Data Set for Manuscript “Day-ahead Market Clearing with Robust
Security-Constrained Unit Commitment Model”
Hongxing Ye, Yinyin Ge, Mohammad Shahidehpour, Zuyi Li
1
∗
Data for Six-bus System
The one-line diagram is shown in Fig.1. The unit data and line data are shown in Table 1 and Table 2,
respectively. Table 3 presents the load and uncertainty information. Column “Base Load” shows the hourly
forecasted load. Assume that the load distributions are 20%, 40%, and 40% for Bus 3, Bus 4, and Bus 5,
respectively. ū1,t and ū3,t in Table 3 are the bounds of the uncertainties at Bus 1 and Bus 3, respectively.
The uncertainty bounds at other buses are 0.
1
G1
G2
L1
2
6
5
4
L3
L2
3
G3
Figure 1: One-line Diagram for 6-bus system
Table 1: Unit Data for the 6-bus System
#
1
2
6
P
min
100
10
10
P
max
220
100
20
P0
a
b
c
Ru
Rd
Cu
Cd
T on
T off
T0
120
50
0
0.004
0.001
0.005
13.5
32.6
17.6
176.9
129.9
137.4
24
12
5
24
12
5
180
360
60
50
40
0
4
3
1
4
2
1
4
3
−2
P min ,P max ,P0 : min/max/initial generation level (MW);
fuel cost ($): aP 2 + bP + c ;
Ru ,Rd : ramping up/down rate (MW/h);
Cu ,Cd : startup/shutdown cost ($);
T on ,T off ,T0 : min on/min off/initial time (h)
∗ This work is supported by the U.S. National Science Foundation Project Number ECCS-1549937. The authors are with
the Robert W. Galvin Center for Electricity Innovation at Illinois Institute of Technology, Chicago, IL 60616, USA. (e-mail:
[email protected]; [email protected]; [email protected]; [email protected]).
1
Table 2: Line Data for the 6-bus System
from
1
1
2
5
3
2
4
to
2
4
4
6
6
3
5
x(p.u.)
0.17 0.258 0.197 0.14 0.018 0.037 0.037
capacity(MW) 200 100 100 100 100 200 200
Table 3: Load and Uncertainty Data for the 6-bus System (MW)
Time (h) Base Load
1
2
3
4
5
6
7
8
9
10
11
12
175.19
165.15
158.67
154.73
155.06
160.48
173.39
177.6
186.81
206.96
228.61
236.1
ū1,t
ū3,t Time (h) Base Load
1.09
2.06
2.98
3.87
4.85
6.02
7.59
8.88
10.51
12.94
15.72
17.71
0.29
0.55
0.79
1.03
1.29
1.6
2.02
2.37
2.8
3.45
4.19
4.72
13
14
15
16
17
18
19
20
21
22
23
24
242.18
243.6
248.86
255.79
256
246.74
245.97
237.35
237.31
232.67
195.93
195.6
ū1,t
ū3,t
19.68
21.32
23.33
25.58
27.2
27.76
29.21
29.67
31.15
31.99
28.16
29.34
5.25
5.68
6.22
6.82
7.25
7.4
7.79
7.91
8.31
8.53
7.51
7.82
Table 4: Marginal Costs at Different Generation Levels ($/MWh)
Gen. 1
P1w
¯
100
124
148
172
196
2
Gen. 2
P̄1w mar. cost P2w
¯
124 14.396
10
148 14.588
28
172
14.78
46
196 14.972
64
220 15.164
82
P̄2w mar. cost P3w
¯
28
32.638
10
46
32.674
12
64
32.71
14
82
32.746
16
100 32.782
18
Gen. 3
P̄3w mar. cost
12
14
16
18
20
17.71
17.73
17.75
17.77
17.79
Traditional SCUC Formulation
The objective of the ISOs/RTOs is to minimize the total operation cost to supply the load. It can be
formulated as
XX
min
CiP (Pi,t ) + CiI (Ii,t ),
(22a)
i
t
CiP (Pit )
where
is the fuel cost function in output level Pit for unit i, and CiI (Iit ) is the cost function related
to the unit status Iit . The objective function (22a) is subject to system generation-load balance constraint,
as formulated in (22b), where the total generation equals the total load.
X
X
Pi,t =
dm,t , ∀t,
(22b)
m
i
where dm,t is the load demand at bus m at time t. The power loss is ignored in this paper. The transmission
line flow constraint is modeled as


X
X
−Fl ≤
Γl,m 
Pi,t − dm,t  ≤ Fl , ∀l, t.
(22c)
m
i∈G(m)
2
inj
In the following context, we also denote the net power injection as Pm,t
.
X
inj
Pm,t
:=
Pi,t − dm,t
(23)
i∈G(m)
The power generation is subject to the unit capacity limits (24a), and unit ramping up/down limits (24b-24c).
Ii,t Pimin ≤ Pi,t ≤ Ii,t Pimax , ∀i, t
(24a)
Pi,t − Pi,(t−1) ≤ riu (1 − yi,t ) + Pimin yi,t , ∀i, t
−Pi,t + Pi,(t−1) ≤ rid (1 − zi,t ) + Pimin zi,t , ∀i, t
(24b)
(24c)
where Ii,t , yi,t , and zi,t are the indicators of the unit being on, started-up, and shutdown, respectively. Units
also respect the minimum on/off time constraints which are related to these binary variables [1].
3
Detailed Formulation for Problem (RSCUC)
(RSCUC)
XX
min
(x,y,z,I,P )∈F
t
CiP (Pi,t ) + CiI (Ii,t )
i
(22b), (22c), (24a) − (24c), minimum on/off time limit
s.t.
and
n
F := (x, y, z, I, P ) : ∀ ∈ U, ∃∆P such that
X
X
m,t , ∀t, ,
∆Pi,t =
(25a)
m
i
Ii,t Pimin ≤ Pi,t + ∆Pi,t ≤ Ii,t Pimax , ∀i, t
(25b)
−Rid (1 − zi,t+1 ) ≤ ∆Pi,t ≤ Riu (1 − yi,t ), ∀i, t
X
inj
∆Pm,t
=
∆Pi,t − m,t , ∀m, t
(25c)
(25d)
i∈G(m)
−Fl ≤
X
o
inj
inj
Γl,m (Pm,t
+ ∆Pm,j
) ≤ Fl , , ∀l, t .
(25e)
m
The basic idea of the above model is to find a robust UC and dispatch for the base-case scenario. The
UC and dispatch solution are immunized against any uncertainty ∈ U. When uncertainty occurs, it
is accommodated by the generation adjustment ∆Pi,t (25a). Generation dispatch is also enforced by the
capacity limits (25b). Equation (25c) models the ramping rate limits of generation adjustment ∆Pi,t . In
fact, the right and left hand sides of (25c) can correspond to a response time ∆T , which is similar to the
10-min or 30-min reserves in the literatures [2]. (25e) stands for the network constraint after accommodating
the uncertainty.
4
Detailed Formulation for Problem (MP) and (SP)
(MP)
XX
min
(x,y,z,I,P,∆P )
t
CiP (Pi,t ) + CiI (Ii,t )
i
S.T. (22b), (22c), (24a)-(24c), minimum on/off time limit
X
X
k
∆Pi,t
=
km,t , ∀t, ∀k ∈ K
(26a)
m
i
k
Ii,t Pimin ≤ Pi,t + ∆Pi,t
≤ Ii,t Pimax , ∀i, t, ∀k ∈ K
(26b)
k
∆Pi,t
(26c)
≤
Riu (1
− yi,t ), ∀i, t, ∀k ∈ K
3
k
−∆Pi,t
≤ Rid (1 − zi,t+1 ), ∀i, t, ∀k ∈ K
X
inj
inj,k
−Fl ≤
Γl,m (Pm,t
+ ∆Pm,t
) ≤ Fl , ∀k ∈ K, ∀l, t
(26d)
(26e)
m
inj,k
∆Pm,t
=
X
k
∆Pi,t
− km,t , ∀m, t, ∀k ∈ K,
(26f)
i∈G(m)
and
(SP) Z:= max
min
(s+ ,s− ,∆P )∈R()
∈U
XX
−
(s+
m,t + sm,t )
m
n
R() := (s+ , s− , ∆P ) :
X
X
−
∆Pi,t =
(m,t + s+
m,t − sm,t ), ∀m, t
(27b)
(27c)
m
i
−Fl ≤
(27a)
t
X
inj
inj
Γl,m Pm,t
+ ∆Pm,t
≤ Fl , ∀l, t
(27d)
X
(27e)
m
inj
∆Pm,t
=
−
∆Pi,t − (m,t + s+
m,t − sm,t )
i∈G(m)
−
s+
m,t , sm,t ≥ 0, ∀m, t
o
(25b), (25c)
(27f)
where K is the index set for uncertainty points ˆ which are dynamically generated in (SP) with iterations.
k
It should be noted that ˆk is the extreme point of U. Variable ∆Pi,t
is associated with ˆk . The objective
−
function in (SP) is the summation of non-negative slack variables s+
m,t and sm,t , which evaluates the viola+
−
tion associated with the solution (x, y, z, I, P ) from (MP). sm,t and sm,t are also explained as un-followed
uncertainties (generation or load shedding) due to system limitations.
5
Lagrangian Function for Problem (RSCED)
L(P, ∆P, λ, α, β, η)
XX
X X
XX
X
CiP (Pi,t ) +
=
λt
dm,t −
Pi,t +
β̄i,t (Pi,t − Iˆi,t Pimax ) + βi,t (Iˆi,t Pimin − Pi,t )
¯
t
t
m
t
i
i
i
XX
u
min
d
min
+
ᾱi,t Pi,t − Pi,t−1 − ri (1 − ŷi,t ) − Pi,t ŷi,t + αi,t Pi,t−1 − Pi,t − ri (1 − ẑi,t ) − Pi,t ŷi,t
¯
t
i
X X X
XX
X
X
X
inj
inj
k
+
η̄l,t
Γl,m Pm,t
− Fl − ηl,t
Γl,m Pm,t
+ Fl
+
λkt
km,t −
∆Pi,t
¯
t
m
m
m
i
l
k∈K t
X XX
k
k
max
k ˆ
min
k
ˆ
+
β̄i,t (Pi,t + ∆Pi,t − Ii,t Pi ) + βi,t (Ii,t Pi − Pi,t − ∆Pi,t )
¯
i
k∈K t
X XX
k
k
u
k
k
d
+
ᾱi,t ∆Pi,t − Ri (1 − ŷi,t ) − αi,t ∆Pi,t + Ri (1 − ẑi,t )
¯
i
k∈K t
X
X
X XX
inj
inj,k
inj
inj,k
k
k
+
η̄l,t
Γl,m (Pm,t + ∆Pm,t ) − Fl − ηl,t
Γl,m (Pm,t + ∆Pm,t ) + Fl
(28)
¯
m
m
k∈K t
l
6
6.1
Proofs for lemmas and theorems
Proof of Lemma 1
u,k
Proof. Consider ˆkm,t > 0, πm,t
< 0. With a small perturbation δ > 0 to ˆkm,t , we replace ˆkm,t with ˆkm,t − δ
u,k
in (RSCED). As the πm,t < 0 , then the optimal value to problem (RSCED) increases. It means that there
4
are violations for the original optimal solution Pi,t to problem (RSCED) with ˆkm,t − δ. Hence, the optimal
solution Pi,t to problem (RSCED) cannot be immunized against the uncertainty ˆkm,t − δ. It contradicts
u,k
with the robustness of the solution Pi,t . Therefore, if ˆkm,t > 0, then πm,t
≥ 0. Similarly, if ˆkm,t < 0, then
u,k
πm,t ≤ 0.
6.2
Proof of Lemma 2
Proof. Assume i ∈ G(m), according to the KKT condition
∂L(P, ∆P, λ, α, β, η)
= 0,
k
∂∆Pi,t
(29)
we have
k
k
k
k
β̄i,t
− βi,t
+ ᾱi,t
− αi,t
− λkt +
¯
¯
X
k
k
(η̄l,t
− ηl,t
)Γl,m = 0.
¯
l
(30)
u,k
k
k
k
k
k
k
Then (17) holds. If πm,t
> 0, then β̄i,t
+ ᾱi,t
> 0 as β̄i,t
, βi,t
, ᾱi,t
, and αi,t
are non-negative. According to
¯
¯
k
the complementary conditions for (7b) and (7d), at least one of (7b) and (7d) is binding. Hence, ∆Pi,t
=
u,k
max
u
ˆ
min{Ii,t Pi
− Pi,t , Ri (1 − ŷi,t )}. Similarly, the other equation holds when πm,t < 0.
6.3
Proof of Theorem 1
Proof. The congestion fee at t is
X
m
X
inj
e
e
πm,t
Pi,t =
πm,t
Pm,t
X
e
πm,t
dm,t −
m
i∈G(m)
inj
k
− ηl,t −
ηl,t
Pm,t
¯
¯
m
l
k∈K
k∈K
X
X
X
k
k
=
η̄l,t +
η̄l,t + ηl,t +
ηl,t (Fl − ∆fl,t )
¯
l
k∈K
k∈K ¯
=
XX
Γl,m η̄l,t +
X
k
η̄l,t
X
The first equality
holds following the definition of net power injection. The second equality holds according
P
P k
inj
to (8) and
P
m m,t = 0. The third equality holds following (19). The sign change of ηl,t and
l ηl,t
¯ FTR holders
¯
in
the
third
equation
is
because
of
the
definition
of
power
flow
direction.
The
credits
to
P
e
e
(m→n) (πm,t − πn,t )FTRm→n can be rewritten as
X X
m→n
l
!
X
k
k
(Γl,n − Γl,m ) η̄l,t − ηl,t +
η̄l,t
− ηl,t
FTRm→n
¯
¯
k∈K
Two cases are considered as follows.
1. If ∆fl,t is non-zero, then η̄l,t and ηl,t must be zeros. In this case, the congestion fee related to l at t is
¯
X
k
k
(η̄l,t
+ ηl,t
)(Fl − ∆fl,t ).
(31)
¯
k∈K
And the credits to FTR holders are
X
e
e
(πm,t
− πn,t
)FTRm→n
(m→n)
=
X XX
(m→n)
l
k
k
(Γl,n − Γl,m )(η̄l,t
− ηl,t
)FTRm→n
¯
k∈K
XX
k
k
≤
(η̄l,t
+ ηl,t
)Fl
¯
l k∈K
5
The first equality holds following η̄l,t = ηl,t = 0. The inequality is true as the amount of FTRm→n
¯
respects
X
−Fl ≤
(Γl,m − Γl,n )FTRm→n ≤ Fl .
m→n
P
k
according to the SFT for FTR market [3–5]. Hence, the maximal credit to FTR holders is k∈K (η̄l,t
+
k
ηl,t )Fl for line l at t. Comparing with the congestion fee in (31), the FTR underfunding is (20).
¯
2. If ∆fl,t is zero, the congestion fee related to l at t is
X
k
k
(η̄l,t + ηl,t )Fl +
(η̄l,t
+ ηl,t
)Fl ,
¯
¯
k∈K
which is the same as the maximal FTR credit. Hence, FTR underfunding is 0 at t for l.
6.4
Proof of Theorem 2
Proof. According to Theorem 1, the FTR underfunding value is (20). Therefore, we need to prove that the
money collected from uncertainty sources can cover the FTR underfunding and credits to generation reserve.
Without loss of generality, we consider the payment collected from uncertainty sources at time t for ˆk
X u,k
πm,t ˆkm,t
m
X
X
k
k
k
=
λkt −
Γl,m η̄l,t
− ηl,t
ˆm,t
¯
m
l
X
XX
X
k
k
k
k k
Γl,m η̄l,t
− ηl,t
(
∆Pi,t
)
=
∆Pi,t
λt −
¯
m
i
l
i∈G(m)
X
k
k
+
(η̄l,t
+ ηl,t
)∆fl,t
¯
l
X X u,k
X
k
k
k
=
πm,t ∆Pi,t
+
(η̄l,t
+ ηl,t
)∆fl,t
¯
m i∈G(m)
l
X u,k
X
k
k
k
=
πmi ,t ∆Pi,t
+
(η̄l,t
+ ηl,t
)∆fl,t
¯
i
l
The first equality holds according to (9). According to (7a), (7f), and (7g), the
the second line can be rewritten as
X
X
XX
k
k
k
Γl,m η̄l,t
(∆Pi,t
+ Pi,t ) − dm,t −
η̄l,t
Fl
m
l
=
=
XX
m
m
k
Γl,m η̄l,t
l
l
X
m
l
k k
Γl,m η̄l,t
ˆm,t in
l
i∈G(m)
XX
P P
k
∆Pi,t
+
X
k
η̄l,t
X
X
+
m
l
i∈G(m)
k
k
Γl,m η̄l,t
(
∆Pi,t
)
i∈G(m)
inj
Γl,m Pm,t
− Fl
X
k
η̄l,t
∆fl,t
l
Hence, the second equality holds. The third equality holds from (9). Therefore,
XX
XX
XX
Ψm,t =
ΘG
ΘTl,t ,
i,t +
m
t
i
t
l
t
the payment collected from uncertainty sources can cover all the credits to the generation and transmission
reserves.
6
6.5
Proof of Competitive Equilibrium
down
Proof. Pi,t and (Qup
) are coupled by constraints (15) and (16). According to (17), we can rewrite
i,t , Qi,t
generation reserve credit as
X u,k
u,up up
u,down down
k
πm,t
Qi,t + πm,t
Qi,t =
πm,t ∆Pi,t
k∈K
=
X
k∈K
=
X
k∈K
k
(β̄i,t
−
k
βi,t
¯
+
k
ᾱi,t
k ˆ
β̄i,t
(Ii,t Pimax
k
+ᾱi,t
(Riu (1
−
k
k
αi,t
)∆Pi,t
¯
!
k
− Pi,t ) + βi,t
(Pi,t − Iˆi,t Pimin )
¯k d
− ŷi,t )) + αi,t
Ri (1 − ẑi,t+1 )
¯
(32)
down
Substituting (32) into problem (PMPi ), we can decouple Pi,t and (Qup
). In fact, we also get all terms
i,t , Qi,t
related to Pi,t in Lagrangian L(P, λ, α, β, η) for problem (RSCED). Since the problem (RSCED) is linear
programming problem. Therefore, the saddle point P̂i,t , which is the optimal solution to (RSCED), is also
the optimal solution to (PMPi ). Consequently, unit i is not inclined to change its generation output level
as it can obtain the maximum profit by following the ISO’s dispatch instruction P̂i,t . Therefore, dispatch
u,k
e
, πm,t
) constitute a competitive partial equilibrium [6].
signal P̂i,t and price signal (πm,t
7
Stoarge Model in Case 2 for IEEE 118-Bus
Et = Et−1 + ρd PtD + ρc PtC , ∀t
0 ≤ Et ≤ E max , ∀t
0 ≤ −PtD ≤ ItD RD , ∀t
0 ≤ PtC ≤ ItC RC , ∀t
ItD + ItC ≤ 1, ∀t
ENT = E0 ,
where Et denotes the energy level, PtD and PtC represent the discharging and charging rates, and ItD and
ItC are the indicators of discharging and charging. As the UMP is the major concern in this section, we use
simplified parameters for storage. The discharging efficiency ρd and charging efficiency ρc are set to 100%.
The capacity E max and initial energy level E0 are set to 30 MWh and 15 MWh, respectively. The maximal
charging rate RD and discharging rate RC are set to 8 MW/h.
References
[1] M. Shahidehpour, H. Yamin, and Z. Li, Market Operations in Electric Power Systems: Forecasting,
Scheduling, and Risk Management, 1st ed. Wiley-IEEE Press, 2002.
[2] Z. Li and M. Shahidehpour, “Security-constrained unit commitment for simultaneous clearing of energy
and ancillary services markets,” IEEE Trans. Power Sys., vol. 20, no. 2, pp. 1079–1088, 2005.
[3] W. W. Hogan, “Financial transmission right formulations.”
[4] “Pjm manual on financial transmission rights,” PJM, Tech. Rep.
[5] W. W. Hogan, “Contract networks for electric power transmission,” Journal of Regulatory Economics,
vol. 4, no. 3, pp. 211–242, 1992.
[6] A. Mas-Colell, M. D. Whinston, J. R. Green et al., Microeconomic theory. Oxford university press New
York, 1995, vol. 1.
7