Newton–Raphson Load Flow

The scalar case first. To solve \(f(x) = 0\), expand about the current estimate:

Replace the curve by its tangent, solve the linear problem exactly, and repeat from the new point.

The vector case is identical with a matrix of derivatives:

The Jacobian is square and, at a well-behaved solution, non-singular. Each iteration is one linear solve.

The order of convergence. Expanding to second order and subtracting the exact root:

Quadratic. The number of correct digits doubles at each step, so once the iteration is near the answer it finishes in two or three more steps whatever the tolerance.

What it costs. Three things Gauss–Seidel does not need:

Against which it needs four iterations rather than forty-three, and — decisively — the count does not grow with the size of the system.

The two failure modes, both real in load flow:

Quadratic convergence is a statement about the neighbourhood of the root. Far from it, Newton can diverge, oscillate, or converge to a different root — which is why the flat start matters as much here as in Set 19.

And the reason it suits load flow particularly. The power-flow equations are quadratic in the voltages, so their second derivatives are constants and the expansion is exact to second order. Newton's method is therefore unusually well-behaved on this problem — better than the general theory promises.

The polar power-flow equations. With \(Y_{ij} = G_{ij}+jB_{ij}\) and \(\theta_{ij} = \theta_i-\theta_j\):

Real equations in real unknowns — which is what makes the conjugate difficulty of Set 19 disappear. The price is two equations per bus instead of one complex one.

The mismatch:

Evaluated at every bus except the slack. Eight entries here: four \(\Delta P\) and four \(\Delta Q\).

At the flat start — every angle zero, magnitudes \(1.06, 1, 1, 1, 1\) — all the sines vanish and

The first four are \(\Delta P_2\) to \(\Delta P_5\); the last four are \(\Delta Q_2\) to \(\Delta Q_5\).

Reading the entries. Three are worth naming:

The largest mismatch is reactive, at bus 2, and it is large because bus 1 is held at 1.06 while bus 2 starts at 1.00 — a 6% difference across the stiffest branch in the network, which drives a large reactive flow.

The convergence test is on this vector, not on the correction:

A mismatch is a physical quantity — the power that fails to balance at a bus — so the tolerance has an engineering meaning. A step size does not.

And the ordering matters for the Jacobian's structure. Grouping all the \(\Delta P\) then all the \(\Delta Q\) gives the block structure of Problem 4; interleaving them bus by bus (\(\Delta P_2, \Delta Q_2, \Delta P_3, \dots\)) gives a matrix with a much better sparsity pattern for elimination. Production codes use the second; textbooks use the first, because the blocks are what make the derivation legible.

The correction equation. The unknowns are the angles and magnitudes; the equations are the real and reactive mismatches:

The dimensions. For \(n\) buses with \(n_{PV}\) voltage-controlled:

Square, as it must be — the same count of equations and unknowns Set 19 established.

Why \(\Delta|V|/|V|\) is often used instead of \(\Delta|V|\). Dividing the voltage columns by \(|V_j|\) and multiplying the correction by the same makes the submatrices more symmetric:

Since \(|V| \approx 1\) the numerical difference is small, but the algebraic symmetry it produces is what makes the decoupling of Problem 13 clean. Both conventions are in use, and results must be scaled consistently.

The Jacobian must be rebuilt every iteration, because it is a matrix of derivatives evaluated at the current estimate. That is the method's chief cost:

The build and the factorisation dominate, and both are avoided in the fast-decoupled method of Problem 14 — which is exactly why that method exists.

And a shortcut worth knowing. The Jacobian need not be exact for the method to converge — only for it to converge quadratically. Holding \(\mathbf{J}\) fixed for several iterations (the "dishonest Newton" method) costs a few extra iterations and saves most of the build and factorisation cost. It is the intermediate step between Newton and fast-decoupled.

The off-diagonals are straightforward, because only one term of the sum contains \(\theta_j\) or \(|V_j|\). Differentiating \(P_i\) with respect to \(\theta_j\), and noting \(\partial\theta_{ij}/\partial\theta_j = -1\):

The other three, by the same route:

Note the pattern: \(\partial Q_i/\partial\theta_j = -|V_j|\,\partial P_i/\partial|V_j|\) and \(\partial Q_i/\partial|V_j| = \dfrac{1}{|V_j|}\partial P_i/\partial\theta_j\). Only two distinct quantities appear, and the four blocks are built from them.

The diagonals require care, because \(\theta_i\) and \(|V_i|\) appear in every term of the sum. Differentiating and re-collecting gives forms in terms of \(P_i\) and \(Q_i\) themselves:

A considerable convenience: the diagonals need no new summation. \(P_i\) and \(Q_i\) were already computed for the mismatch, so each diagonal costs two multiplications.

The relative sizes, and this is the point that matters:

Because \(Q_i\) and \(P_i\) are small compared with \(B_{ii}\), and because \(|B| \gg |G|\) in a transmission network. The diagonal blocks are dominated by \(B\) and the off-diagonal blocks by \(G\) — which is the entire justification for decoupling.

Checking on the five-bus system at the flat start. For bus 2, \(B_{22} = -32.4150\) and \(G_{22} = 10.8333\):

A ratio of 3.2, matching this network's \(x/r = 3\). On a real 400 kV network the ratio would be 10 or more, and the decoupling correspondingly better.

At the flat start every \(\theta_{ij} = 0\), so \(\cos\theta_{ij} = 1\) and \(\sin\theta_{ij} = 0\). The formulas collapse:

For \(j \ne i\). Every off-diagonal element is a scaled entry of \(\mathbf{Y}_{\text{bus}}\).

One element in full. \(\partial P_2/\partial\theta_1\), with \(B_{21} = +15.0\), \(|V_2| = 1.0\), \(|V_1| = 1.06\):

Except that bus 1 is the slack, so this column does not appear in the Jacobian at all. The retained columns are buses 2 to 5.

The \(8\times8\) matrix, rows and columns ordered \(2,3,4,5\) then \(2,3,4,5\):

Three structural observations. First, \(\mathbf{J}_1\) and \(\mathbf{J}_4\) are nearly identical — compare \(33.400\) with \(31.430\), \(-30.000\) with \(-30.000\). Both are essentially \(-\mathbf{B}\) scaled by voltages.

Second, the off-diagonal blocks are about a third the size of the diagonal ones: 10.533 against 33.400 at the (2,2) position. That ratio is \(G/B = r/x = 1/3\) for this network, and it is the number that decides whether decoupling will work.

Third, the sparsity is inherited. Eight of the 64 entries are zero, at exactly the positions where \(\mathbf{Y}_{\text{bus}}\) is zero — buses 2–4, 3–5 within each block and their mirrors. The Jacobian has the same sparsity pattern as \(\mathbf{Y}_{\text{bus}}\), four times over, which is why the ordering machinery of Set 17 applies to it unchanged.

The system to solve is \(\mathbf{J}\Delta\mathbf{x} = \Delta\mathbf{f}\) with the \(8\times8\) Jacobian of Problem 5 and the mismatch of Problem 2. Solving:

Applying it:

Against the exact answer, and against Gauss–Seidel:

One Newton iteration has every angle within 0.32° of the answer. Two Gauss–Seidel sweeps have bus 5's angle at \(-3.7^\circ\) against a true \(-6.15^\circ\) — 40% of the way.

The magnitudes have overshot. Every one is now above its final value: 1.05504 against 1.04744, 1.03176 against 1.02418. Newton's linear model, applied to a quadratic function, over-corrects — and the next iteration brings them back.

The remaining mismatch:

A reduction by a factor of 11 in one iteration.

The cost comparison for that step. One Newton iteration required building 64 derivatives and factorising an \(8\times8\) matrix; one Gauss–Seidel sweep required about 24 complex operations. Newton's step is perhaps thirty times the cost — and it achieved more than twenty sweeps' worth of progress.

The Jacobian is rebuilt at the new point and the correction equation solved again. After iteration 2:

Every magnitude within \(7\times10^{-5}\) of the answer and every angle within 0.003°. The overshoot of iteration 1 has been corrected.

After iteration 3:

Correct to five decimal places — the solution Set 19 reached after 43 sweeps.

The full mismatch history:

Five entries, of which the last is machine precision. A production tolerance of \(10^{-4}\) would stop at iteration 4.

Where the iterations go. The tabulated bus voltages stop changing visibly after iteration 3, and iterations 4 and 5 move only the sixth decimal place. In engineering terms the answer exists after three iterations; the fourth confirms it.

The comparison, stated once:

On five buses the total costs are comparable — which is why Gauss–Seidel is the method taught first and why the comparison must be made at scale to mean anything.

The step counts at scale are the argument. Gauss–Seidel's 43 becomes several hundred at a thousand buses; Newton's four stays four. The per-step cost ratio of 30 is a constant; the step-count ratio grows without limit.

The test. Quadratic convergence means \(e_{k+1} = Ce_k^{2}\), so the ratio \(e_{k+1}/e_k^{2}\) should be roughly constant:

Constant to within a factor of two over five orders of magnitude of error. That is a clear demonstration; a linearly convergent method would give ratios differing by the same five orders.

Contrast with Gauss–Seidel, where the corresponding test was on \(e_{k+1}/e_k\) and gave a constant 0.78. Applying the quadratic test to that data would give ratios growing without bound — which is how the two orders are distinguished from the numbers alone.

Digits, not orders of magnitude, is the useful framing:

The count of correct digits doubles at each step. That is the operational meaning of quadratic convergence, and it is why the last iteration is always nearly free of information about how many are needed.

The consequence for the tolerance. Because the digit count doubles, the difference between a loose and a tight tolerance is almost never more than one iteration:

Eight orders of magnitude for two extra iterations. There is no reason to use a loose tolerance with a Newton solver, and there is every reason to with a Gauss–Seidel one — where the same range costs 36 extra sweeps.

The constant \(C\) is bounded by the second derivative over the first, and for the power-flow equations both are known: the equations are quadratic, so the second derivatives are constants and \(C\) depends only on the Jacobian's conditioning. A well-conditioned network gives a small \(C\) and fast convergence; a network near collapse gives a large one, which is the mechanism of Problem 18.

And a diagnostic. If the ratio \(e_{k+1}/e_k^{2}\) is not roughly constant, something is wrong: the Jacobian may be inconsistent with the mismatch function (a coding error), or the iteration may be far from the root, or a limit may be switching between iterations. The test costs two divisions and localises all three.

The converged Jacobian:

The comparison, element by element:

Every element has moved, and none by more than about 10%.

The symmetry has been lost. At the flat start \(J_1(3,4) = J_1(4,3) = -30.000\); at the solution they are \(-31.510\) and \(-31.388\). The Jacobian is not symmetric away from a flat profile, because the voltage magnitudes multiplying each element differ.

The sparsity is unchanged. The same eight zeros, at the same positions — they are set by the topology, which does not move. That is why the ordering computed once at the start remains valid for every iteration.

The 10% observation is the important one. If the Jacobian moves by only 10% over the whole solution, it need not be rebuilt every iteration:

The last is the "dishonest Newton" method, and it is the direct ancestor of the fast-decoupled load flow — which goes further and uses a Jacobian that was never correct at any iteration.

The determinant is the other useful output. Here it is \(2.47\times10^{10}\) and the condition number is 22.9 — a comfortably well-conditioned matrix. Problem 19 tracks both as the load rises, and they are what signals the approach of voltage collapse.

The observation. At a PV bus, \(|V_i|\) is known — so \(\Delta|V_i|\) is not an unknown — and \(Q_i\) is not specified, so \(\Delta Q_i\) is not an equation.

The Jacobian shrinks. Delete the row for \(\Delta Q_i\) and the column for \(\Delta|V_i|\):

For the five-bus system with bus 2 as PV: \(7\times7\) instead of \(8\times8\). One equation and one unknown removed together, so the system stays square.

That is the whole modification. No rescaling, no special case, no discarded information. The remaining equations are solved exactly as before, and the deleted \(Q_i\) is recovered afterwards from the converged voltages:

The contrast with Gauss–Seidel is instructive:

Newton handles the constraint by removing it from the problem; Gauss–Seidel handles it by violating and then projecting. The first is both faster and cheaper.

Why the difference. Gauss–Seidel solves each bus's equation in isolation and cannot express a constraint that couples the bus's \(Q\) to the rest of the network. Newton solves all the equations simultaneously, so a constraint is simply a smaller simultaneous system. Constraints are cheap in a simultaneous method and expensive in a sequential one — a general principle, and the reason tap changers, area interchange and other controls are all easier to add to Newton.

The practical consequence. A real network is 20–30% PV buses, so the Jacobian is 20–30% smaller than the worst case and the factorisation correspondingly cheaper. Adding voltage control to a Newton load flow makes it faster, which is the opposite of what one would expect.

The unconstrained PV solution requires \(Q_2 = 0.15065\) pu, obtained in four iterations:

Still quadratic, still four iterations — the deleted row and column change nothing about the rate.

The limit binds, since 0.15065 exceeds 0.10. The bus must be converted:

The Jacobian grows from \(7\times7\) back to \(8\times8\), and the specified \(Q_2\) becomes the limit value.

The limited solution:

Bus 2 settles at 1.0425 against its 1.045 target — the same answer Gauss–Seidel reached in Set 19, by a different route.

The structural change is the difficulty. Converting a bus between PV and PQ changes the Jacobian's dimensions, and therefore invalidates any factorisation and any ordering computed for the previous structure:

This is the one place where Gauss–Seidel's sequential structure is genuinely more convenient.

The standard remedies keep the dimensions fixed:

The second is the most common: solve with all PV buses free, check the reactive outputs, convert the violators, and solve again. Two or three passes usually settle it, and each is only four iterations.

Limit oscillation afflicts Newton exactly as it does Gauss–Seidel: a bus that limits, drops in voltage, un-limits, rises, and limits again. Newton is if anything more vulnerable, because it converges so fast that the flip happens between successive full solutions rather than being averaged out over many sweeps. Hysteresis and latching are used for the same reasons.

The rectangular form. Write \(V_i = e_i + jf_i\). Then

No trigonometric functions anywhere — the equations are polynomials in \(e\) and \(f\).

The comparison:

The rectangular form's advantage is real. The equations being exactly quadratic, the Taylor expansion terminates:

exactly, with \(\mathbf{H}\) constant. Including the second-order term gives a method with cubic convergence, and the constant Hessian makes it affordable. This is the basis of several specialised solvers.

The PV bus is where it loses. A voltage magnitude constraint is \(e_i^{2}+f_i^{2} = |V_i^{\text{sp}}|^{2}\) — a nonlinear equation that must be added to the system, rather than a variable that can be deleted from it:

Since a real network is 20–30% PV buses, this is not a marginal consideration.

And the second reason polar won: the decoupling of Problem 13. \(P\) depends chiefly on \(\theta\) and \(Q\) on \(|V|\) — a statement about polar variables that has no rectangular equivalent, because \(e\) and \(f\) both mix magnitude and angle. The fast-decoupled method exists only in polar coordinates, and it is what production software runs.

Where rectangular survives. In optimal power flow and state estimation, where the constant Hessian makes second-order methods practical and where PV buses are handled by other means anyway. It is a specialist's formulation rather than an obsolete one.

The approximation. Problem 4 showed \(\mathbf{J}_2\) and \(\mathbf{J}_3\) scale with \(G\) and the diagonal blocks with \(B\). Setting the small blocks to zero:

Two independent half-size systems in place of one full one.

The immediate saving. Factorising two \(m/2\) systems costs a quarter of one \(m\) system for dense matrices and about half for sparse ones — plus the two off-diagonal blocks need never be formed at all.

The result on the five-bus system:

Eight iterations against Newton's four — and the convergence is now linear, the ratio settling near 0.09.

The convergence order has changed, and that is the essential point. Newton's quadratic rate requires the exact Jacobian; an approximate one gives at best linear convergence. What is bought is a cheaper iteration; what is sold is the rate.

The trade, quantified for this network:

Roughly a wash on five buses. On a large sparse system the per-iteration saving is bigger and the decoupled method wins — but not by much, which is why the fast-decoupled refinement of Problem 14 was needed to make the idea worthwhile.

And the answer is identical to five decimals. Approximating the Jacobian changes how the iteration travels, not where it arrives — the fixed point is defined by the mismatch equations, which were not approximated. That distinction is the whole safety of the method: a wrong Jacobian slows convergence but cannot corrupt the answer.

The further approximations, on top of decoupling. In \(\mathbf{J}_1\) take \(\cos\theta_{ij} \approx 1\), \(G_{ij}\sin\theta_{ij} \ll B_{ij}\), \(Q_i \ll B_{ii}|V_i|^{2}\):

With the voltage magnitudes divided out, both matrices become constant.

The two matrices differ, as Set 17 established:

Because shunts and taps matter for voltage and not for angle. The asymmetry is empirical — Stott and Alsac found it worked and then justified it.

The result:

Seven iterations — fewer than the exact decoupled method's eight, which is at first sight absurd.

Why the cruder method converges faster. The exact decoupled Jacobian includes the \(Q_i\) and \(P_i\) terms in its diagonals, which move as the iteration proceeds and can overshoot. The constant \(\mathbf{B}'\) and \(\mathbf{B}''\) are smoother, and the fast-decoupled step happens to be better damped. This is not a general result — the two are usually within an iteration of each other — but it illustrates that a more accurate Jacobian is not automatically a better one.

The cost is where the method wins:

Two real matrices, factorised once, and every iteration thereafter is two forward-and-back substitutions. That is the whole method, and it is three to five times faster than full Newton on a large system.

The answer is identical to five decimals — 1.04744, 1.02418, 1.02357, 1.01794 pu at the same angles. Again: the Jacobian chooses the path, the mismatch defines the destination.

The pattern is inherited. Every Jacobian element for \(i \ne j\) contains \(G_{ij}\) or \(B_{ij}\) as a factor, so

The Jacobian is \(\mathbf{Y}_{\text{bus}}\)'s sparsity pattern replicated four times — which is why eight of its 64 entries are zero, matching \(\mathbf{Y}_{\text{bus}}\)'s six zeros in the non-slack part.

The fill for a large network:

The same \(4/n\) behaviour as \(\mathbf{Y}_{\text{bus}}\), on a matrix four times the area.

The ordering carries across unchanged. The elimination order computed for \(\mathbf{Y}_{\text{bus}}\) by the schemes of Set 17 minimises fill-in for the Jacobian too, because the graphs are the same. It is computed once when the network is read and re-used for every iteration of every case.

Which is why the interleaved ordering is preferred. Grouping the variables bus by bus — \(\Delta\theta_2, \Delta V_2, \Delta\theta_3, \Delta V_3, \dots\) — makes the Jacobian block-tridiagonal in \(2\times2\) blocks with exactly \(\mathbf{Y}_{\text{bus}}\)'s pattern:

The second eliminates as if it were \(\mathbf{Y}_{\text{bus}}\) with scalar entries replaced by \(2\times2\) blocks — the same ordering, the same fill, and better cache behaviour.

The cost of a Newton iteration, assembled:

The factorisation dominates, and its exponent is the reason Newton scales at all. Without sparse ordering it would be \(O(n^{3})\) and the method would be unusable above a few hundred buses.

And this is what the fast-decoupled method avoids. Its two constant matrices are factorised once, so the \(O(n^{1.4})\) term appears once in the whole solution rather than four times. That is where its speed comes from — not from the halved matrix size, which is a secondary effect.

The four stages:

Stages 1 and 2 also need \(2e\) sine and cosine evaluations, which on older hardware dominated everything and on modern hardware do not.

The dominant term is the factorisation, and its exponent depends entirely on the ordering:

Which is why Set 17's ordering discussion belongs to this set as much as to that one.

The totals, with the iteration count taken as 4 regardless of size:

Taking the per-iteration cost as \(n^{1.4}\). The whole growth is in the iteration's cost; the count is flat.

Against Gauss–Seidel, whose per-sweep cost is \(O(n)\) but whose count grows:

Taking a Newton iteration as 30 Gauss–Seidel sweeps at five buses and scaling both. The ratio moves steadily in Newton's favour, and beyond about a thousand buses Gauss–Seidel simply stops converging — which is a discontinuity rather than a trend.

The memory comparison now favours nobody:

All negligible on any modern machine, and all decisive in 1965 — which is the whole history of the subject in one table.

And the observation that ends the argument. A Newton load flow on a 10 000-bus system takes well under a second. The computational question was settled decades ago, and what remains difficult is everything around it: limit handling, ill-conditioned data, and cases that have no solution.

The measured results, all to a mismatch tolerance of \(10^{-6}\) and all giving the same answer to five decimals:

The convergence ratios distinguish the three linear methods:

All three are linear, and the difference between 0.78 and 0.06 is the difference between 43 iterations and 7. Being linear is not the same as being slow — the constant matters more than the order.

Where each belongs:

A control room runs all four, at different points in the same workflow.

The one that is not on the list is the exact decoupled method. It is dominated on both counts — more iterations than Newton and a more expensive iteration than fast-decoupled — and exists chiefly as the intermediate step in the derivation.

The robustness ordering is different from the speed ordering:

Which is why a heavily stressed case is confirmed with Newton even when fast-decoupled would be quicker.

And the historical sequence is worth noting, because it is not a sequence of better ideas but of different constraints: Gauss–Seidel in 1956 because memory was scarce; Newton in 1967 because sparse factorisation had been invented; fast-decoupled in 1974 because the physical decoupling was finally recognised and exploited. Each was the right answer to its decade's limitation.

Scaling the loads at buses 3, 4 and 5:

Newton reaches \(k = 3.16\); Gauss–Seidel gave up at 3.0. And Set 19 established that no solution exists beyond 3.16.

So Newton fails where the system fails, and Gauss–Seidel fails earlier. That is the practically important difference: a Newton failure is evidence about the network, a Gauss–Seidel failure is not.

The failure mode is also different. At \(k = 3.2\) the Newton mismatch history is

It descends, stalls, and then jumps. The iterate has been thrown across the solution surface into a region where the linear model is meaningless — and the voltages it reports are \(-0.774\) and \(-0.415\) pu, which are not physical.

Compare the approach to the limit. At \(k = 3.16\) the mismatch falls as

Linear, with a ratio near 0.25 — the quadratic convergence has been lost. That is the diagnostic: near the nose the Jacobian approaches singularity, the second-order term stops being negligible relative to the first, and Newton degrades to a linear method taking eleven iterations instead of four.

The three signatures of an approaching collapse, all visible in the solver:

A load flow that takes eleven iterations where it usually takes four is reporting something about the system, and it should be read that way.

The remedies, when a solution is believed to exist:

The third is what a voltage-stability study does, and it is how the 3.16 figure was obtained — not by solving at 3.2 and failing, but by tracing the curve from 1.0 upward until no further step was possible.

The measurements, taken at the converged solution for each load level:

The determinant falls by three and a half orders of magnitude and the condition number rises by a factor of 47.

Why the determinant vanishes at the nose. The nose is where two solution branches merge, so two roots of the system coincide. At a double root the function is tangent to zero rather than crossing it, and the derivative — the Jacobian — is singular:

Exactly, not approximately. The point of voltage collapse and the point of Jacobian singularity are the same point.

The behaviour near the nose is not linear, which limits the determinant's usefulness:

The collapse of the determinant happens in the last 5% of the loading. A determinant at half its base value indicates nothing much; one at a thousandth indicates imminent trouble — but by then the margin is already small.

Better indices exist, for exactly this reason:

The last is the practically useful one: the eigenvector of the smallest eigenvalue of the reduced \(\mathbf{J}_4\) shows which buses participate in the collapse mode, and therefore where to install the remedy.

The voltage at the nose is 0.60 pu, and no system would be operated within sight of it. A practical criterion stops far earlier:

So the voltage criterion binds at less than half the collapse loading, and the margin to collapse is a stability measure rather than an operating one.

And the Jacobian's dual role is the point of this problem. The same matrix that makes the solver work also measures the system's distance from collapse — not by coincidence, but because both are statements about the sensitivity of the power-flow equations to a perturbation. A network is hard to solve for the same reason it is close to failing.

The five treatments of one network:

The results, side by side:

The linear model got the flows within 4% and the angles within 8%, for a fraction of a per cent of the effort — and told us nothing at all about voltages, losses or reactive power.

What each analysis alone revealed:

Five findings, no two of which come from the same calculation.

And the one fact that appears in all five. The \(0.01+j0.03\) tie between buses 3 and 4 shows up as the largest off-diagonal in \(\mathbf{Y}_{\text{bus}}\), as a near-zero between-bus impedance in \(\mathbf{Z}_{\text{bus}}\), as a joint voltage collapse in the fault study, as two identical voltages in the load flow, and as the largest entries in the Jacobian. A single physical feature, visible in every representation — which is what one would expect, and a reassurance that the five calculations describe the same network.

The workflow, in the order a study would run it:

Each stage uses the previous stage's data structures, and each is cheaper than the one after it. That ordering is not a convention but an optimisation, and it is what a planning department's software does automatically.

The network, finally, in one paragraph. A five-bus 220 kV system carrying 145 MW with 4.6 MW of loss; voltages from 1.018 to 1.060 pu; the heaviest circuit at 89 MVA and the binding contingency the loss of either circuit from the slack bus; fault levels of 489 to 833 MVA; a net reactive surplus of 17 MVAr requiring the slack machine to run under-excited; and a margin of 3.16 times its present load before voltage collapse. Every number in that sentence came from one of the five sets, and none of them from the network data alone.