Appendix
Answers to Selected Problems
Chapter 2
dC
= 50q 2 + 2000q
dq
Prot = Revenue Cost = 25q 2
2.1 Revenue = q
2.2 a. q max = 200
b.
max = 2000 $/widget
c. Maximum surplus = $200 000. It cannot be realized because it is unlikely that
producers will sell for nothing.
d.
q = 100
Gross consumers surplus = $150 000
Revenue = $100 000
Net consumers surplus = $50 000
e. q = 80
Revenue = $96 000
f. = 1
g.
Gross consumers surplus = 2000q 5q 2
Net consumers surplus = 5q 2
h.
Gross consumers surplus = 200 000 0.05 2
Net consumers surplus = 200 000 200 + 0.05 2
2.3 a. q = 120
= 800 $/widget
b.
Consumers gross surplus: $168 000
Consumers net surplus: $72 000
Fundamentals of Power System Economics Daniel Kirschen and Goran Strbac
2004 John Wiley & Sons, Ltd ISBN: 0-470-84572-4
APPENDIX: ANSWERS TO SELECTED PROBLEMS
266
Producers revenue: $96 000
Producers prot: $36 000
Global welfare: $108 000
2.4 a. = 900 $/widget
q = 110
Consumers net surplus: $60 500
Producers prot: $46 750
Global welfare: $107 250
b.
= 600 $/widget
q = 80
Consumers net surplus: $80 000
Producers prot: $16 000
Global welfare: $96 000
c. = 1100 $/widget
q = 90
Consumers net surplus: $40 500
Producers prot: $20 250
Tax revenue: $40 500
Global welfare: $101 250
2.5
q = 200
2.6 11
12
22
21
q = 10 000/
200
50
150
200
100
100
100
150
50
1/3
66.6
200
50
= 0.120
= 0.048
= 0.108
= 0.160
2.8 a.
Since the average production cost must be lower than the price, we get 65
y 155; y opt = 110
b.
The xed cost is so high that there is no range of production at which the rm
would make a prot.
APPENDIX: ANSWERS TO SELECTED PROBLEMS
267
Chapter 3
3.2
Pool Price
($/MWh)
a.
NSPCo
SAlCo
16
Produces 200 MWh Receives $3200
from the pool
Consumes 200 MWh Pays $3200 to
the pool
18
Produces 200 MWh Receives $3600
from the pool Pays $400 to SAlCo
Consumes 200 MWh Pays $3600 to
the pool Receives $400 from NSPCo
13
Produces 200 MWh Receives $2600
from the pool Receives $600 from
SAlCo
Consumes 200 MWh Pays $2600 to
the pool Pays $600 to NSPCo
b.
18
Produces 50 MWh Receives $900 from Consumes 200 MWh Pays $3600 to
the pool Pays $400 to SAlCo
the pool Receives $400 from NSPCo
c.
13
Produces 200 MWh Receives $2600
from the pool Receives $600 from
SAlCo
Consumes 100 MWh Pays $1300 to
the pool Pays $600 to NSPCo
3.3
Company
Prot ($)
Red
Green
Blue
Yellow
Magenta
Purple
650
1280
1325
515
287.50
125
3.4 a. The supply curve is piecewise constant and is as follows in tabular form:
Company
Amount
(MWh)
Cumulative
Amount (MW)
Price
($/MWh)
Blue
200
0200
10.5
Red
100
200300
12.5
Blue
200
300500
13
Green
50
500550
13.5
100
550650
14
Green
50
650700
14.5
Blue
100
700800
15
Green
50
800850
15.5
Red
50
850900
18
Red
APPENDIX: ANSWERS TO SELECTED PROBLEMS
268
b.
Forecast
Load
(MW)
Demand
(MW)
Price
($/MWh)
Blue
Production
(MWh)
Blue
Revenue
($)
Red
Production
(MWh)
Red
Revenue
($)
Green
Production
(MWh)
Green
Revenue
($)
400
400
13.00
300
3900
100
1300
600
600
14.00
400
5600
150
2100
50
700
875
875
18.00
500
9000
225
4050
150
2700
Forecast
Load
(MW)
Demand
(MW)
Price
($/MWh)
Blue
Production
(MWh)
Blue
Revenue
($)
Red
Production
(MWh)
Red
Revenue
($)
Green
Production
(MWh)
Green
Revenue
($)
400
348
13.00
248
3224
100
1300
600
546
13.50
400
5400
100
1350
46
621
875
813
15.50
500
7750
200
3100
113
c.
1751.50
3.5 a.
Item
Energy
bought
(MWh)
Industrial customer
Energy
sold
(MWh)
Price ($)
Expenses ($)
Revenue ($)
50
19.00
950.00
Other customers
1150
21.75
25 012.50
Future contract
200
21.00
4200.00
Put option
200
23.50
4700.00
Long-term contract
600
20.00
12 000.00
Future contract
100
22.00
2200.00
Call option
150
20.50
3075.00
Generation
300
21.25
6375.00
Spot market
purchase
450
21.50
9675.00
150 MW Call option
fee
1.00
150
200 MW Put option
fee
1.00
200
300 MW Call option
fee
1.00
300
Prot
Balance
887.50
1600
1600
34 862.50
34 862.50
APPENDIX: ANSWERS TO SELECTED PROBLEMS
b.
269
If the spot price increases to $23.47, the cost of the 450 MWh purchase on
the spot market would offset the prot. The 20.50 $/MWh call option and the
23.50 $/MWh put option would still be in the money. The 24.00 $/MWh call
option would still be out of the money.
3.6 Since the market operator accepted 175 MW of bids, using the supply curve, we
determine that the spot market price was 21.00 $/MWh.
a.
Item
Energy
bought
(MWh)
Energy
sold
(MWh)
Expenses ($)
Future T4
600
20.00
12 000.00
Nuclear unit
400
16.00
6400.00
Gas-red unit
200
18.00
3600.00
Forward T1
Revenue ($)
50
21.00
1050.00
Long-term T3
350
20.00
7000.00
Forward T5
100
22.00
2200.00
Exercise Put option T6
250
23.50
5875.00
50
21.00
1050.00
Residential customers
300
25.50
7650.00
Commercial customers
200
25.00
5000.00
Spot sale T9
Balancing spot purchase
100
21.00
2100
Fee option T6
2.00
500.00
Fee option T7
2.00
400.00
Fee option T8
2.00
200.00
Prot
Balance
b.
Price ($)
4625.00
1300
1300
29 825.00
29 825.00
Borduria Energys decit for that period would increase from 100 MW to
500 MW. The spot price would increase from 21.00 to 28.00 $/MW. The
cost of spot purchases would increase from $2000 to $14 000 but the cost of
operating the nuclear power plant would drop to zero. Borduria Energy would
therefore incur a loss of $975.00
Chapter 4
4.1 Cheapo Electrons makes a $1738.50 loss. The breakeven rate is 25.52 $/MWh.
4.2 The unit makes an operational prot of $690.07.
APPENDIX: ANSWERS TO SELECTED PROBLEMS
270
4.3 The unit makes an operational prot of $688.00
4.4 The unit should be brought on-line at the beginning of Period 3 and shutdown at
the end of Period 5. Its operational prot would then be $976.43.
4.5 The unit should be brought on-line at the beginning of Period 3 and shutdown at
the end of Period 6. Its operational prot would then be $680.43.
4.6 PA = 95.3 MW; PB = 74.2 MW; PC = 180.5 MW. Total hourly cost = 1927.15 $/h
4.7 PA = 85 MW; PB = 66 MW; PC = 160 MW; Market purchase: 39 MW
Total hourly cost = 1911.20 $/h
4.8 PA = 110 MW; PB = 86 MW; PC = 210 MW; Market sale: 56 MW
Prot from the sale: $33.03
4.9 PA = 100 MW; PB = 80 MW; PC = 210 MW; Market sale: 40 MW
Prot from the sale: $27.23
4.10 PA = 25 MW; PB = 30 MW; D = 55 MW; = 65 $/MWh; A = $725; B
= $1, 020
4.11 PA = 26.33 MW; PB = 31.33 MW; D = 55 MW; = 57.66 $/MWh; A =
$694; B = $982
4.12 Prot: $5235; Efciency that reduces prot to zero: 66.33%
Chapter 5
5.1
5.3
5.4
5.5
350 MW
a: 600 MW; b: 300 MW; c: 500 MW; d: 660 MW; e: 640 MW; f: 759 MW.
92.3 MW
106.5 MW
Chapter 6
6.1
F1 2 (MW)
F1 3 (MW)
F2 3 (MW)
Feasible?
Set 1
120
20
80
Yes
Set 2
400
400
No
Set 3
80
180
220
Yes
6.2 a. A = 80 $/MWh; B = 35 $/MWh; PA = 2000 MW; PB = 1000 MW;
FAB = 0
b.
A = B = 53 $/MWh; PA = 1100 MW; PB = 1900 MW; FAB = 900 MW
c. A = B = 65 $/MWh; PA = 1500 MW; PB = 1500 MW; FAB = 500 MW
APPENDIX: ANSWERS TO SELECTED PROBLEMS
d.
271
A = B = 57 $/MWh; PA = 900 MW; PB = 2100 MW; FAB = 1100 MW
e. A = 62 $/MWh; B = 47 $/MWh; PA = 1400 MW; PB = 1600 MW; FAB =
600 MW
6.3
Case:
EA ($)
160 000
106 000
130 000
114 000
124 000
EB ($)
35 000
53 000
65 000
57 000
47 000
RA ($)
160 000
58 300
97 500
51 300
86 800
RB ($)
35 000
100 700
97 500
62 700
75 200
The generator at B and the demand at A benet from the line because it increases
the price at B and lowers the price at A.
6.4 $9000. The congestion surplus is equal to zero when the ow is equal to zero
and when it is equal to the unconstrained value of 900 MW.
6.5 PA = 0 MW; PB = 0 MW; PC = 120 MW; PD = 400 MW
1 = 2 = 3 = 10 $/MWh
6.6 F21 = 120 MW; F31 = 280 MW; F32 = 200 MW
Line 13 is overloaded by 30 MW.
6.7 Method 1:
PA = 0 MW; PB = 48 MW; PC = 72 MW; PD = 400 MW
F21 = 102 MW; F31 = 250 MW; F32 = 182 MW
Increase in cost: $240
Method 2:
PA = 80 MW; PB = 0 MW; PC = 40 MW; PD = 400 MW
F21 = 150 MW; F31 = 250 MW; F32 = 150 MW
Increase in cost: $160
Method 2 is preferable because it is cheaper.
6.8 1 = 13.33 $/MWh; 2 = 12.00 $/MWh; 3 = 10.00 $/MWh
6.9 PA = 63.33 MW; PB = 10 MW; PC = 6.67 MW; PD = 400 MW
1 = 15 $/MWh; 2 = 12 $/MWh; 3 = 10 $/MWh
APPENDIX: ANSWERS TO SELECTED PROBLEMS
272
6.10 FBA = 730 MW; PA = 1270 MW; PB = 1783 MW; Losses = 53 MW
A = 58.10 $/MWh; B = 50.67 $/MWh;
Surplus: $2727
6.11
100.00
1000
900
90.00
Flow
80.00
800
700
70.00
MCA
60.00
(MW)
600
50.00 ()
500
MCB
400
300
30.00
200
20.00
10.00
100
Losses
0
0
0.0001
0.0002
0.0003
K = R/V2
0.00
0.0004
Figure P6.11 Losses and optimal ow
6.12
40.00
Y11 1 y13 31 = y12 2 + y13 3
y21 1 = Y22 2 + y23 3
1 = 13.33 $/MWh
31 = 5.33 $/MWh
6.13 Slack bus at bus 1:
y21 1 = Y22 2 + y23 3
y31 1 + y31 31 = Y33 3 + y32 2
Slack bus at bus 2:
Y11 1 y13 31 = y12 2 + y13 3
y31 1 + y31 31 = Y33 3 + y32 2
6.14 K = {1, 2, 3}; U = ; 31 and 21 are unknown.
Choose bus 1 as the slack.
y21 21 = Y22 2 + y21 1 + y23 3
y31 31 = Y33 3 + y31 1 + y32 2
21 = 1.67 $/MWh
31 = 7.00 $/MWh
0.0005
APPENDIX: ANSWERS TO SELECTED PROBLEMS
273
6.16 62.5 MW of owgate rights on branch 3-1.
6.17 37.5 MW of owgate rights on branch 2-1 and 62.5 MW of owgate rights on
branch 3-1.
Chapter 7
7.1
7.2
7.3
7.4
7.5
12.14%; 32.28 $/MWh
11.17%
14.13%; 12.33%;
Yes, because the IRR is 12.49%.
The investment is higher if technology A is adopted, but the Incremental Internal
Rate of Return on the additional investment is 14.13%, which is higher than the
Minimum Acceptable Rate of Return.
7.6 The plant should continue operating because it continues to generate a positive
cash ow of $32 524 128 per year. Borduria Power would not have built the plant
because it would not have achieved its MARR.
7.7 If the plant has 20 years of expected life left, Borduria Power should repair it
because the Internal Rate of Return on the investment required for the repair is
12.17%, which is above the MARR used by the company. If the plant has only
15 years left, the IRR is only 10.51% and the plant should be closed down.
7.8 Minimum price: 78.80 $/MWh. Average production cost: 37.70 $/MWh.
Chapter 8
8.3 8.00 $/MWh
8.4
Short run marginal value of transmission
50
40
30
20
10
0
0
100
200
300
400
500
600
700
10
Transmission capacity (MW)
800
900
1000
274
8.5
8.6
8.7
8.8
8.9
APPENDIX: ANSWERS TO SELECTED PROBLEMS
T = 45 0.05 FBA
12.00 $/MWh
660 MW
750 MW; $78 750 000. The two amounts are identical.
500 MW; $90 000 000 versus $52 500 000
1000 MW; $35 000 000 versus $105 000 000
