# We can work on Data Analysis

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
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 + 6.23𝑃1𝑡 + 0.006𝑃1𝑡2) ∗ 𝐼1𝑡+(210 + 6.4𝑃2𝑡 + 0.007𝑃2𝑡2) ∗ 𝐼2𝑡 + (190 + 5.9𝑃3𝑡 + 0.005𝑃3𝑡2) ∗ 𝐼3𝑡]
𝑡=1

• ∑4 [200 𝐼1𝑡(1 − 𝐼1(𝑡−1)) + 150𝐼2𝑡(1 − 𝐼2(𝑡−1))]
𝑡=1

S.t.

𝑃11𝐼11 + 𝑃21𝐼21 + 𝑃31𝐼31 = 500
𝑃12𝐼12 + 𝑃22𝐼22 + 𝑃32𝐼32 = 700
𝑃13𝐼13 + 𝑃23𝐼23 + 𝑃33𝐼33 = 800
𝑃14𝐼14 + 𝑃24𝐼24 + 𝑃34𝐼34 = 400

𝑃𝑚𝑎𝑥,1𝐼11 + 𝑃𝑚𝑎𝑥,2𝐼21 + 𝑃𝑚𝑎𝑥,3𝐼31 ≥ 500 + 50 = 550
𝑃𝑚𝑎𝑥,1𝐼12 + 𝑃𝑚𝑎𝑥,2𝐼22 + 𝑃𝑚𝑎𝑥,3𝐼32 ≥ 700 + 70 = 770
𝑃𝑚𝑎𝑥,1𝐼13 + 𝑃𝑚𝑎𝑥,2𝐼23 + 𝑃𝑚𝑎𝑥,3𝐼33 ≥ 800 + 80 = 880
𝑃𝑚𝑎𝑥,1𝐼14 + 𝑃𝑚𝑎𝑥,2𝐼24 + 𝑃𝑚𝑎𝑥,3𝐼34 ≥ 400 + 40 = 440

∑4 (190 + 5.9𝑃3𝑡 + 0.005𝑃3𝑡2)𝐼3𝑡

≤ 2,000

𝑡=1

100𝐼1𝑡 ≤ 𝑃1𝑡 ≤ 400𝐼1𝑡
80𝐼2𝑡 ≤ 𝑃2𝑡 ≤ 400𝐼2𝑡
40𝐼3𝑡 ≤ 𝑃3𝑡 ≤ 200𝐼3𝑡

𝐼10 = 1, 𝐼20 = 1, 𝐼30 = 0, 𝐼1𝑡, 𝐼2𝑡, 𝐼3𝑡= {0, 1}, t = 1,2,3,4

LR version of the original UC problem

Φ = Min ∑4 [(200 + 6.23𝑃1𝑡 + 0.006𝑃1𝑡2) ∗ 𝐼1𝑡 +(210 + 6.4𝑃2𝑡 + 0.007𝑃2𝑡2) ∗ 𝐼2𝑡 + (190 + 5.9𝑃3𝑡 + 0.005𝑃3𝑡2) ∗
𝑡=1
𝐼3𝑡] + ∑4 [350 𝐼1𝑡(1 − 𝐼1(𝑡−1)) + 100𝐼2𝑡(1 − 𝐼2(𝑡−1))]
𝑡=1

• λ1[500-(𝑃11𝐼11 + 𝑃21𝐼21 + 𝑃31𝐼31 )] + λ2[700-𝑃12𝐼12 + 𝑃22𝐼22 + 𝑃32𝐼32)]
• λ3[800-(𝑃13𝐼13 + 𝑃23𝐼23 + 𝑃33𝐼33)] + λ4[400-(𝑃14𝐼14 + 𝑃24𝐼24 + 𝑃34𝐼34)]
• μ1 [550-( 𝑃𝑚𝑎𝑥,1𝐼11 + 𝑃𝑚𝑎𝑥,2𝐼21 + 𝑃𝑚𝑎𝑥,3𝐼31)] + μ2 [770-( 𝑃𝑚𝑎𝑥,1𝐼12 + 𝑃𝑚𝑎𝑥,2𝐼22 + 𝑃𝑚𝑎𝑥,3𝐼32 )] + μ3 [880- (𝑃𝑚𝑎𝑥,1𝐼13 + 𝑃𝑚𝑎𝑥,2𝐼23 + 𝑃𝑚𝑎𝑥,3𝐼33)] + μ4[440-(𝑃𝑚𝑎𝑥,1𝐼14 + 𝑃𝑚𝑎𝑥,2𝐼24 + 𝑃𝑚𝑎𝑥,3𝐼34)]
• π[2000-∑4 (190 + 5.9𝑃3𝑡 + 0.005𝑃3𝑡2) 𝐼31)]
𝑡=1

S.t.

100𝐼1𝑡 ≤ 𝑃1𝑡 ≤ 400𝐼1𝑡
80𝐼2𝑡 ≤ 𝑃2𝑡 ≤ 400𝐼2𝑡
40𝐼3𝑡 ≤ 𝑃3𝑡 ≤ 200𝐼3𝑡
𝐼10 = 1, 𝐼20 = 1, 𝐼30 = 0, 𝐼1𝑡, 𝐼2𝑡, 𝐼3𝑡= {0, 1}, t = 1,2,3,4

Iteration 1

Let 𝛌𝟏 = 7.5, 𝛌𝟏 = 10, 𝛌𝟏= 11, 𝛌𝟏 = 7, which as same as iteration 12 of Problem 1
𝟏 𝟐 𝟑 𝟒

And Let 𝛍𝒕 = 0, π𝑡= 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
𝑃 = λ1−b1 = 7.5−6.23 =106

11 2𝐶1

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

𝑃 = λ2−b1 = 10−6.23 =314

12 2𝐶1

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

𝑃 = λ3−b1 = 11−6.23 =398

13 2𝐶1

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

𝑃 = λ4−b1 = 7−6.23 == 100

14 2𝐶1

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
𝑃 = λ1−b2 = 7.5−6.4 =80

21 2𝐶2

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

𝑃 = λ2−b2 = 10−6.4 =257

22 2𝐶2

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

𝑃 = λ3−b1 = 11−6.4 =329

33 2𝐶1

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

𝑃 = λ4−b1 = 7−6.4

== 80

14 2𝐶1

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
𝑃 = λ1−b3 = 7.5−5.9 =160

31 2𝐶3

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

𝑃 = λ2−b3 = 10−5.9 =200

32 2𝐶3

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

𝑃 = λ3−b3 = 11−5.9 =200

33 2𝐶3

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

𝑃 = λ4−b3 = 7−5.9

= 110

34 2𝐶3

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``

𝐼1𝑡 1 1 1 0
𝐼2𝑡 0 1 1 0
𝐼3𝑡 0 1 1 0
𝑃1𝑡 106 314 398 0
𝑃2𝑡 0 257 329 0
𝑃3𝑡 0 200 200 0

Hour 2 does not satisfy the Reserve constraints. So all Pit=0, Iit=0. Iteration continues.

Iteration 2

Let 𝛌𝟏 = 7.5, 𝛌𝟏 = 10, 𝛌𝟏= 11, 𝛌𝟏 = 7,
𝟏 𝟐 𝟑 𝟒

𝛌𝟐 = 7.5 + 0.002(500-0)=8.5

𝛌𝟐 = 10 + 0.002(700-0)=11.4

𝛌𝟐 = 11 + 0.002(800-0)=12.6

𝛌𝟐 = 7 + 0.002(400-0)=7.80

And Let 𝝁𝟏= 0,

𝛍𝟐 = 0 + 0.002(550-0)=1.1

𝛍𝟐 = 0 + 0.0005(770-0)= 1.54

𝛍𝟐 = 0 + 0.0005(880-0)= 1.76

𝛍𝟐 = 0 + 0.002(440-0)=0.88

π𝑡= 0,

∑4 (190 + 5.9𝑃3𝑡 + 0.005𝑃3𝑡2)𝐼3𝑡 = 0 ≤ 2,000
𝑡=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   hour4``

I1 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   Hour4``

lambda1 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
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   hour4``

I1 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   hour4``

P1 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   Hour4``

lambda1 9.5 12.8 14.2 8.6
mui orig -93.525 -131.945 -148.54 -74.02
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   hour4``

I1 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   hour4``

P1 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   Hour4``

lambda1 9.5 12.8 14.2 8.6
mui orig -95.725 -135.015 -152.06 -75.78
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   hour4``

I1 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   hour4``

P1 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. The post Data Analysis appeared first on UKCustomPapers.