Linearized optimal power flow

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1 Linearized optimal power flow. Some introductory comments The advantae of the economic dispatch formulation to obtain minimum cost allocation of demand to the eneration units is that it is computationally very fast and reasonably easy to solve. However, it suffers from two main drawbacks:. It is somewhat inaccurate due to loss approximation and due to implicit assumption that transmission has infinite capacity;. It provides no information about line flows. The optimal power flow (OF) compensates for these drawbacks by replacin the one power balance equation with the nodal power flow equations. The enerator limits are retained as inequality constraints, but we add to the inequality constraints the constraints on line flows. We desire to brin network constraints into the problem in order to study the influence of transportation (transmission) constraints on the most economic distribution of eneration. There are a number of ways to solve the problem, includin the followin. Linearized Nonlinear, eneralized reduced radient Nonlinear, penalty function approach. In these notes, we will focus on the linearized approach. It is very attractive to do so because linear optimization problems may be solved usin linear prorammin (L). L has been under development now for over 5 years, and extremely efficient alorithms have been developed so that linear prorams, even very lare ones, may be solved quickly. In addition, there are available software packaes which make L solutions relatively easy to develop. We will not have time to present L in this course, but there are courses that do (see IE and IE 5). In addition, the appendix of Chapter 6 in W&W overviews a method.

2 The OF has been a tool for many years in the power enineerin community, and has been effective in solvin the followin kinds of problems (see Section. of W&W): Minimizin dispatch cost while meetin transmission constraints (this is the classical OF) Security-constrained OF (corrective and preventative) Voltae-var optimization Maximum power transfer Loss minimization ut it is the advent of electricity markets that has made OF a central, essential part of runnin power systems. There are two comments to be made here: Most electricity markets utilize the simultaneous co-optimized linear prorammed security-constrained OF (SC-L-SCOF). We will focus only on the L-OF in these notes. However, the main difference between L-SCOF and L-OF is that security constraints associated with branch overload are present in the former and not in the latter. The main difference between SC-L-SCOF and L-SCOF is that the ancillary services (reserves) are co-optimized in the former but not in the latter. Electricity markets may allow both offers (to sell) and bids (to buy). We will allow only the former in these notes, but introduction of the latter provides nothin conceptually new. My treatment complements that of Section. in W&W. In linearizin the OF, to maintain simplicity, we will assume that each enerator makes only a sinle-price offer. Thus, each unit is represented in the objective function by a constant times the eneration level for that unit. This approach (s i ), and an alternative (s i, s i, s i ), are illustrated in Fi. below (similar to Fi..6 in W&W).

3 C i $/hr s i s i s i s i i i i i,min Fi. i (MW) So here is the formal statement of our problem: Subject to: min s k k k enerator buses ' ( D A), max,max k enerator buses, max where k k, k k dk, k,... N 6 5 k is index on the units. These are DC power flow equations to represent the network. However, we must include all nodal injections, N and all anles θ θ N in this set of equations. These are equation to et line flows. Aain, we need to include all anles θ θ N in this set of equations. These are the limits on the line flows. Notice that there is only one circuit ratin, but it must be enforced as a limit if the flow is in one direction or in the other. 5 These are the limits on the linear cost curve variables. 6 This relates variables used in the cost curves ( kj ) to variables used in DC power flow equations ( k ).

4 efore we proceed with formulatin this problem, we must make one very important observation. In the DC power flow problem, we found that if we included an equation for all buses in the network, usin all anles in the network, that there was a dependency in our set of equations, and the matrix was not invertable. Now, however, we will include all N equations for our network because we want a Larane multiplier for each bus, in order to know the locational marinal price (LM) of enery. This implies that we must also include all of the anles as our unknowns. This will not create the same problem of matrix sinularity that we had before because our overall system of equations will include (), (), and (6). In fact, this system will be under-constrained (there will be more unknowns than equations). This is acceptable because we are not tryin to solve them (because they are underconstrained, there are many solutions); rather, we are tryin to minimize a function that is subject to them.(with appropriate convexity requirements, there is only one solution that will do this). Therefore we will include in = θ all DC power flow equations, one for each node, as a function of all anles in the network. So be aware that in all of what follows, the vector includes the reference bus injection, the vector θ includes the reference bus anle, the matrix is N N (i.e., it does not have a row and column eliminated correspondin to the reference bus), and adjacency (node-arc) matrix D is M N (i.e, we do not eliminate the column correspondin to the reference bus). To solve this problem as a linear proram, we need to be able to write it in a form that a standard L solver will handle. This means equality and inequality constraints must be written as a function of a sinle vector of unknowns (the solution vector ).

5 Special note on Larane multipliers and dual variables: We concluded in our notes on asics of Optimization that: It is important to understand the sinificance of μ and λ. The optimal values of the LaGranian Multipliers are in fact the rates of chane of the optimum attainable objective value f(x) with respect to chanes in the riht-hand-side elements of the constraints. Economists know these variables as shadow prices or marinal values. This information can be used not only to investiate chanes to the oriinal problem but also to accelerate repeat solutions. The marinal values λ j or μ k indicate how much the objective f(x) would improve if a constraint b j or c k, respectively, were chaned. It is also true that, for Ls: The coefficients of the slack variables in the objective function expression of the final tableau (dual variables) ive the improvement in the objective for a unit increase in the riht-hand-sides of the correspondin constraints. Thus we see that the dual variables we were discussin for Ls are the same as the Larane multipliers. Special note on Larane multipliers for equality constraints: We saw in our notes called LSimplex that equality constraints may be converted into two inequality constraints via the followin approach: h( x) c h( x) c h( x) c and we may then reverse the sin of the second inequality, resultin in: h( x) c h( x) c h( x) c This means we may include all of our equality constraints h(x)=c from our eneral form () in our inequality constraints (x)< b. 5

6 However, we should also comment about Larane multipliers for equality constraints. As stated in [ i, p. 6-7], ecause an equality constraint is represented as two inequality constraints, we will et two dual variables, both nonneative, havin the same coefficients but opposite sins. Instead of treatin these two variables separately, the difference can be treated as a variable unrestricted in sin. Hence the dual of an equality constraint is a variable unrestricted in sin. Reference [i] ives a nice example illustratin this concept. Our main take-away is that equality constraints also contribute dual variables. Special note on L solver: The below formulation was developed for use in Matlab. However, this formulation may also be implemented in CLEX. To do so, you need to realize that CLEX is quite flexible in handlin input constraints. The below example illustrates this flexibility: Equality constraints Less-than inequality constraint Less-than, reater-than inequality constraints Greater-than inequality constraint You will find a reat deal of additional information on how to use 6

7 CLEX at: and Formulation of the linearized OF We define the followin solution vector for our problem as: x where is the vector of eneration increments [ k ] T for all bus k that has eneration. is the vector of line flows [ b b bm ] T, M is # of branches. θ is the vector of bus anles, in radians [θ θ θ N ], N is # of buses. We want to capture all of our equality constraints in a sinle matrix relation. There are two kinds of equality constraints: one due to line flows ( D A) () and the other due to injections: ' () We rewrite these slihtly modified to make them more convenient to embed in a sinle matrix equation. ( D A) () () 7

8 ' (5) The inequality constraints are straihtforward iven our definition of the solution vector. The only issue here is what to use as constraints on the anles? Clearly, all anles must reside between π radians and +π radians. Therefore, inequality constraints will be: Example:,min,max,max,max We illustrate usin an example that has units connected to different buses in a bus network supplyin load at different buses. This system is very similar to that of Fi.. in your text. The one-line diaram for the example system is iven in Fi.. We will modify the load so that it has a total of.787 per unit (or 7.87 MW), with per unit load at bus ( d =.) and.787 per unit load at bus ( d =.787). (6) 8

9 y =-j y =-j y =-j d =pu y =-j y =-j d =.787pu Fi. : One line diaram for example system The offers and correspondin min and max eneration limits are as follows: with K, 5 ( ) s ( ) s ( ) s K, K, 5 8 s =.7 $/MWhr s =. $/MWhr s =.5 $/MWhr A modification is necessary at this point in that we need to chane the eneration variables to per-unit. This is necessary in order to brin in the transmission equations, which are in per-unit. Thus, the decision variables,, and, are all divided by. We can compensate for this in terms of obtainin the correct evaluation for the cost by multiplyin the offers by. Thus, we will use: 9

10 s =7 $/pumwhr s = $/pumwhr s =5 $/pumwhr Objective function: We explicitly write the solution vector; then we will be able to write the objective function immediately. 5 x Now we see that the objective function is iven by:

11 5 5 7 ) ( x Z Equality constraints: The equality constraints are iven in eqs. () and (5), repeated here for convenience: ) ( A D () ' (5) We will build all of these equality constraints into a matrix form of A eq x=b eq. We bein by notin dimensions. Columns: Since the solution vector x is, A eq must have columns in order to pre-multiply x. Rows: Since there are 5 branches, eq. () will contribute 5 rows to A eq. Since there are buses, eq. (5) will contribute rows to A eq. So A eq will have total of 9 rows. Therefore, the dimensions of A eq will be 9. We bein with the line flow equations, eq. (). The D matrix is:

12 D The node-arc incidence matrix, A, is: - - A The D A product required by eq. () is then iven by: - - A D So based on eq. () and the solution vector, we can see that these elements will occupy the upper riht hand corner of A eq. So that will take care of the last columns in the first 5 rows. ut what about the first 8 columns? These are the elements in the line flow equations that multiply the variables,,,,,,, 5. Since we do not use the eneration variables within the line flow equations, the first columns of these top 5 rows will be zeros. The next 5 columns in these top 5 rows (columns -8) will also be zeros, except the one element in each of

13 these rows that multiplies the correspondin line flow variable, and that element will be -. Finally, with respect to these top 5 equations, eq. () indicates that the riht-hand-side will be for each of them. Thus, we can now write down all elements in the first 5 rows of our matrix, as follows: 5 eq x A Now we need to write the last equations. These are the DC power flow equations correspondin to eq. (5). Aain, we must remember that the solution vector contains all anles, and therefore the DC power flow matrix needs to be a. This aumented DC power flow matrix is iven below:

14 ' So based on eq. (5) and the solution vector, we can see that this matrix will occupy the lower riht hand side of the A eq matrix. So that will take care of the last columns in the bottom rows. The resultin matrix appears as: 5 eq x A Once aain, we need to consider the first eiht columns. Columns -8 correspond to the line flow variables, which do not appear in the DC power flow equations, so these will be zero.

15 5 5 eq x A The first three columns multiply the eneration variables,, and. However, the DC power flow equations, eq. (5), require the neative of the injections for all buses, and the injections are the eneration minus the load, i.e., k - dk. We do not have load variables dk included in the solution vector. In addition, we do not have eneration for bus (it is just a load bus) included in the solution vector. So what do we do? The answer to this lies in reconizin that the variables we do not have in the solution vector, d, d, d, d, and, are not (when the electricity market does not allow demand bids) variables at all! In fact, they are known quantities, constants, iven by: d =, d =., d =.787, d =, = Since these are constants, they can o to the riht-hand side. ut with what sin? Eq. (5) ' (5)

16 6 indicates that the injection, if modeled on the left-hand-side, should be neative, i.e., on the left-hand side, k - dk should be neative. So we should see on the left-hand-side k + dk. ut now we will take the load term onto the riht-hand-side by subtractin it from both sides. Thus, we see that the load term should show up on the riht-handside as a neative number. This same loic also shows us that the elements multiplyin the eneration terms in the A eq matrix should be -. So we are now prepared to complete the matrix relation for the equality constraints eq x A Inequality constraints: The inequality constraints are simple, as iven in what follows:

17 Solution by Matlab: The code for solvin this linear proram usin Matlab is iven below:

18 %Load is system load plus losses Load=.787; %uild objective function vector. c=[7 5 ]'; %uild Aeq matrix for equality constraints. Aeq=[ - -; - - ; - - ; - - ; - - ; ; - - -; ;]; %uild riht-hand side of equality constraint. beq=zeros(9,); beq(7)=-; beq(8)=-.787; %uild upper and lower bounds on decision variables. L=[ pi -pi -pi -pi]'; U=[ pi pi pi pi]'; [X,FVAL,EXITFLAG,OUTUT,LAMDA]=LINROG(c,A,b,Aeq,beq,L,U); 8

19 The solution vector x is iven by: x , Z=75.8 The solution is provided on the one-line diaram of Fi.. =.5pu =.87pu = =.97 =.955 d =pu =. =.8 =.5pu d =.787pu Fi. : Result in terms of eneration levels and flows 9

20 One can easily check to see that the power is conserved at the buses. The objective function that Matlab provides, is iven above as Z=75.8 $/hr. It is of interest to compare this solution with a solution obtained from an economic dispatch (implyin no representation of transmission). This problem will be Subject to: min This is a L that we can use Matlab or CLEX to solve; however, it is actually a rather trivial solution. You can see that and are the more expensive units, and so we will take as little of those units as possible. Therefore we will assume =.5 =.5 This means that = =.87. The fact that this value of is feasible (between its limits) means that this must be the answer to the problem, since this is the maximum amount of the cheapest unit that we can have. Compare this to the solution we obtained in our LOF, and you see it is the same!!!

21 Should we expect the solutions to be the same, iven the network representation in the LOF approach? The answer is NO, if the flows are at the branch capacities. ut if the flows are all within the branch capacities (and we are not representin losses), then the transmission system has no impact on the dispatch, and in this case, the solution will be identical to the solution that we obtained in the economic dispatch scenario. The LOF solution is for the case where all flows are within their branch capacities, since we have modeled all branch capacities to be 5 pu. In per-unit, this is effectively infinite branch capacity. ASE CASE: Now we will investiate the Larane multipliers for this loadin level, assumin infinite capacity lines. These Larane multipliers, which are the same as the dual variables, are iven in Table. Table : Larane multipliers for d =., d =.787 and infinite transmission capacity ($/per unit-hr) Equality constraints Lower bounds Upper bounds Equation Value* Variable value variable value θ θ. θ θ. θ θ. θ θ.

22 Larane multipliers on the last four equality constraints are very interestin to us, since they ive the improvement in the objective function if we increase the riht-hand-side of the correspondin equation by unit. These are the so-called nodal prices, i.e., the LMs, iven in $/per unit-hr. We see that the numbers are all $/per unit-hr, and if we divide this by the power base of MVA, we et $./MW-hr. We also see that all Larane multipliers on the lower bounds are all zero, with the exception of the ones on and, which are 96 and, respectively, with units of $/per unit-hr. Convertin to $/MW-hr, these values are.96 and., respectively, indicatin the amount of improvement we can expect if we increase the correspondin riht-hand-side of these inequalities by unit. Since these equations look like, for example, - <-.5, an increase in the riht-hand-side corresponds to a decrease in the lower limit. Thus, for, if we move the lower limit from 5 MW to 9 MW, we can expect to improve the objective function by 96 cents per hour. Larane multipliers for all other lower bounds, and for all upper bounds, are zero, since none of the other inequality constraints are bindin. CASE : d =., d =.787, and infinite transmission capacity We now make a sliht modification, by chanin the loadin of bus from. to., an increase of. per unit or MW. The chanes necessary to the Matlab code are: %uild riht-hand side of equality constraint. It will be vector of zeros %except for element in first row, which is load-sum of minimum eneration beq=zeros(9,); beq(7)=-.; beq(8)=-.787; The solution is almost the same as in the base case, except for the objective function now evaluates to Z=77.9. The previous

23 objective function was Z=75.8, an increase of. $/hr, confirmin our understandin of the meanin of the nodal prices. We find that a MW increase in the loadin anywhere in this network has the same effect on the objective function, to increase it by.$/hr. Thus, nodal prices have no locational variation in this network. Any network with infinite transmission capacity will have this attribute (assumin no losses), since the transmission system effectively makes the entire system look like a sinle bus. Case : d =., d =.787, and. capacity constraint on In this case, we maintain loadin at the same levels as the base case, where we saw the flows as in Fi. 5: =.5pu =.87pu = =.97 =.955 d =pu =. =.8 =.5pu d =.787pu Fi. 5: ase case flows We observe the flow on branch, =.. So let s consider that this branch has capacity of.. This means that upper and lower bounds on should be chaned from -5,5 to -.,.. The resultin solution is iven in Fi. 6:

24 =.5pu =.8pu = =.97 =.97 d =pu =. =.59 =.98pu d =.787pu Fi. 6: Cases flows In comparin the previous two diarams, we observe that The flow on branch is constrained to. as desired. The flows all over the network have chaned. The eneration levels at buses and have chaned. Thus, the activation of a transmission constraint has chaned the dispatch. This will affect the enery prices! This conclusion can be verified by lookin at the Larane multipliers, iven in the table below.

25 Table : Larane multipliers for d =., d =.787 and infinite transmission capacity except for and. capacity constraint on ($/per unit-hr) Equality constraints Lower bounds Upper bounds Equation Value* Variable value variable value θ. θ. θ. θ. θ. θ. θ. θ. Some comments about the Larane multipliers in Table :. Generation limits: We still see a non-zero Larane multiplier on the lower limit, as before, but the Larane multiplier on the lower limit has become zero, reflectin that had to increase and come off of its lower limit to compensate for the decrease in necessary to redispatch around the constraint.. ranch limits: The Larane multiplier on the upper bound is 86, and after dividin by to chane from per-unit to MW, it is.86 $/MW-hr, reflectin the improvement in objective function that can be obtained from increasin the branch limit by MW (from. per-unit to. per-unit).. ranch flows: The Larane multiplier on the flow (the equality constraint) is.86* =86, and after dividin by to chane from per-unit to MW, it is.86 $/MW-hr, reflectin the improvement in objective function that can be obtained from 5

26 increasin the flow in this branch by MW (from. per-unit to. per-unit). Observe that the Larane multiplier on the flow is the same as the Larane multiplier on the branch limit.. Nodal prices: The Larane multipliers on the equality constraints correspondin to the nodes are the nodal prices. Without transmission constraints, these prices were all the same, at. $/MW-hr, a price set entirely by the enerator at bus since it was the bus enerator that responded to any load chane. ut now they are all different, with only bus price at. $/MW-hr. This difference reflects that, because of the transmission constraint, a load increase at one bus will incur a different cost than a load increase at another bus. Comment # is worth investiatin further. Let s increase the load at the hihest price bus, bus #, from.787 to.887 per unit, an increase of MW. The resultin dispatch and flows are shown in Fi. 7. In order to ain intuition into what has happened, we have repeated Fi. 6 just below it so as to provide a convenient comparison. 6

27 =.5pu =.778pu = -. 5 =. =. d =pu =. =.665 =.59pu d =.887pu Fi. 7: Flows for case with MW increase in d =.5pu =.8pu = =.97 =.97 d =pu =. =.59 =.98pu d =.787pu Fi. 6: Case flows The comparison shows that in order to supply an additional MW at bus, the eneration levels of different units had to be modified. Specifically, Unit was decreased from.8 to.778, a decrease of.5 per unit (.5 MW) and Unit was increased from.98 per unit to.59 per unit, an increase of.5 7

28 (.5 MW). Thus, Unit was increased enouh to supply the increased load at bus and the decreased eneration at bus. Question: Why did we not just increase Unit or increase Unit by MW? Answer: ecause the resultin flow on branch would exceed its capacity!!! In fact, it is not possible to supply additional load at bus with only a sinle unit increase. We will always have to compensate for the load AND redispatch to compensate for the additional flow on the branch. As a result, the nodal price at bus is a function of the eneration costs at those buses that are used in the particular redispatch that achieves the minimum cost. Althouh we only used two different units here, there could be situations where more than two units must be redispatched to supply an additional MW at a bus when conestion is present. How many units must be redispatched depends on the bids of the units; the location of the units relative to the conested line (or lines). It will find the minimum cost redispatch, and so it will tend to utilize the less expensive units. However, it will also tend to use the units that have to move the least amount relative to the solution before the additional MW was added. Thus, it could be the case that a unit bid at $ ets moved up instead of a unit bid at $ if the first unit has location which required that it only move. MW and the second unit has a location which requires that it move.5 MW. [ i ] S.Zionts, Linear and Inteer rorammin, rentice-hall, 97. 8

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