 1138

UtilitiesBuyingEnergy

DecrementalCost ($/MWh)

MWhfor Purchase

Buyer's TotalDecrease in Cost ($)

1 86.4 20 1728

1 82.8 20 16561 79.2 20 15841 75.6 20 15121 72 20 14401 68.4 20 13681 64.8 20 12961 61.2 20 12241 57.6 20 1152

Net pool savings = 12960 $ 8526 $ = 4434 $

The broker sets up transactions as shown in the following table

Transaction Savings Computation

TotalTransactionSavings ($)

3 sells 20 MWh to 1 20 MWh (86.4 38.7 $/MWh = 9543 sells 20 MWh to 1 20 MWh (82.8 40.3 $/MWh = 8503 sells 20 MWh to 1 20 MWh (79.2 41.9 $/MWh = 7462 sells 20 MWh to 1 20 MWh (75.6 44.9 $/MWh = 6142 sells 20 MWh to 1 20 MWh (72 47.3 $/MWh = 4942 sells 20 MWh to 1 20 MWh (68.4 49.7 $/MWh = 374

2 sells 20 MWh to 1 20 MWh (64.8 52.1 $/MWh = 2542 sells 20 MWh to 1 20 MWh (61.2 54.5 $/MWh = 1342 sells 20 MWh to 1 20 MWh (57.6 56.9 $/MWh = 14

Total 4434


18/21

The rates and total payments are easily computed under the split-savings arrangement. These areshown in the following table

TransactionPrice

($/MWh)Total

Costs ($)3 sells 20 MWh to 1 62.55 = 1251

3 sells 20 MWh to 1 61.55 = 12313 sells 20 MWh to 1 60.55 = 12112 sells 20 MWh to 1 60.25 = 12052 sells 20 MWh to 1 59.65 = 11932 sells 20 MWh to 1 59.05 = 11812 sells 20 MWh to 1 58.45 = 11692 sells 20 MWh to 1 57.85 = 11572 sells 20 MWh to 1 57.25 = 1045

Total 10743

Utility 3 receives $3693 from Utility 1 and Utility 2 receives $7050. Note that each participant

benefits: Utility 3 receives $1275 above its costs; Utility 2 receives $942 above its costs; and, Utility 1saves $2217

c. The system savings have been reduced.

d.The optimal allocations savings based on the Shapley Value are:

Coalition utilities

Coalition Savings

Independent 0

1-2 9062-3 01-3 4216

1-2-3 5122

Shapley value C= s1!ns !n !

Utility 1

Coalition Marginal Contribution Weight

1 0 0 = 0 11! 31 !3 !

= 13

1-2 906 0 = 906 21! 32!

3 !=

1

6

1-3 4216 0 = 4216 1

6

1-2-3 5122 0 = 5122 1

3


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Shapley value = 1=1

30

1

6906

1

64216

1

35122

1=$2561 /h

Utility 2

Coalition Marginal Contribution Weight 2 0 0 = 0 11!31 !

3 !=

1

3

2-1 906 0 = 906 21! 32!

3 !=

1

62-3 0 1

6

2-1-3 5122 906 =4216 1

3

Shapley value = 2=1

30

1

6906

1

60

1

34216

2=$ 1707/ h

Utility 3

Coalition Marginal Contribution Weight 3 0 0 = 0 11!31 !

3!

=1

3

3-1 4216 0 = 4216 21! 32!

3 !=

1

63-2 0 0 = 0 1

6

3-1-2 5122 4216 = 906 1

3

Shapley value = 3=1

30

1

64216

1

60

1

3906

1=$2561 /h 2=$1707/ h 3=$1004.66/h


20/21

Problem 6

j

PG j=k

PDk

PGA +PGB = 200

A42

0.8B27

0.55=200

1.552.51.8249.1=200

=$98.24 /MWh

PGA = 70.3MWPDX= 129.53MW

Now considering an intertie capacity of 40 MW

For Area A:

A42

0.8=175

A=$182/MWh

PGA = 175 MWPDX= 135MW

For Area B:

B27

0.5540=65

B27

0.5540=65

B=$40.75 /MWh

PGB = 25MWPDY= 65 MW

Problem 7

j

PG j=k

PD k

PGA +PGB =PDX +PDY


21/21

3502=2502.86629

=$104.85/MWh

PGA = 264.55 MW PGB = 209.7MW

PDX= 145.5MW PDY = 329.13MW

Now considering an intertie capacity of 70 MW

For Area A:

3A50=250A70

A=$92.5/MWh

PGA = 227.5 MW

PDX= 157.5MW

For Area B:

2B70=6292.86B

B=$115.02/MWh

PGB = 230.04MWPDY= 300.04 MW

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