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The accounting method charges operations enough to pay for their work and save credits for later work. A nonnegative balance from an initially empty account makes every prefix's total charge an upper bound on its actual cost. Choosing the balance to be $\Phi$ gives the potential method; the ledger for dynamic array doubling illustrates this equivalence.^[ref]

// Destination: computer-science.bigb

Amortized analysis bounds the total cost of every permitted sequence of operations from a specified initial state. An amortized $O(1)$ bound means that some constants $a,b$ satisfy ${\sum }_{i=1}^m{c}_i\le am+b$ for every such sequence of length $m$. The cost model and allowed operations are part of the claim; Big O notation alone does not specify them.

This distinction, and the credit and potential method viewpoints, are developed in R. Endre Tarjan's 1985 paper Amortized Computational Complexity, sections 1–2, pp. 306–308. The dynamic array doubling example below shows how an expensive operation can pay for itself through the cheap operations that precede it.

In the potential method, an operation's amortized cost is ${\hat{c}}_i=c_i+{\Phi }_i-{\Phi }_{i-1}$. It can exceed or fall below its actual cost $c_i$. The endpoint potentials determine how the total charge compares with the total actual cost.

The potential method assigns a number $\Phi (D)$ to each state $D$ of a data structure. Let $D_i$ be the state after operation $i$, ${\Phi }_i=\Phi (D_i)$, and $c_i$ its actual cost. Its amortized cost is the actual cost plus the change in potential. Summing and cancelling adjacent potentials gives a telescoping sum:If ${\Phi }_0=0$ and every reachable ${\Phi }_i\ge 0$, the sum of amortized costs bounds the actual total for every prefix. A nonzero initial potential contributes the explicit additive term ${\Phi }_0$. Potential is proof bookkeeping; the algorithm need not store it.^[ref]

For dynamic array doubling, chooseThe potential method now starts at $\Phi (0,1)=0$. After every nonempty reachable state, $C<2n$, so the maximum equals $2n-C$. There are just three cases for append:

For every $n\ge 1$, the total charge is $2+3(n-1)=3n-1$, and the telescoping sum provesThis agrees exactly with $T(n)=n+C_n-1$. The unmodified choice $2n-C$ has initial value $-1$, so calling it nonnegative at the empty state would be incorrect. It also works if that initial constant is retained in the identity; alternatively, $2n-C+1$ starts at zero and is nonnegative on these reachable states. The maximum above keeps a literal nonnegative credit balance from the beginning. The proof covers append alone; allowing reserve or arbitrary deletion changes the reachable states and requires a new argument.