Problem 1:
The generator data are given in Table 1 and the load and reserve data are given in Table 2. The fuel consumption functions of the generating units are quadratic H(P) = af + bf * P + cf * P2 (MBtu). The fuel prices are all 1 $/MBtu. The unit shutdown costs and the system losses are assumed to be zero. Unit 3 has a fuel contract of 2000 MBtu. The initial Lagrangian multipliers for power balance and reserve requirements are given in Table 3. The initial multiplier for Unit 3’s fuel constraint is zero. The adjustment steps of multipliers are given in Table 4. Set J=10,000 if the ED is infeasible based on a given commitment. Use the LR method to solve the UC problem. Obtain two different feasible solutions and show the corresponding relative duality gaps.
Table 1: Generator data
Unit af (MBtu) bf (MBtu/MW) cf (MBtu/MW2) Pmin (MW) Pmax (MW) Min ON
(h) Min OFF
(h) Startup Cost ($) Initial Status
(h)/(MW)
1 200 6.23 0.006 100 400 2 2 200 ON 4 /300
2 210 6.4 0.007 80 400 2 1 150 ON 4 /200
3 190 5.9 0.005 40 200 1 2 0 OFF 4 /0
Table 2: Load and reserve data
Hour Load (MW) Reserve (MW)
1 500 50
2 700 70
3 800 80
4 400 40
Table 3: Initial and
Hour (Power balance) (Reserve requirements)
1 7.5 0
2 10 0
3 11 0
4 7 0
Table 4: Adjustment steps of Lagrangian multipliers
Lagrangian Multiplier k1 k2
(Power balance) 0.002 0.001
(Reserve requirements) 0.002 0.0005
(Fuel constraint for unit 3) 0.0005 0.0001
Solution:
Objective Function
Min ∑4 [(200 + 6.23?1? + 0.006?1?2) ∗ ?1?+(210 + 6.4?2? + 0.007?2?2) ∗ ?2? + (190 + 5.9?3? + 0.005?3?2) ∗ ?3?]
?=1
- ∑4 [200 ?1?(1 − ?1(?−1)) + 150?2?(1 − ?2(?−1))]
?=1
S.t.
?11?11 + ?21?21 + ?31?31 = 500
?12?12 + ?22?22 + ?32?32 = 700
?13?13 + ?23?23 + ?33?33 = 800
?14?14 + ?24?24 + ?34?34 = 400
????,1?11 + ????,2?21 + ????,3?31 ≥ 500 + 50 = 550
????,1?12 + ????,2?22 + ????,3?32 ≥ 700 + 70 = 770
????,1?13 + ????,2?23 + ????,3?33 ≥ 800 + 80 = 880
????,1?14 + ????,2?24 + ????,3?34 ≥ 400 + 40 = 440
∑4 (190 + 5.9?3? + 0.005?3?2)?3?
≤ 2,000
?=1
100?1? ≤ ?1? ≤ 400?1?
80?2? ≤ ?2? ≤ 400?2?
40?3? ≤ ?3? ≤ 200?3?
?10 = 1, ?20 = 1, ?30 = 0, ?1?, ?2?, ?3?= {0, 1}, t = 1,2,3,4
LR version of the original UC problem
Φ = Min ∑4 [(200 + 6.23?1? + 0.006?1?2) ∗ ?1? +(210 + 6.4?2? + 0.007?2?2) ∗ ?2? + (190 + 5.9?3? + 0.005?3?2) ∗
?=1
?3?] + ∑4 [350 ?1?(1 − ?1(?−1)) + 100?2?(1 − ?2(?−1))]
?=1
- λ1[500-(?11?11 + ?21?21 + ?31?31 )] + λ2[700-?12?12 + ?22?22 + ?32?32)]
- λ3[800-(?13?13 + ?23?23 + ?33?33)] + λ4[400-(?14?14 + ?24?24 + ?34?34)]
- μ1 [550-( ????,1?11 + ????,2?21 + ????,3?31)] + μ2 [770-( ????,1?12 + ????,2?22 + ????,3?32 )] + μ3 [880- (????,1?13 + ????,2?23 + ????,3?33)] + μ4[440-(????,1?14 + ????,2?24 + ????,3?34)]
- π[2000-∑4 (190 + 5.9?3? + 0.005?3?2) ?31)]
?=1
S.t.
100?1? ≤ ?1? ≤ 400?1?
80?2? ≤ ?2? ≤ 400?2?
40?3? ≤ ?3? ≤ 200?3?
?10 = 1, ?20 = 1, ?30 = 0, ?1?, ?2?, ?3?= {0, 1}, t = 1,2,3,4
Iteration 1
Let ?? = 7.5, ?? = 10, ??= 11, ?? = 7, which as same as iteration 12 of Problem 1
? ? ? ?
And Let ?? = 0, π?= 0,
For Unit 1
Hour0 Hour1 Hour2 Hour3 Hour4
132.7958 132.7958 -392.204 -259.408 -748.037 -1007.45 183 ON +2 -824.446
183
0 0 -548.037 0 ON +1 383
0
132.7958 -259.408 OFF -1 -1007.45
0 -548.037 0 383 0 OFF -2 -259.408
0 0 0 0
Forward Search Hour 1
? = λ1−b1 = 7.5−6.23 =106
11 2?1
2∗0.006
TC (on+2, 0 → on+2, 1) = 200+6.23106+0.006106^2-7.5*106= 133
TC (on+2, 0 → off-1, 1) = 0
V (on+2, 1) = 133
V (off-1, 1) = 0
Hour 2
? = λ2−b1 = 10−6.23 =314
12 2?1
2∗0.006
TC (on+2, 1 → on+2, 2) = 200+6.23314+0.006314^2-10*314= -392
TC (on+2, 1 → off-1, 2) = 0
TC (off-1, 1 → off-2, 2) = 0
V (on+2, 2) =V(on+2, 1) + TC (on+2, 1 → on+2, 2) = -259.4 V (off-1, 2) = V(on+2,1) + TC (on+2, 1 → off-1, 2) = 132.8 V (off-2, 2) = V(off-1, 1) + TC (off-1, 1 → off-1, 2)} = 0
Hour 3
? = λ3−b1 = 11−6.23 =398
13 2?1
2∗0.006
TC (on+2, 2 → on+2, 3) = 200+6.23398+0.006398^2-11*398= -748
TC (off-2, 2 → on+1, 3) = -748+200= -548
TC (on+2, 2 → off-1, 3) = 0
TC (off-1, 2 → off-2, 3) = 0
TC (off-2, 2 → off-2, 3) = 0
V (on+2, 3) = V(on+2, 2) + TC (on+2, 2 → on+2, 3) = -1007.4
V (on+1, 3) = V(off-2, 2) + TC (off-2, 2 → on+1, 3) = -548 V (off-1, 3) = V(on+2,2) + TC (on+2, 2 → off-1, 3) = -259.4
V (off-2, 3) = min {V(off-1, 2) + TC (off-1, 2 → off-2, 3), V(off-2, 2) + TC (off-2, 2 → off-2, 3)} = 0
Hour 4
? = λ4−b1 = 7−6.23 == 100
14 2?1
2∗0.006
TC (on+2, 3 → on+2, 4) = 200+6.23100+0.006100^2-7*100= 183
TC (on+1, 3 → on+2, 4) = 183
TC (off-1, 3 → off-2, 4) = 0
TC (off-2, 3 → off-2, 4) = 0
V (on+2, 4) = min {V(on+2, 3) + TC (on+2, 3 → on+2, 4), V(on+1, 3) + TC (on+1, 3 → on+2, 4)} = -824.4
V (on+1, 4) = V(off-2, 3) + TC (off-2, 3 → on+1, 4) = 183+200=383 V (off-1, 4) = V(on+2,3) + TC (on+2, 3 → off-1, 4) = -1007.4
V (off-2, 4) = min {V(off-1, 3) + TC (off-1, 3 → off-2, 4), V(off-2, 3) + TC (off-2, 3 → off-2, 4)} = -259.4 Backward Search
V(off-1, 4) → V(on+2, 3) → V(on+2, 2) → V(on+2, 1)
For Unit 2
Hour0 Hour1 Hour2 Hour3 Hour4
166.8 166.8 -252.857 -86.0571 -545.714 -648.571 206.8 ON +2 -441.771
-545.714
-102.857 -395.714
ON +1 0 0 0 0 270.7429
-102.857 -395.714 356.8 OFF -1 -648.571
0 0 0 0 -86.0571 0
Forward Search Hour 1
? = λ1−b2 = 7.5−6.4 =80
21 2?2
2∗0.007
TC (on+2, 0 → on+2, 1) = 210+6.480+0.00780^2-7.5*80= 167
TC (on+2, 0 → off-1, 1) = 0
V (on+2, 1) = 166.8
V (off-1, 1) = 0
Hour 2
? = λ2−b2 = 10−6.4 =257
22 2?2
2∗0.007
TC (on+2, 1 → on+2, 2) = 210+6.4257+0.007257^2-10*257= -253
TC (on+2, 1 → off-1, 2) = 0
TC (off-1, 1 → on+1, 2) = -252.9+150 = -103
V (on+2, 2) =V(on+2, 1) + TC (on+2, 1 → on+2, 2) = -86.1 V (on+1, 2) = V(off-1,1) + TC (off-1, 1 → on+1, 2) = -102.9
V (off-1, 2) = min {V(off-1, 1) + TC (off-1, 1 → off-1, 2), V(on+2, 1) + TC (on+2, 1 → off-1, 2)} = 0
Hour 3
? = λ3−b1 = 11−6.4 =329
33 2?1
2∗0.007
TC (on+2, 2 → on+2, 3) = 210+6.4329+0.007329^2-11*329= -546
TC (on+1, 2 → on+2, 3) = -546
TC (off-1, 2 → on+1, 3) = -545.7+150= -396
TC (on+2, 2 → off-1, 3) = 0
TC (off-1, 2 → off-1, 3) = 0
V (on+2, 3) = min {V(on+2, 2) + TC (on+2, 2 → on+2, 3), V(on+1, 2) + TC (on+1, 2 → on+2, 3)} = -648.6
V (on+1, 3) = V(off-1, 2) + TC (off-1, 2 → on+1, 3) = -395.7
V (off-1, 3) = min {V(on+2, 2) + TC (on+2, 2 → off-1, 3), V(off-1, 2) + TC (off-1, 2 → off-1, 3)} = -86.1
Hour 4
? = λ4−b1 = 7−6.4
== 80
14 2?1
2∗0.007
TC (on+2, 3 → on+2, 4) = 210+6.480+0.00780^2-7*80= 207
TC (on+1, 3 → on+2, 4) = 207
TC (off-1, 3 → on+1, 4) = 357
TC (on+2, 3 → off-1, 4) = 0
TC (off-1, 3 → off-1, 4) = 0
V (on+2, 4) = min {V(on+2, 3) + TC (on+2, 3 → on+2, 4), V(on+1, 3) + TC (on+1, 3 → on+2, 4)} = -441.8
V (on+1, 4) = V(off-1, 3) + TC (off-1, 3 → on+1, 4) = 270.7
V (off-1, 4) = min {V(off-1, 3) + TC (off-1, 3 → off-1, 4), V(on+2, 3) + TC (on+2, 3 → off-1, 4)} = -648.6
Backward Search
V(off-1, 4) → V(on+2, 3) → V(on+1, 2) → V(off-1, 1)
For Unit 3
Hour0 Hour1 Hour2 Hour3 Hour4
62 -430 -430 -630 -1060 129.5 ON +2 -930.5
0 0 0
62 -430
OFF-1 -1060
62
-430 -630 129.5
0 0 0
OFF -2 -430
0 0 0 0 0 0 0
Forward Search Hour 1
? = λ1−b3 = 7.5−5.9 =160
31 2?3
2∗0.005
TC (off-2, 0 → on+1, 1) = 190+5.9160+0.005160^2-7.5*160= 62
TC (off-2, 0 → off-2, 1) = 0
V (on+1, 1) = 62
V (off-1, 1) = 0
Hour 2
? = λ2−b3 = 10−5.9 =200
32 2?3
2∗0.005
TC (on+2, 1 → on+2, 2) = 190+5.9200+0.005200^2-10*200 = -430 TC (on+2, 1 → off-1, 2) = 0
TC (off-2, 1 → off-2, 2) = 0
TC (off-1, 1 → on+1, 2) = -430
V (on+1, 2) = min {V(on+1, 1) + TC (on+1, 1 → on+1, 2), V(off-2, 1) + TC (off-2, 1 → on+1, 2)} = -430
V (off-2, 2) = V(off-2,1) + TC (off-2, 1 → off-2, 2) = 0 V (off-1, 2) = V(on+1, 1) + TC (on+1, 1 → off-1, 2)= 62
Hour 3
? = λ3−b3 = 11−5.9 =200
33 2?3
2∗0.005
TC (on+2, 2 → on+2, 3) = 190+5.9200+0.005200^2-11*200= -630
TC (off-2, 2 → on+1, 3) = -630
TC (on+2, 2 → off-1, 3) = 0
TC (off-2, 2 → off-2, 3) = 0
V (on+1, 3) = min {V(on+1, 2) + TC (on+1, 2 → on+2, 3), V(off-2, 2) + TC (off-2, 2 → on+1, 3)} = -1060
V (off-1, 3) = V(on+1, 2) + TC (on+1, 2 → off-1, 3) = 0
V (off-2, 3) = min {V(off-2, 2) + TC (off-2, 2 → off-2, 3), V(off-1, 2) + TC (off-1, 2 → off-2, 3)} = 0
Hour 4
? = λ4−b3 = 7−5.9
= 110
34 2?3
2∗0.005
TC (on+1, 3 → on+1, 4) = 190+5.9110+0.005110^2-7*110= 130
TC (off-2, 3 → on+1, 4) = 130
TC (on+2, 3 → off-1, 4) = 0
TC (off-1, 3 → off-1, 4) = 0
V (on+1, 4) = min {V(on+1, 3) + TC (on+1, 3 → on+2, 4), V(off-2, 3) + TC (off-2, 3 → on+1, 4)} = -930.5
V (off-1, 4) = V(on+1, 3) + TC (on+1, 3 → off-1, 4) = -1060
V (off-2, 4) = min {V(off-2, 3) + TC (off-2, 3 → off-2, 4), V(off-1, 3) + TC (off-1, 3 → off-2, 4)} = -430
Backward Search
V(off-1, 4) → V(on+1, 3) → V(on+1, 2) → V(off-2, 1)
Therefore,
Hour 1 Hour 2 Hour 3 Hour 4?1? 1 1 1 0
?2? 0 1 1 0
?3? 0 1 1 0
?1? 106 314 398 0
?2? 0 257 329 0
?3? 0 200 200 0
Hour 2 does not satisfy the Reserve constraints. So all Pit=0, Iit=0. Iteration continues.
Iteration 2
Let ?? = 7.5, ?? = 10, ??= 11, ?? = 7,
? ? ? ?
?? = 7.5 + 0.002(500-0)=8.5
?? = 10 + 0.002(700-0)=11.4
?? = 11 + 0.002(800-0)=12.6
?? = 7 + 0.002(400-0)=7.80
And Let ??= 0,
?? = 0 + 0.002(550-0)=1.1
?? = 0 + 0.0005(770-0)= 1.54
?? = 0 + 0.0005(880-0)= 1.76
?? = 0 + 0.002(440-0)=0.88
π?= 0,
∑4 (190 + 5.9?3? + 0.005?3?2)?3? = 0 ≤ 2,000
?=1
π2= 0,
For Unit 1
Hour0 Hour1 Hour2 Hour3 Hour4
-454.704 -454.704 -1524 -1978.7 -2092 -4070.7 -254.704 ON +2 -4325.41
-254.704
0 0 -1892 0 ON +1 -509.408
0
-454.704 -1978.7 OFF -1 -4070.7
0 -1892 0 -54.7042 0 OFF -2 -1978.7
0 0 -454.704 0
For Unit 2
Hour0 Hour1 Hour2 Hour3 Hour4
-387.5 -387.5 -1298.86 -1686.36 -1854 -3540.36 -212 ON +2 -3752.36
-1854
-1148.86 -2091.5
ON +1 0 0 0 0 -1748.36
-1148.86 -1704 -62 OFF -1 -3540.36
0 0 -387.5 0 -1686.36 0
For Unit 3
Hour0 Hour1 Hour2 Hour3 Hour4
-350 -1018 -1368 -1302 -2670 -166.5 ON +2 -2836.5
0 0 0
-350 -1368
OFF-1 -2670
-350
-1018 -1302 -166.5
0 0 0
OFF -2 -1368
0 0 0 0 0 -350 0
Therefore, from dynamic programming, we can have the following dispatch:
hour1 hour2 hour3 hour4I1 1 1 1 0
I2 1 1 1 0
I3 1 1 1 1
P1 189.1667 400 400 0
P2 150 357.1429 400 0
P3 200 200 200 190
Phi 0 29156.8 Phi 18242.53
In the last hour, the reserve constraints cannot meet, so all Pit=0, Iit=0.
Iteration 3
Hour1 Hour2 Hour3 Hour4lambda1 9.5 12.8 14.2 8.6
mui orig 2.2 3.08 3.52 1.76
mui 2.2 3.08 3.52 1.76
Loadlambda 4750 8960 11360 3440 (Load+Reserve)mui 1210 2371.6 3097.6 774.4
Fuel of unit 3 0 0 0 0
Pi orig -0.124
Pi 0
F or Unit 1
Hour0 Hour1 Hour2 Hour3 Hour4
-1125.54 -1125.54 -2700 -3825.54 -3436 -7261.54 -738.038 ON +2 -7999.58
-738.038
0 0 -3236 0 ON +1 -1663.58
0
-1125.54 -3825.54 OFF -1 -7261.54
0 -3236 0 -538.038 0 OFF -2 -3825.54
0 0 -1125.54 0
For Unit 2
Hour0 Hour1 Hour2 Hour3 Hour4
-1013.21 -1013.21 -2462 -3475.21 -3198 -6673.21 -666.857 ON +2 -7340.07
-3198
-2312 -4061.21
ON +1 0 0 0 0 -3992.07
-2312 -3048 -516.857 OFF -1 -6673.21
0 0 -1013.21 0 -3475.21 0
For Unit 3
Hour0 Hour1 Hour2 Hour3 Hour4
-770 -1606 -2376 -1974 -4350 -502 ON +2 -4852
0 0 0
-770 -2376
OFF-1 -4350
-770
-1606 -1974 -502
0 0 0
OFF -2 -2376
0 0 0 0 0 -770 0
Therefore, from dynamic programming, we can have the following dispatch:
hour1 hour2 hour3 hour4I1 1 1 1 1
I2 1 1 1 1
I3 1 1 1 1
P1 272.5 400 400 197.5
P2 221.4286 400 400 157.1429
P3 200 200 200 200
Phi 0 35963.6 Phi 15771.95
From economic dispatch, we can have the solution:
hour1 hour2 hour3 hour4P1 255 359.1 395 200
P2 205 295 330 160
P3 40 45.9 75 40
Total Cost 21693.53
RDG 0.37545
The RDG is greater than0.2, so we continue iterations.
Iteration 4
Hour1 Hour2 Hour3 Hour4lambda1 9.5 12.8 14.2 8.6
mui orig -93.525 -131.945 -148.54 -74.02
mui 0 0 0 0
Loadlambda 4750 8960 11360 3440 (Load+Reserve)mui 0 0 0 0
Fuel of unit 3 434 471.3441 660.625 434
Pi orig -3.1E-06
Pi 0
For Unit 1
Hour0 Hour1 Hour2 Hour3 Hour4
-245.538 -245.538 -1468 -1713.54 -2028 -3741.54 -34.0375 ON +2 -3775.58
-34.0375
0 0 -1828 0 ON +1 -79.575
0
-245.538 -1713.54 OFF -1 -3741.54
0 -1828 0 165.9625 0 OFF -2 -1713.54
0 0 -245.538 0
For Unit 2
Hour0 Hour1 Hour2 Hour3 Hour4
-133.214 -133.214 -1230 -1363.21 -1790 -3153.21 37.14286 ON +2 -3116.07
-1790
-1080 -1773.21
ON +1 0 0 0 0 -1176.07
-1080 -1640 187.1429 OFF -1 -3153.21
0 0 -133.214 0 -1363.21 0
For Unit 3
Hour0 Hour1 Hour2 Hour3 Hour4
-330 -990 -1320 -1270 -2590 -150 ON +2 -2740
0 0 0
-330 -1320
OFF-1 -2590
-330
-990 -1270 -150
0 0 0
OFF -2 -1320
0 0 0 0 0 -330 0
Therefore, from dynamic programming, we can have the following dispatch:
hour1 hour2 hour3 hour4I1 1 1 1 1
I2 1 1 1 1
I3 1 1 1 1
P1 272.5 400 400 197.5
P2 221.4286 400 400 157.1429
P3 200 200 200 200
Phi 0 28510 Phi 18841.21
From economic dispatch, we can have the solution:
hour1 hour2 hour3 hour4P1 255 359 395 200
P2 205 295 330 160
P3 40 46 75 40
Total Cost 21693.11
RDG 0.151365
The RDG is smaller than 0.2, we get the first feasible solution. Iteration continues to find the second feasible solution.
Iteration 5
Hour1 Hour2 Hour3 Hour4lambda1 9.5 12.8 14.2 8.6
mui orig -95.725 -135.015 -152.06 -75.78
mui 0 0 0 0
Loadlambda 4750 8960 11360 3440 (Load+Reserve)mui 0 0 0 0
Fuel of unit 3 434 471.98 660.625 434
Pi orig 0.000303
Pi 0.000303
For Unit 1
Hour0 Hour1 Hour2 Hour3 Hour4
-245.538 -245.538 -1468 -1713.54 -2028 -3741.54 -34.0375 ON +2 -3775.58
-34.0375
0 0 -1828 0 ON +1 -79.575
0
-245.538 -1713.54 OFF -1 -3741.54
0 -1828 0 165.9625 0 OFF -2 -1713.54
0 0 -245.538 0
For Unit 2
Hour0 Hour1 Hour2 Hour3 Hour4
-133.214 -133.214 -1230 -1363.21 -1790 -3153.21 37.14286 ON +2 -3116.07
-1790
-1080 -1773.21
ON +1 0 0 0 0 -1176.07
-1080 -1640 187.1429 OFF -1 -3153.21
0 0 -133.214 0 -1363.21 0
For Unit 3
Hour0 Hour1 Hour2 Hour3 Hour4
-330 -990 -1320 -1270 -2590 -150 ON +2 -2740
0 0 0
-330 -1320
OFF-1 -2590
-330
-990 -1270 -150
0 0 0
OFF -2 -1320
0 0 0 0 0 -330 0
Therefore, from dynamic programming, we can have the following dispatch:
hour1 hour2 hour3 hour4I1 1 1 1 1
I2 1 1 1 0
I3 1 1 1 1
P1 272.5 400 400 197.5
P2 221.4286 400 400 0
P3 200 200 200 200
Phi 0 28509.4 Phi 18840.61
From economic dispatch, we can have the solution:
hour1 hour2 hour3 hour4P1 250 320 360 305.6934
P2 200 270 300 0
P3 50 110 140 94.30657
Total Cost 21316.24
RDG 0.131399
The RDG is smaller than 0.2, we get the second feasible solution. Stop here.
Sample Solution
