Economic Load Dispatch

Part 4 of this book solved the load flow. Given the generation at every plant and the demand at every bus, Chapters 19 and 20 returned the voltages, the angles and the flows, and told us whether the resulting operating point was acceptable. What that machinery never asked is where the generation schedule came from. It was an input.

Chapter 30 supplied the other half of the input: the load, hour by hour, given to the operator and not chosen by him. Put the two together and a gap appears between them. At three o'clock this afternoon the system will be asked for, say, \(1000\) MW. There are twenty machines synchronised and capable of producing it. There is an enormous number of ways to divide \(1000\) MW among twenty machines, every one of them satisfies the power balance, and a load flow would pronounce most of them perfectly feasible. They do not cost the same. Some of them cost several percent more than others, and several percent of a national fuel bill is a very large sum of money.

Economic load dispatch is the problem of choosing among those feasible schedules. Its statement is short: for a set of generators already running and synchronised, find the outputs \(P_1,\dots,P_N\) that minimise the total cost of fuel while meeting the demand exactly and respecting each machine's operating limits.

Three things are deliberately outside this statement, and it is worth naming them now so that the boundaries of the chapter are clear.

Which machines are running is not decided here. Dispatch takes the set of synchronised units as given. Choosing that set — deciding hours ahead which boilers to light and which to bank — is unit commitment, and it is the subject of Chapter 32. The two problems nest: commitment must repeatedly ask "if I commit this set, what will dispatch cost?", so every commitment study runs an economic dispatch inside it many thousands of times.

Fixed costs do not appear. Chapter 30 separated the cost of a power system into a part that depends on the plant installed and a part that depends on the energy produced. Capital charges, interest, depreciation, insurance, salaried staff — all of these are incurred whether the machine is loaded to \(30\) MW or \(300\) MW, so they cannot influence the split of a given demand. Only the costs that vary with output belong in \(F_i(P_i)\), and in a thermal or nuclear station that means fuel, with a small allowance for output-dependent operation and maintenance.

Dispatch is a fast loop. The demand of Chapter 30 moves continuously, so the answer computed at three o'clock is stale by five past. In a modern control centre the dispatch is re-solved every two to five minutes, and between solutions the automatic generation control of Chapter 33 moves the machines along their governor characteristics to follow the residual error. Dispatch sets the targets; AGC chases them.

A thermal unit is tested by holding it at a steady output and measuring the fuel consumed. Repeating the test across the operating range produces the input–output curve: heat input in kcal/h (or Btu/h, or tonnes of coal per hour) plotted against electrical output in MW. Multiplying the vertical axis by the delivered price of fuel converts it into the fuel cost curve \(F_i(P_i)\) in Rs/h, and it is this curve that dispatch uses.

Two features of the measured curve matter. It does not pass through the origin — a machine held at zero output still burns fuel to keep the boiler hot, the auxiliaries turning and the turbine on turning gear — and it bends upward, because a turbine is most efficient near its design point and its efficiency falls away as it is pushed harder. Over the normal operating band both features are captured well by a quadratic:

Here \(a_i\) is the no-load cost in Rs/h, \(b_i\) has units of Rs/MWh and \(c_i\) of Rs/MW²h. The condition \(c_i>0\) is not cosmetic; it makes \(F_i\) convex, and Section 31-3 will show that convexity is precisely what guarantees the rule we are about to derive finds a minimum rather than some other stationary point.

The quantity that governs the whole subject is the slope of this curve, not its height and not its average.

The distinction between incremental and average cost is the single most common source of error in this subject, so it deserves a table. Take a unit with \(F = 250 + 8.0P + 0.004P^{2}\) Rs/h and an operating range of \(50\) to \(250\) MW.

The two columns move in opposite directions over most of the range. Average cost falls as the no-load cost \(a_i\) is spread over more megawatt-hours; incremental cost rises because the machine grows less efficient as it is loaded. They meet at one point, and only one. Setting \(d(F/P)/dP = 0\) on \(F/P = a/P + b + cP\) gives \(-a/P^2 + c = 0\), so the average cost is least at

That average and incremental cost coincide at the point of best efficiency is a general fact, not an accident of these numbers: a function's average equals its marginal exactly where the average is stationary. It will reappear in Chapter 32, where the full-load average cost is the number used to rank machines for commitment, precisely because commitment is a decision about whole machines and not about marginal megawatts.

Real machines depart from the smooth quadratic in one important way. A steam turbine admits steam through several governor valves opened in sequence, and each time a new valve begins to open the throttling loss jumps, putting a small ripple — the valve-point loading effect — on the incremental cost line. A unit fired on two fuels shows a step where the fuels change over. Both destroy convexity locally and both are ignored in the classical treatment that follows; they are the reason production-grade dispatch programs use piecewise-linear incremental cost curves rather than a single quadratic.

Before any calculus, an argument that needs none. Two units share a fixed demand \(P_D\), so \(P_1+P_2=P_D\) and any change must satisfy \(dP_2=-dP_1\). Suppose at the present split unit 1 has the lower incremental cost, \(dF_1/dP_1 < dF_2/dP_2\). Move a small amount \(\Delta P\) of output from unit 2 to unit 1. Unit 1's cost rises by \((dF_1/dP_1)\Delta P\); unit 2's falls by \((dF_2/dP_2)\Delta P\). The net change is

The total cost has gone down, and it will keep going down for as long as the two incremental costs differ, because the transfer is always profitable in the direction of the cheaper margin. The process can only stop when there is no difference left to exploit. That is the whole idea; everything that follows is bookkeeping to extend it to \(N\) units and to handle the cases where the transfer runs into a wall.

The formal derivation uses a Lagrange multiplier. We minimise \(F_T=\sum_i F_i(P_i)\) subject to the single equality constraint \(\phi = P_D - \sum_i P_i = 0\), so we form the Lagrangian

The first set of \(N\) equations says that every unit's incremental cost equals the same number \(\lambda\); the last equation is the power balance we started with. Together they are \(N+1\) equations in the \(N+1\) unknowns \(P_1,\dots,P_N,\lambda\).

Stationarity alone does not prove a minimum, and here the second-order test is easy. With \(P_N\) eliminated by the constraint, the Hessian of \(F_T\) in the remaining \(N-1\) variables has entries \(F_i''\delta_{ij}+F_N''\), and since every \(F_i''=2c_i>0\) that matrix is positive definite. The cost surface is a bowl, the stationary point is its bottom, and it is unique. This is why the assumption \(c_i>0\) in Section 31-2 mattered: strip it away — allow the valve points and fuel changeovers back in — and the surface acquires local minima that the equal-\(\lambda\) rule cannot distinguish.

The multiplier \(\lambda\) is not an artefact of the algebra. Differentiate the optimal cost with respect to the demand and the constraint terms cancel, leaving \(dF_T^{*}/dP_D = \lambda\). The multiplier is the system incremental cost — the rupees per hour it costs to serve one more megawatt, evaluated at the optimum. It is the number displayed on the wall of every load despatch centre, the number that decides whether an import from a neighbouring region is worth taking, and, in a market, the number that clears the energy price. Chapter 30 argued that a sound tariff should reflect the cost of causing the next unit of consumption; \(\lambda\) is that cost, computed.

For quadratic cost curves the coordination equations can be solved in closed form. Invert each incremental cost relation \(\lambda = b_i + 2c_iP_i\) to express the output demanded of unit \(i\) by a given system incremental cost,

and substitute into the power balance \(\sum_i P_i = P_D\):

One evaluation gives \(\lambda\), and back-substitution gives every \(P_i\). The formula has an instructive shape: \(\lambda\) is a weighted average of the intercepts \(b_i\) plus a term proportional to demand, and the weights \(1/2c_i\) are the flatness of each incremental cost line. A unit with a flat incremental cost — small \(c_i\) — is given a large weight and takes a large share of any change in demand, which is exactly the behaviour one would want from a large base-load machine.

The closed form is a luxury that disappears the moment anything is added to the problem: a generator limit, a loss term, a piecewise-linear cost curve, a valve point. What survives all of these is the observation that the power balance is a scalar equation in the scalar \(\lambda\). Define the mismatch

Every \(P_i(\lambda)\) is non-decreasing in \(\lambda\), so \(g\) is non-decreasing and a root-finder cannot get lost. Bracket the root with a low and a high trial value, then bisect or interpolate. This is the λ-iteration, and it is the backbone of practical dispatch codes.

One refinement makes the iteration converge in very few steps. Between two consecutive values of \(\lambda\) at which no unit hits a limit, every \(P_i(\lambda)\) is linear and therefore \(g\) is linear too, with slope \(\sum_{i\,\text{free}}1/2c_i\) over the currently free units. Linear interpolation between two brackets that share the same set of free units therefore lands on the root exactly, not approximately. In practice a few bisections identify which units are at their limits, and one interpolation then finishes the job — a pattern Example 3 displays step by step.

No machine can be loaded arbitrarily. The upper limit \(P_{i,\max}\) is set by the turbine and generator ratings and by the boiler's maximum continuous rating. The lower limit \(P_{i,\min}\) is less obvious but no less real: below roughly \(30\) to \(40\) per cent of rating a large boiler cannot maintain stable combustion and the correct steam temperature at the superheater outlet without oil support, and the drum-level and flame-stability margins vanish. A hydro set has its own lower limit set by cavitation and rough-running zones. So each unit is confined to

The equal-\(\lambda\) rule may demand an output outside this band, and when it does the rule as stated is simply wrong. The correct generalisation comes from attaching multipliers to the inequalities as well. Writing the Lagrangian with multipliers \(\mu_i^{\max}\ge0\) on \(P_i-P_{i,\max}\le0\) and \(\mu_i^{\min}\ge0\) on \(P_{i,\min}-P_i\le0\), the stationarity condition becomes \(dF_i/dP_i-\lambda+\mu_i^{\max}-\mu_i^{\min}=0\), together with the complementary slackness statements that \(\mu_i^{\max}\) can be non-zero only when \(P_i=P_{i,\max}\), and \(\mu_i^{\min}\) only when \(P_i=P_{i,\min}\). Sorting the three possible cases gives the Kuhn–Tucker conditions for dispatch.

The computational recipe follows immediately and is the classical one taught in every course. Solve the unconstrained problem; if any unit violates a limit, fix the most badly violating unit at the limit it crossed, subtract its fixed output from the demand, and re-solve the reduced problem for the remaining free units. Repeat until no violation remains, then verify the inequality conditions above at each fixed unit. The verification is not a formality: it is possible, when a unit is fixed prematurely, for the re-solved \(\lambda\) to move back across that unit's incremental cost, in which case the unit must be released and the process restarted.

The λ-iteration of Section 31-4 handles all of this automatically, which is why practical codes prefer it. Clamping \(P_i(\lambda)\) to the band at step 3 makes each \(P_i(\lambda)\) a saturating ramp: flat at \(P_{i,\min}\) for small \(\lambda\), linear in the middle, flat at \(P_{i,\max}\) for large \(\lambda\). The sum of such ramps is a continuous non-decreasing piecewise-linear function of \(\lambda\), the root of \(g\) is unique, and the clamped solution automatically satisfies the Kuhn–Tucker inequalities — a unit clamped high has \(b_i+2c_iP_{i,\max}\le\lambda\), which is exactly the condition required.

Everything so far treated the network as a copper plate. That is defensible for the units of a single station, all connected to the same busbar through a few metres of busduct. It is indefensible for a system whose plants are hundreds of kilometres apart, because the megawatt that leaves a remote plant is not the megawatt that arrives at the load.

Consider the extreme case. Plant \(A\) is sited at the coal pit, \(400\) km from the load centre, and generates at an incremental cost of \(9\) Rs/MWh. Plant \(B\) stands beside the load and costs \(10\) Rs/MWh. The equal-\(\lambda\) rule of Section 31-3 says load \(A\) until its incremental cost climbs to meet \(B\)'s. But if \(12\) per cent of \(A\)'s extra output is dissipated as \(I^{2}R\) on the way, then producing one extra megawatt at \(A\) delivers only \(0.88\) MW to the load; to deliver a full megawatt \(A\) must generate \(1/0.88 = 1.136\) MW, at a cost of \(1.136\times 9 = 10.23\) Rs. Plant \(A\), cheaper at the busbar, is dearer at the load. The merit order has reversed, and no amount of care with the fuel cost curves would reveal it.

So the loss must enter the optimisation, and it enters through the constraint rather than the objective. The power balance becomes

with \(P_L\) the total real power loss of the transmission network, which depends on how the generation is distributed and not merely on its total. Two features of this constraint distinguish it sharply from the copper-plate case. It is non-linear, so the closed-form solution of Section 31-4 is gone. And it couples the units to each other through the network, so a change at plant 1 alters the loss attributable to plant 2.

Chapter 16 built the bus admittance matrix and Chapters 19 and 20 solved the load flow, which computes \(P_L\) exactly for a given schedule. Using a full load flow inside a dispatch iteration is what a modern optimal power flow does; the classical treatment instead replaces the network by a compact algebraic surrogate — the loss formula of the next section.

The surrogate we want is an expression for \(P_L\) written purely in terms of the plant outputs \(P_1,\dots,P_N\), with no voltages and no angles in it. The figure above shows the smallest network that generates one. Two plants feed a common node \(t\) through lines of resistance \(R_1\) and \(R_2\), and the node feeds the load through a line of resistance \(R_3\) that carries both currents. The three-phase loss is

Now make the assumption that turns this into an expression in powers: take \(I_1\) and \(I_2\) to be in phase, so that \(|I_1+I_2|^{2}=(I_1+I_2)^{2}\). Expanding and grouping,

Each plant current is related to its output by \(P_i = \sqrt3\,V_i I_i\cos\phi_i\), so \(I_i = P_i/(\sqrt3\,V_i\cos\phi_i)\). Substituting and cancelling the factors of three,

The structure generalises. For \(N\) plants, collecting the outputs into a column vector \(\mathbf{P}=[P_1\;P_2\;\cdots\;P_N]^{T}\),

The general form used in practice carries two further sets of terms, \(P_L = \sum_i\sum_j P_iB_{ij}P_j + \sum_i B_{0i}P_i + B_{00}\), where the linear and constant terms absorb the effect of loads that are not proportional to the total, and of the reference-bus injection. The purely quadratic form above is the one every textbook problem uses and the one we shall use.

It is essential to be honest about what has been assumed to get here, because the B-coefficients are the weakest link in the classical theory:

In consequence a utility computes several \(\mathbf{B}\) matrices — one per season, per load level, per major network configuration — from base-case load flows, and the control centre selects among them. The error in the resulting loss is typically a few per cent of the loss, which is a fraction of a per cent of the demand, and that is accurate enough for dispatch.

One algebraic property of the quadratic form will be used repeatedly. Because \(P_L\) is a homogeneous quadratic in the \(P_i\), Euler's theorem for homogeneous functions gives

This is a free arithmetic check on any B-coefficient calculation, and Example 5 uses it as one.

Return to the optimisation with the loss term restored. Minimise \(F_T=\sum_iF_i(P_i)\) subject to \(P_D + P_L(\mathbf{P}) - \sum_iP_i = 0\). The Lagrangian is

Rearranged, the condition reads

The derivative \(\partial P_L/\partial P_i\) is the incremental transmission loss (ITL) of plant \(i\): the extra megawatts of loss created system-wide by one extra megawatt generated at that plant, with all other plants held fixed. Its reciprocal-style companion is given a name of its own.

The interpretation is the one the numerical illustration of Section 31-6 already suggested. If generating at plant \(i\) adds loss, \(\partial P_L/\partial P_i>0\), then \(L_i>1\) and the plant's incremental cost is inflated before comparison: to deliver a megawatt to the load the plant must generate \(L_i\) megawatts, so it should be charged \(L_i\) times its busbar cost. A plant electrically close to the load has a small ITL and a penalty factor near unity. And a plant so placed that increasing its output actually relieves the network — because it displaces power that was flowing a long way through a heavily loaded corridor — has a negative ITL, hence \(L_i<1\), and its incremental cost is discounted. Negative incremental losses are common at plants embedded in load centres and they are one reason such plants are valuable out of all proportion to their size.

Two limiting cases keep the algebra honest. If a plant is at the load bus its ITL is zero and \(L_i = 1\). And as \(\partial P_L/\partial P_i\to 1\) the penalty factor diverges: at that point an extra megawatt generated is entirely dissipated on the way, the plant can deliver nothing to the load however hard it is pushed, and no finite fuel cost could justify loading it further. Real systems never approach that condition, but the divergence is a useful reminder that \(L_i\) is a delivery efficiency turned inside out.

The coordination equations are non-linear and coupled: \(P_i\) appears in its own incremental cost and, through the loss formula, in every other plant's penalty factor. There is no closed-form solution, but the λ-iteration structure of Section 31-4 carries over unchanged. Substituting the quadratic cost and the quadratic loss into \(dF_i/dP_i=\lambda(1-\partial P_L/\partial P_i)\) gives, for each plant,

For a fixed \(\lambda\) this is a set of \(N\) linear equations in \(P_1,\dots,P_N\), and that is the key to the whole algorithm: the non-linearity is confined to the outer loop on \(\lambda\). Solve the linear system, evaluate \(P_L\) from the loss formula, and test the power balance.

An equivalent and historically common arrangement iterates on the plant outputs directly: hold the penalty factors at their values from the previous iteration, solve the resulting lossless-looking problem with modified intercepts and slopes, then recompute the penalty factors from the new schedule. This converges more slowly than the λ-iteration but requires no matrix solution, and it is the form in which most hand-worked examples — including Example 4 — are set out.

Two remarks on where this leads. First, the loss formula is a stand-in for a load flow, and the obvious improvement is to stop standing in: compute \(\partial P_L/\partial P_i\) from the Jacobian of a Newton–Raphson load flow (Chapter 20) at each iteration. That is exactly what the optimal power flow does, and it additionally enforces bus voltage limits and line flow limits that the classical dispatch cannot see at all. Second, since the penalty factors depend on the schedule, a system operator running a security-constrained dispatch is solving a problem in which the network constraints of Part 4, the machine limits of this chapter and the commitment decisions of Chapter 32 are all coupled — which is why the industry solves them together as a single mixed-integer programme rather than in the sequence this book presents them.

Work in Rs, MW and MWh throughout, and check every answer against the optimality conditions — equal \(\lambda\) for free units, the Kuhn–Tucker inequalities at limits, and \(L_i\,dF_i/dP_i=\lambda\) when losses are present.