Economic Load Dispatch

The form and its three terms. Each is a quadratic in output:

Since \(C\) is in thousand Rs/h and \(P\) in MW, the slope \(dC/dP\) comes out in thousand Rs per MWh, which is rupees per kWh — the natural unit for the whole chapter.

Why \(a\) is not zero. A synchronised unit at zero output still burns fuel:

These are notional — no unit can actually run at zero output — but the constant matters, because it is the cost that disappears when a unit is shut down. That is what makes unit commitment (Problem 18) a different problem from dispatch, and it is the reason the constant is carried through even though it plays no part in the dispatch itself.

Why \(c > 0\), and why it must be for the theory to work:

Physically, a turbine's efficiency falls at high output as steam velocities and throttling losses rise. Mathematically, convexity is what guarantees that the equal-incremental-cost condition of Problem 5 gives a minimum rather than any other stationary point, and that a local optimum is global. Problem 17 examines what happens when the real curve is not convex.

The output range of the station:

With all three units synchronised, any demand between 180 and 620 MW can be met. Below 180 MW at least one unit must be shut down — again Problem 18 — and above 620 MW the station cannot help.

The cost at the extremes, which shows how different the three units are:

Unit 2 is much the cheapest per unit of energy at full load, and unit 1 the dearest — a small old machine against a large modern one. It is tempting to conclude that unit 2 should be loaded first and unit 1 last, and Problem 4 shows why that conclusion, although it happens to be roughly right here, is reached by invalid reasoning.

Where the coefficients come from in practice:

The last is what most utilities actually use, updated annually. The fit is good to a few per cent over the normal operating range, and considerably worse near the minimum, where the physical curve steepens more sharply than a quadratic can follow.

Heat rate is heat input per unit of electrical output:

It is the reciprocal of efficiency in disguise, and it is the quantity engineers actually quote because it can be measured directly from a fuel-flow meter and a wattmeter.

Efficiency follows from the thermal equivalent of a kilowatt-hour:

A heat rate of 2400 kcal/kWh is an efficiency of 35.8%; one of 2000 kcal/kWh is 43%. Modern supercritical coal units reach 2100–2200 kcal/kWh and combined-cycle plant about 1500.

Across the load range:

The heat rate falls, reaches a minimum, and rises again — a shallow U. The unit is least efficient at low load, not at high load, which surprises people who expect efficiency to fall monotonically.

The point of best efficiency is found by minimising HR:

Best efficiency at 70% of rating is entirely typical. The no-load term \(\alpha\) dominates at low output and the curvature \(\gamma\) at high output, and they balance at \(\sqrt{\alpha/\gamma}\).

The crucial distinction: heat rate against incremental heat rate.

The incremental heat rate is higher than the average, and it rises monotonically with output while the average has a minimum. That difference is the entire subject of Problem 4: dispatch is governed by the incremental quantity, and reasoning from the average gives wrong answers.

From heat to cost. The cost curve is the heat curve times the fuel price:

A fuel price of Rs 3835/Mkcal corresponds to coal of 4000 kcal/kg at Rs 15.34/kg — imported coal. Note that every coefficient scales with the fuel price, so a change in fuel cost scales the whole curve and, crucially, scales the incremental cost too. That is what Problem 15 exploits.

Typical heat rates, for calibration:

Note the reciprocal relation to the plant-mix logic of Set 26: the plant with the best heat rate is not always the cheapest to run, because the fuel price differs. A combined-cycle unit at 1500 kcal/kWh burning gas at Rs 6000/Mkcal costs Rs 9.00/kWh in fuel; a coal unit at 2400 kcal/kWh burning coal at Rs 2500/Mkcal costs Rs 6.00/kWh. Efficiency and cost are different questions.

Differentiate each cost curve:

Three straight lines. The quadratic cost curve gives a linear incremental cost curve, which is exactly why the quadratic form is used — it makes the dispatch equations linear and solvable in closed form.

The range each unit covers:

Three observations follow immediately, and each anticipates a later result.

Observation 1 — unit 2's range lies entirely below unit 1's.

Whatever the demand, unit 2 should be fully loaded before unit 1 is loaded above its minimum. The two ranges do not overlap, so no equal-incremental-cost solution involving both units strictly between their limits can exist.

Observation 2 — units 2 and 3 overlap, and units 1 and 3 overlap:

In those two bands of \(\lambda\) the corresponding pair of units genuinely shares load at equal incremental cost. Outside them one or the other is pinned at a limit.

Observation 3 — unit 2's line is much the flattest (\(2c = 0.009\) against 0.030 for unit 1):

Unit 2 responds three times as much as unit 1 to a given change in \(\lambda\). A flat incremental cost curve means a unit that takes most of any load change — and it is the large modern units, with their low \(c\), that do the regulating.

The station's aggregate incremental cost curve is obtained by adding the units horizontally:

Monotonically increasing in \(\lambda\) — which is the property that makes the lambda-iteration of Problem 8 always converge. The curve is piecewise linear with a break wherever a unit reaches a limit.

What incremental cost means physically:

Not the average cost of its output, and not the cost of the last MWh it produced. It is a forward-looking marginal quantity, and it is the only cost concept that answers the question "where should the next megawatt come from?".

The two quantities are different functions, and behave differently:

The average has a minimum at \(P = \sqrt{a/c}\); the incremental rises monotonically. They are equal only at the point of minimum average cost, and nowhere else.

Tabulated for the three units:

The average exceeds the incremental everywhere in this range, because every unit is operating below its minimum-average-cost output. The gap narrows as output rises, and the two would cross at \(\sqrt{a/c}\) — 152.8, 333.3 and 223.6 MW respectively, all outside the units' ranges.

The logical error in the average-cost rule:

But loading it fully means taking its last megawatt at IC = 8.90, while unit 3's first megawatt above minimum costs only 8.16. Those last megawatts of unit 2 are dearer than the first megawatts of unit 3, so the allocation is not optimal — even though unit 2's average cost is lower at every output.

A worked counter-example. Take \(P_D = 350\) MW with only units 2 and 3 running:

The average-cost rule loads unit 2 to 290 MW (leaving unit 3 at its 60 MW minimum) and costs Rs 8.45 thousand per hour more. The optimum backs unit 2 down to 264 MW and raises unit 3 to 86 MW, at which point both units have \(\text{IC} = 8.5760\) Rs/kWh.

Verify the optimum satisfies the rule:

Equal to four decimal places, and the total is 350 MW ✓. Note that unit 2's average cost at 264 MW is 9.28 against unit 3's at 86 MW of 12.54 — the average costs are wildly unequal at the optimum, which is exactly the point.

The general principle, stated as an exchange argument:

Only the incremental costs appear. The average costs, the no-load costs and the units' sizes are all irrelevant to whether a small transfer is beneficial — and the optimum is precisely the point at which no such transfer helps, which requires all the incremental costs to be equal.

Where the average-cost rule is nonetheless used, and legitimately:

The second row is genuinely an average-cost question, because shutting a unit down removes its no-load cost entirely — a discrete change, not a marginal one. That is unit commitment, and it is why Problem 18 uses different reasoning from Problems 5–11.

Suppose an allocation is optimal and consider transferring a small amount \(\Delta P\) from unit \(j\) to unit \(i\). The demand constraint is preserved, and the cost changes by:

to first order in \(\Delta P\). Nothing else changes: the constant terms are untouched, the other units are untouched, and the total output is unchanged.

The argument. If \(\text{IC}_i < \text{IC}_j\), then \(\Delta C < 0\) for positive \(\Delta P\) — the transfer reduces the cost, so the original allocation was not optimal. Contradiction.

Calling the common value \(\lambda\):

The criterion:

\(n\) equations from the criterion plus one demand constraint, in \(n+1\) unknowns (\(n\) outputs and \(\lambda\)). Determinate, and for quadratic cost curves, linear.

The qualification that makes it correct. The argument required the transfer to be possible in both directions. If unit \(j\) is at its lower limit it cannot give up output; if unit \(i\) is at its upper limit it cannot take any:

Read the second line physically: a unit is at its maximum precisely because it is cheap — we would take more if we could. And the third: a unit is held at its minimum because it is dear — we would take less if the minimum allowed. These are the Karush–Kuhn–Tucker conditions, arrived at without invoking them.

Why convexity is essential. The exchange argument shows only that the optimum is a stationary point:

If some \(c_i\) were negative the same condition could identify a maximum, and if the cost curve had several local minima the criterion would be satisfied at each. Problem 17 examines exactly that situation, which arises physically from turbine valve points.

What the criterion does not depend on:

The first row is worth dwelling on. Two units with identical \(b\) and \(c\) but no-load costs of 100 and 1000 are dispatched identically — because once both are running, the no-load cost is sunk. It reappears only when the question is whether to run them at all.

The result is more general than thermal units. The same criterion governs:

The last row is the general economic principle of which economic dispatch is one instance: at an optimum, the marginal return per unit of resource is equal across all uses. Power engineers call it equal \(\lambda\); economists call it equimarginal allocation; the mathematics is identical.

State the problem:

Ignore the inequality limits for now; they are restored in Problem 9. One equality constraint means one multiplier.

Form the Lagrangian:

The sign of the multiplier term is a convention; taking \((P_D-\sum P_i)\) makes \(\lambda\) come out positive, which is convenient because it will turn out to be a price.

Set the partial derivatives to zero:

The first recovers the exchange argument of Problem 5; the second recovers the constraint. \(n+1\) equations in \(n+1\) unknowns.

Solve for quadratic costs. With \(C_i = a_i+b_iP_i+c_iP_i^2\):

A closed-form answer, valid whenever no limit binds. Note its structure: \(\lambda\) is a weighted average of the \(b_i\) plus a term proportional to the demand, with the weights \(1/2c_i\) — so the flattest unit dominates.

Evaluate for the three units of Problem 1 at \(P_D = 450\) MW, ignoring limits:

And the outputs: \(P_1 = 18.23\), \(P_2 = 316.33\), \(P_3 = 115.44\) MW. Two of these violate the limits — \(P_1\) is below 40 and \(P_2\) above 300 — which is why Problems 9 and 10 exist.

The interpretation of \(\lambda\). Differentiate the optimal cost with respect to the demand:

using the fact that all the incremental costs equal \(\lambda\) and the outputs sum to \(P_D\). So:

The result:

This is a general property of Lagrange multipliers — each is the sensitivity of the optimal objective to a relaxation of its constraint — and it is what makes \(\lambda\) useful far beyond the dispatch itself. It is the correct price for a spot market, the correct value to place on a megawatt of loss saved, and the correct threshold for deciding whether to buy an import. Problem 13 develops all three.

Restoring the inequality limits requires the Karush–Kuhn–Tucker conditions:

The complementary-slackness condition in the last line says a multiplier is non-zero only when its constraint is active. Working through the three cases reproduces exactly the inequalities of Problem 5 — \(\text{IC} \le \lambda\) at a maximum, \(\ge \lambda\) at a minimum — which is reassuring but adds nothing the exchange argument had not already given.

Apply the closed form with two units:

The outputs:

Both within limits (80–300 and 60–200) ✓, and summing to 350 MW ✓. Exact round numbers, which is a happy accident of the chosen data.

Verify the criterion directly:

An average of 10.08 Rs/kWh — but note again that neither unit's average cost is anywhere near that, and neither needs to be.

The two-unit result in a memorable form. Eliminating \(\lambda\):

Checking: \(P_2 = [0.016(350)+(7.20-6.20)]/[2(0.0045+0.0080)] = (5.60+1.00)/0.025 = 264\) ✓. Two terms: a share of the demand weighted by the other unit's curvature, plus a fixed offset set by the difference in the linear coefficients.

How the allocation shifts with demand:

While both units are free, unit 2 takes a fixed 64% of every increment, because \(\Delta P_2/\Delta P_D = c_3/(c_2+c_3) = 0.0080/0.0125 = 0.64\). The sharing ratio is constant, set entirely by the curvatures — not by the units' sizes, their limits or their linear coefficients.

Two limiting cases worth recognising:

The second case is a unit with a linear cost curve — constant incremental cost. Such a unit is either fully loaded or at its minimum, never in between, because its incremental cost cannot adjust to match \(\lambda\). That is exactly what a merit-order despatch assumes, and it is why merit order and equal-\(\lambda\) dispatch coincide when all the cost curves are linear.

Why an iterative method is needed at all. The closed form of Problem 6 fails as soon as a limit binds, because the set of units at their limits is not known in advance:

This is a nonlinear function of \(\lambda\) — piecewise linear, with a kink wherever a unit hits a limit. There is no general closed form, but the structure makes iteration trivial.

The key property — monotonicity:

The total output is a non-decreasing function of \(\lambda\). So the equation \(\sum P_i(\lambda) = P_D\) has a solution, it is essentially unique, and any bracketing method finds it. Convergence is guaranteed — a property remarkably few power-system algorithms enjoy.

The algorithm:

Bisection halves the interval each time, so it needs about \(\log_2(\text{range}/\varepsilon)\) iterations — around 20 for a 0.01 MW tolerance on a 500 MW system. The secant method converges in 4–6 but can fail at a kink; bisection never fails, and 20 evaluations of a linear formula cost nothing.

Demonstration at \(P_D = 500\) MW. Bracket \([6.20,\ 12.10]\):

Twenty bisections take the mismatch below 0.01 MW. Note that unit 2 is pinned at 300 MW throughout — the algorithm handles the limit automatically, with no special logic.

Verify the converged answer analytically. With unit 2 at its maximum, units 1 and 3 must supply 200 MW at a common \(\lambda\):

Matching the iteration to five figures. And the consistency check on unit 2: \(\text{IC}_2 = 8.90 \le \lambda = 9.739\) ✓ — a unit at its maximum must have incremental cost below \(\lambda\), which it does.

The secant method, for comparison, using the projected gradient:

Since the aggregate curve is piecewise linear, this converges in exactly one step within a segment and needs one extra step for each limit crossed. Starting from \(\lambda = 9.15\) it reaches the answer in three iterations against bisection's twenty — and this is the method real dispatch software uses, with a bisection fallback for robustness.

Scale in practice:

Plain lambda iteration remains fast enough for real-time use at any scale, which is why it survives inside the automatic generation control loop even where a full optimal power flow runs every five minutes alongside it.

The three cases, established in Problem 5 and re-derived from the KKT conditions in Problem 6:

The inequalities point in the directions common sense demands, which is a useful mnemonic: a unit is pinned at its ceiling because it is a bargain, and at its floor because it is expensive.

The systematic procedure:

Fixing the most-violated unit first is what makes the procedure terminate reliably. Step 6 is not optional: a unit fixed early can turn out, once other limits bind, to belong in the free set after all.

Worked at \(P_D = 450\) MW. Round 1, from Problem 6:

Two violations. Unit 2 exceeds its maximum by 16.33 MW; unit 1 falls short of its minimum by 21.77 MW. Unit 1 is the more violated, so fix \(P_1 = 40\).

Round 2 — units 2 and 3 supply \(450-40 = 410\) MW:

Fix \(P_2 = 300\) as well.

Round 3 — only unit 3 remains free, supplying \(450-40-300 = 110\) MW:

Within limits ✓. The system incremental cost is set by unit 3 alone — it is the only marginal unit.

Step 6 — check the fixed units:

Both inequalities hold, confirming the solution. Note how close unit 2's inequality is: 8.90 against 8.96. Only 0.06 Rs/kWh of margin — a slightly lower demand would free unit 2 from its limit.

Where the limits come from:

The lower limit is the more interesting engineering constraint and the more expensive one. Below about 40% load a coal boiler cannot sustain a stable flame on coal alone and must be supported with oil, which is both costly and a fire risk — so the "technical minimum" is a hard operating constraint, not a preference. Reducing it from 55% to 40% is a substantial retrofit and is now commonly undertaken specifically to accommodate variable renewable generation.

Round 1 — ignore limits:

Unit 2 is violated by 43.18 MW and unit 1 by 13.71 MW, so unit 2 is fixed first this time.

Round 2 — units 1 and 3 supply 200 MW:

Both within limits, and summing with unit 2's 300 MW to exactly 500 MW ✓. Unit 1 is now just above its minimum — it has come off the limit that bound it at \(P_D = 450\).

The complete result:

Note the two free units share the same incremental cost, 9.7391, while their average costs differ by 60% (17.45 against 11.03). That is the equal-incremental-cost rule doing exactly what Problem 4 said it would.

Reading the result. Four observations, each carrying information:

The second row is a capacity-planning signal emerging from an operating calculation: a unit persistently at its maximum is telling the planner that more of it would be worth building — which is the link back to Set 26.

The value of unit 2's constraint. Its KKT multiplier is:

One more megawatt of capability on unit 2 would save Rs 839 per hour, or Rs 73.5 lakh a year if the condition persisted. That is the economic value of relieving the constraint — an uprate, a condenser clean, or a lower cooling-water temperature — and it is available directly from the dispatch without any further calculation.

The dispatch as demand varies through the day. Using the load curve of Set 26 scaled to this station:

Unit 2 does all the load-following until it saturates at 300 MW, after which unit 3 takes over and unit 1 finally comes off its floor. Units 1 and 3 sit at their minimums for two-thirds of the day — which raises the question of whether they should be running at all, and that is Problem 18.

Apply the procedure of Problem 9 at each demand. The full schedule:

Five distinct regimes, separated by the demands at which a limit becomes active or inactive.

The five regimes:

The transition at 317.8 MW is where \(\lambda\) reaches unit 3's minimum incremental cost of 8.16; the one at 446.3 MW is where unit 2 saturates at 8.90; the one at 496.3 MW is where \(\lambda\) reaches unit 1's 9.70; and the last, at 563.3 MW, is where unit 3 saturates at 10.40.

The \(\lambda\) curve is the station's supply curve. Plotting \(\lambda\) against \(P_D\):

The slope is a staircase, and it jumps upward whenever a unit saturates and downward whenever one comes free. The steepest region above 563 MW, where unit 1 alone is free, is where a single small unit carries every increment and the price is most sensitive — exactly the condition under which real electricity markets show price spikes.

The average cost curve is U-shaped:

Falling from 13.70 at 200 MW to 10.4263 at 564 MW, then rising to 10.458 at 600 MW. At low demand the no-load costs of three synchronised units are spread over little output; at high demand the expensive unit 1 must be loaded. The station is cheapest per unit at 91% of its capability — which is also, exactly, the point at which unit 3 saturates.

Where \(\lambda\) and the average cost cross is worth noting:

Marginal cost passes through the minimum of average cost, exactly as in elementary economics — and for the same reason. If the marginal unit is cheaper than the average, adding output pulls the average down; if dearer, it pushes it up.

What the schedule is used for in an operating control room:

The participation factors are what automatic generation control uses between full dispatch calculations: the AGC allocates each second's small correction in proportion to \(1/2c_i\) over the free units, and the full dispatch is recomputed every few minutes to correct the drift.

The optimum, from Problem 10:

Four alternative rules:

Annual figures assume the condition persists for 8760 hours, which it does not — but the comparison is what matters. The worst rule costs Rs 12.5 lakh a year more, or 2.7%.

The second row is the interesting one. Simple merit order by the linear coefficient \(b_i\) — load the unit with the lowest \(b\) first — costs only Rs 0.04 thousand per hour more than the true optimum:

Essentially exact. The reason is that the limits are doing most of the work here: with unit 2 at its maximum and unit 1 near its minimum, there is very little freedom left for the optimisation to exploit. Merit order gives 40/300/160 against the optimal 41.30/300/158.70 — a difference of 1.3 MW.

When does merit order fail badly? Precisely when the units are free:

At 400 MW, merit order loads unit 2 to its maximum of 300 and leaves unit 3 at its 60 MW minimum — but the optimum backs unit 2 down to 270.4 and raises unit 3 to 89.6. The error is 0.25%, some three hundred times larger than at 500 MW, though still small in absolute terms.

Why the savings are modest in absolute terms. The general result:

The cost penalty of a misallocation is quadratic in the error, because the optimum is a stationary point. A 20 MW misallocation on unit 1 costs \(0.015\times400 = 6\) thousand Rs/h — small. That quadratic flatness is why crude rules perform well, and why economic dispatch was for decades regarded as a refinement rather than a necessity.

Why it is nonetheless worth doing. Three reasons:

The second row is the practical argument. A system burning Rs 50 000 crore of fuel a year saves Rs 500 crore from a 1% improvement, for the cost of a computer program written once. Few investments in the industry have a comparable return.

A caution about the annual figures. The table above extrapolates one hour to a year, which overstates the saving:

Since the excess varies from 0.00% at 500 MW to 0.25% at 400 MW, the true annual figure depends on the load duration curve. Weighting by the daily schedule of Problem 11 gives an average excess of about 0.1% for merit order — roughly Rs 5 lakh a year for this station, and the honest number to quote. The pro-rata and equal-percentage rules, at 2%, are twenty times worse.

The market interpretation. In a competitive pool, every generator is paid the system marginal price for every megawatt-hour, whatever its own cost:

Note this is the fuel revenue only; a real settlement includes capacity and ancillary payments.

Compare with the cost:

The generators collectively lose Rs 3.66 lakh an hour. This is not an error, and understanding why is the point of the problem.

Unit by unit:

Only the saturated unit makes money. Units 1 and 3, both operating at \(\text{IC} = \lambda\), lose — and they lose exactly their no-load costs plus a share of the curvature.

The explanation:

for a free unit. So a unit is profitable at marginal-cost pricing only when \(c_iP_i^2 > a_i\), i.e. when \(P_i > \sqrt{a_i/c_i}\) — which, from Problem 2, is exactly the output of minimum average cost. Checking: unit 1 needs 152.8 MW and has 41.3; unit 3 needs 223.6 and has 158.7. Neither reaches it, so both lose.

The general result, which is the central fact about marginal-cost pricing:

A generator earns a positive margin only in the hours when it is inframarginal — when it is cheaper than the price-setting unit. Unit 2 earns Rs 1.57 lakh an hour here precisely because it is at its maximum and the price is set by someone dearer.

How real markets close the gap:

The first row is the theoretically clean answer and the practically troublesome one: in a well-functioning energy-only market the peak-hour prices must be high enough, often enough, to recover all fixed costs — which means occasional prices of many times the fuel cost. Political tolerance for that is limited, which is why capacity mechanisms exist almost everywhere.

Three other uses of \(\lambda\), all of which follow from \(\lambda = dC^*/dP_D\):

The second is the bridge to the next chapter: once transmission losses are included, a megawatt of generation at a remote bus delivers less than a megawatt to the load, and \(\lambda\) is the price at which that shortfall is valued.

Recompute the dispatch at neighbouring demands around the base case of 500 MW:

The ratio converges to \(\lambda = 9.73913\) as the increment shrinks, from above on one side and below on the other ✓. This is the numerical confirmation of the Lagrangian result of Problem 6.

Why the finite-difference value exceeds \(\lambda\) for an increase. Because \(\lambda\) itself rises with demand:

Matching the +10 MW row exactly, using \(d\lambda/dP_D = 1/95.833 = 0.010435\) from Problem 11. The second-order term is what the marginal price omits.

The −50 MW case is different, and shows a limitation:

The quadratic estimate is wrong by 1.3%, because between 500 and 450 MW the dispatch crosses the regime boundary at 496.3 MW: unit 1 returns to its minimum, and \(d\lambda/dP_D\) changes from 0.01043 to 0.0160. A Taylor expansion cannot cross a kink, and any use of \(\lambda\) for a large increment must check whether the limit set changes.

Pricing a block of demand response. An industrial consumer offers to shed 20 MW for four hours in the evening:

Below \(\lambda = 9.739\) by 1.4%, partly from the curvature and partly because the block crosses the 496.3 MW regime boundary. So the utility can offer up to Rs 9.598/kWh — and paying \(\lambda\) itself would overpay by 1.4%, which for a 20 MW block is worth the extra calculation.

The four-hour value:

If such an event occurs on 60 evenings a year, the contract is worth Rs 4.61 crore annually — before counting the capacity benefit of a 20 MW lower peak, which from Set 26 would be worth a further Rs 0.4 crore. Demand response is priced from \(\lambda\), and that is the whole of its economics.

When \(\lambda\) is not the right price for a finite block:

The last row can produce enormous values. If shedding 20 MW allows a whole unit to be shut down, the saving includes that unit's no-load cost and its start-up cost — which can make the block worth several times \(\lambda\). This is exactly why demand response is most valuable at the margin of a commitment decision, and why its value is so lumpy.

How a fuel price enters the cost curve. From Problem 2, \(C = F\times H\), so a change in fuel price scales every coefficient:

Both the intercept and the slope of the incremental cost line scale together. This is often got wrong: raising \(b\) alone shifts the line without tilting it, which corresponds to no physical change at all.

Unit 2's fuel up 20% (\(k = 1.2\)): its curve becomes \(600+7.44P_2+0.0054P_2^2\), with \(\text{IC}_2 = 7.44+0.0108P_2\) spanning 8.30 to 10.68:

Unit 2 comes off its maximum — its incremental cost now reaches 10.68 at full load, above the system \(\lambda\), so it is no longer worth loading fully. The other two units take up the 44 MW it sheds.

The cost increase is less than proportional, which is the value of re-dispatching:

Re-optimising after the price change recovers 3.8% of the increase. This is the general envelope-theorem result: the optimal cost responds less than proportionally to any parameter change, because the allocation adapts — but the recovery is second-order, so it is always modest.

Unit 2's fuel down 20% (\(k = 0.8\)): \(\text{IC}_2 = 4.96+0.0072P_2\), spanning 5.54 to 7.12:

The dispatch does not move at all. Unit 2 was already at its maximum; making it cheaper cannot make it produce more. The cost falls by exactly \(0.2\times2765.00 = 553.00\) — 20% of unit 2's own cost, to the rupee — and the system price is entirely unaffected.

The asymmetry is the lesson:

Only marginal plant sets the price. A change in the fuel cost of a unit at a limit changes the bill but not the price — which is why, in a market, a cheap saturated generator captures the full benefit of a fall in its own fuel cost while a marginal one passes it straight to consumers.

Two more sensitivities, for completeness:

The first is striking: making the expensive unit 1 cheaper by 20% raises its output from 41 to 90 MW and drops \(\lambda\) from 9.74 to 8.96. A 20% change in one unit's fuel has moved the system price by 8%, because that unit was marginal.

The general sensitivity, for a free unit \(j\):

The price moves in proportion to the unit's own share of the free capacity. A large flat unit (small \(c\)) dominates the price; a small steep one barely affects it. In a national system with hundreds of units, no single fuel contract moves \(\lambda\) perceptibly — but a change in the price of a whole fuel type moves it a great deal.

Why the piecewise-linear form is used. Three reasons, all practical:

The second is decisive for large systems. A quadratic dispatch with a full AC network is a hard nonlinear program; a piecewise-linear one with a DC network is a linear program of a few hundred thousand variables, which solves in seconds.

Segment unit 2's range 80–300 MW into four equal blocks and take the average incremental cost over each:

The segment incremental cost equals the exact incremental cost at the midpoint — exactly, not approximately. That is a property of quadratics: the average slope over an interval is the slope at its centre. So the piecewise approximation is a set of chords touching the true incremental cost line at four points.

Dispatch with piecewise-linear costs is pure merit order. Each segment of each unit is a block with a fixed price; load the cheapest blocks first:

Ranked strictly by price, block 4 (8.1575) precedes block 3 (8.2400). The merit stack interleaves the units' segments, which is exactly what equal-\(\lambda\) dispatch does continuously — merit order over segments is the discrete form of the equal-incremental-cost rule.

The price is set by the marginal block, and this is where the two formulations differ:

A piecewise-linear dispatch gives a step price curve — the price stays at 7.6625 for an entire 55 MW block, then jumps to 8.1575. Real markets show exactly this behaviour, and it is why spot prices sit on plateaux and then move abruptly.

The approximation error, assessed by comparing the two dispatches at 400 MW:

Four segments give an error of five thousandths of one per cent. Even two segments cost only 0.15%. The error falls as the square of the segment width, for the same reason the misallocation penalty of Problem 12 was quadratic — so doubling the number of segments quarters the error.

How many segments are needed in practice:

Note that market rules usually cap the number of segments, not to limit the accuracy but to limit the scope for strategic bidding — a generator with a hundred segments can shape its offer curve far more finely than one with five, and that flexibility is worth money to it and costs the consumer.

One requirement the segments must satisfy:

This is the discrete statement of convexity, and market rules enforce it. A non-monotonic offer curve would allow a generator to be dispatched to a locally cheap block while skipping a dearer one below it — which breaks the merit-order logic entirely and turns the linear program into a mixed-integer one.

The physical cause. A large steam turbine admits steam through four to eight nozzle groups, each with its own valve:

So the heat rate does not vary smoothly. It sags to a local minimum each time a valve reaches full opening, then climbs as the next valve is cracked open — one ripple per valve across the load range.

The standard model adds a rectified sinusoid to the smooth curve:

with \(e\) the ripple amplitude and \(f\) setting its period. Taking \(e = 30\) thousand Rs/h and \(f = 0.045\) per MW for unit 2 gives a period of \(2\pi/f = 139.6\) MW — about 1.6 ripples across its 220 MW range, corresponding to a two-valve machine.

The rippled curve:

The ripple is at most 1.3% of the cost — small in itself, but its derivative is not small, and that is what matters.

Why it destroys the argument. Differentiating:

Against a smooth incremental cost spanning 6.92 to 8.90, a ripple of \(\pm1.35\) is enormous. The incremental cost curve is no longer monotonic — it rises, falls, and rises again — so the equation \(\text{IC}(P) = \lambda\) can have several solutions, and the cost function is no longer convex.

The consequences:

The last row is the practical killer: lambda iteration relies on monotonicity for its guaranteed convergence, and with valve points that guarantee is gone. The method may oscillate or land on a poor local optimum.

What is done about it. Four approaches, in ascending order of sophistication:

The first row is what most utilities actually do, and it is defensible: the ripple is a genuine physical effect but the fitted smooth curve passes through the middle of it, so the error is bounded by half the ripple amplitude. The fourth row is the subject of hundreds of papers, most of which compare algorithms on the same three-unit and thirteen-unit test systems — a substantial literature built on a 0.3% effect.

The second approach is the one used in practice where it matters:

A unit run continuously at base load is set at a valve point deliberately, and the regulating duty is given to another machine. This is why "valve-point loading" appears in operating instructions, and why a base-load unit's output is often a peculiar number like 273 MW rather than a round 275.

Why it is a different problem. Dispatch takes the committed set as given and optimises continuously; commitment optimises over a discrete set:

The last two rows are why commitment is hard. With 100 units over 24 hours there are \(2^{2400}\) possible schedules, and the time coupling means they cannot be evaluated hour by hour.

Enumerate for three units. With only seven combinations, exhaustive search is trivial:

Unit 2 alone serves up to 300 MW — its own maximum. From 300 to 450 MW, unit 2 and unit 3 suffice, and unit 1 is not needed at all. Above 450 MW all three must run, since \(300+200 = 500\) is the two-unit ceiling and reserve considerations bite before then.

The savings are an order of magnitude larger than dispatch savings:

Six times more. And the reason is exactly the no-load cost: shutting down units 1 and 3 removes \(350+400 = 750\) thousand Rs/h of fuel that was buying nothing. Getting the commitment right is worth far more than getting the dispatch right, which is why modern system operation solves them together and why the dispatch problem is, in a sense, the easy half.

The simple decision rule for whether to keep a unit running, ignoring time coupling:

In words: the unit is worth running only if its own cost is less than what the remaining units would charge for the same output. At \(P_D = 300\), unit 3 at its 60 MW minimum costs 860.80, while unit 2 would supply that 60 MW for about 435 — so unit 3 should be shut down, as the table says.

What the simple rule omits — start-up cost. Restarting a shut-down unit is expensive:

In thousand Rs, for a unit of this size. Taking a warm start of 8000 for unit 3, and a saving of 364.60 thousand Rs/h while it is off:

The break-even down time:

So shutting unit 3 down is worthwhile only if it will stay off for more than 22 hours. Since the demand at 300 MW lasts only the six overnight hours of the Set 26 load curve, the unit should not be shut down at all — the hour-by-hour analysis of the table above is misleading, and this is precisely the time coupling that makes commitment a scheduling problem rather than an hourly one.

The other time-coupled constraints:

The last is often decisive and rarely modelled explicitly: a large drum boiler is designed for a few thousand starts over forty years, and cycling a base-load unit daily consumes that budget in a decade. The cost is real but appears only as a shortened plant life, which is why it is usually converted into a notional start-up cost rather than a constraint.

How the problem is actually solved at scale:

MILP won because commercial solvers improved by roughly a factor of a million between 1990 and 2015 — about half from algorithms and half from hardware. A security-constrained unit commitment over 24 hours with 1000 units and 10 000 network constraints now solves to within 0.1% of optimal in under twenty minutes, which was inconceivable when Lagrangian relaxation was devised.

The categories of must-run constraint:

The third and fourth rows have grown enormously in importance. As renewable generation displaces synchronous plant, system operators increasingly run thermal units purely for inertia and short-circuit level — paying fuel for a service that has nothing to do with energy.

How a must-run constraint enters the dispatch. It simply raises the unit's lower limit:

Nothing else in the method changes. The unit is pinned, the remaining demand is dispatched among the others, and the cost of the constraint is the difference between the constrained and unconstrained optima.

Worked example — reserve. Require 50 MW of spinning reserve at \(P_D = 500\) MW. Reserve must come from headroom on synchronised units:

In the base dispatch the headroom is \((120-41.30)+(300-300)+(200-158.70) = 78.7+0+41.3 = 120\) MW — comfortably above 50, so the constraint is not binding. It becomes binding only above 570 MW.

Worked example — voltage support. Suppose unit 1 must run at a minimum of 80 MW rather than 40 for local voltage support at \(P_D = 500\):

Rs 23 440 an hour, or Rs 2.05 crore a year if permanent. Note that \(\lambda\) falls from 9.74 to 9.12: forcing an expensive unit up displaces a cheaper one, and the marginal unit — now unit 3 alone — is further down its curve.

That is a general and counter-intuitive result:

Because the forced generation displaces marginal generation, leaving less for the free units to supply. In a market this produces the familiar situation where the spot price is depressed by out-of-merit running, while the total system cost — recovered through uplift charges — rises. Consumers see a low price and a large uplift bill, and the two move in opposite directions.

Cogeneration is the constraint that changes the problem most fundamentally:

Such a unit has no dispatch freedom at all in electrical terms and enters as a fixed injection. An extraction-condensing set has partial freedom, described by a two-dimensional feasible region in the \((P,Q)\) plane rather than a line — and its dispatch requires a two-dimensional generalisation of the equal-\(\lambda\) rule, with a separate multiplier for the heat balance.

How much of a real dispatch is actually free? A representative breakdown:

Only a minority of the fleet is actually dispatchable in the sense this chapter assumes. That does not diminish the theory — the free fraction is where all the marginal decisions are made, and it is what sets the price for everything else.

Step 1 — the load. Scaling Set 26's daily curve so the peak is 500 MW:

Load factor \(= 8300/(500\times24) = 0.6917\). All demands lie within the station's 180–620 MW range with all three units committed.

Step 2 — the commitment decision. From Problem 18, unit 1 is not needed below 450 MW and unit 3 not below 300 MW. But the start-up analysis showed a 22-hour break-even:

Both shut-down options lose money, because the overnight trough is not long enough. All three units stay synchronised for the whole day — which is exactly the answer a naive hourly analysis would have got wrong.

Step 3 — the hourly dispatch:

Rs 9.40 crore per day, for 8300 MWh — an average of Rs 11.325 per kWh. Note this exceeds every individual period's average except the two cheapest, because the low-demand hours carry three units' no-load costs on very little output.

Step 4 — the energy split:

Unit 1 produces 11.6% of the energy and incurs 19.0% of the cost — the expensive unit is over-represented in the bill, which is the whole point. Unit 2 is the reverse: 58.8% of the energy for 48.4% of the cost.

Step 5 — the marginal unit through the day, which is what a market would price:

The price ranges from 7.10 to 9.74 Rs/kWh across the day — a factor of 1.37, which is modest because this station's units are similar. On a real system with hydro, nuclear and gas turbines the daily price range is a factor of three to ten.

Step 6 — the verification checks. Six identities, each independent of what it checks:

The fifth is the one that most often fails in a hand calculation and is easiest to check: at 450 MW, unit 1 at its 40 MW minimum has \(\text{IC} = 9.70 \ge \lambda = 8.96\) ✓ and unit 2 at its 300 MW maximum has \(\text{IC} = 8.90 \le 8.96\) ✓.

Step 7 — the value of the exercise. Comparing with three cruder policies over the whole day:

Rs 1.4 crore a year for one small station, from a calculation that takes microseconds. Scaled to a national fleet of two hundred such stations, the figure is measured in hundreds of crores — which is the answer to anyone who asks whether a 0.06% improvement is worth the trouble.

Step 8 — what has been assumed away, and what Set 28 restores:

The first two are the subject of the next chapter, and they change the criterion in a specific way: with losses, a megawatt generated at a remote bus delivers less than a megawatt to the load, so the equal-incremental-cost rule becomes equal penalised incremental cost, and the whole load-flow machinery of Part 4 re-enters the economics.