In Asymptotic Analysis and Big O Notation, we introduced a method of quantifying the running time of a procedure in terms of its worst-case performance on all inputs of a given size. Such an analysis is helpful, for example, in comparing the relative performance of two procedures that perform the same task. Yet in many instances, the worst-case running time of a procedure does not accurately reflect the true running time in practice. In particular, in the study of data structures, we are often interested in the time to perform a sequence of operations, rather than the worst-case time to perform a single operation in isolation. In this note, we describe a method of analysis called amortized analysis that considers the running time of a sequence of operations with the aim of understanding the average running time of the operations when averaged over the entire sequence.

For a concrete example, consider the following two implementations of the SimpleStack interface:

Both implementations use an array, Object[] contents, to store the contents of the stack. In both cases, when the array is full, a call to the push method copies contents to a larger array thereby increasing the capacity of the data structure. The only difference between the two implementations is how this capacity increase is performed. ArraySimpleStackOne simply increases the capacity by one to make room for the single new item being pushed to the stack. Here is the code for the increaseCapacity() method for ArraySimpleStackOne:

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private void increaseCapacity() {

// create a new array with larger capacity
Object[] bigContents = new Object[capacity + 1];

// copy contents to bigContents
for (int i = 0; i < capacity; ++i) {
bigContents[i] = contents[i];
}

// set contents to refer to the new array
contents = bigContents;

// update this.capacity accordingly
capacity = capacity + 1;
}


On the other hand, ArraySimpleStackTwo doubles the capacity of contents each time the array’s size needs to be increased:

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private void increaseCapacity() {

// create a new array with larger capacity
Object[] bigContents = new Object[2 * capacity];

// copy contents to bigContents
for (int i = 0; i < capacity; ++i) {
bigContents[i] = contents[i];
}

// set contents to refer to the new array
contents = bigContents;

// update this.capacity accordingly
capacity = 2 * capacity;
}


If the stack has size $$n$$, both implementations of increaseCapacity() have running time $$O(n)$$.

The push method for the two implementations of SimpleStack are identical:

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public void push(E x) {
if (size == capacity) {
increaseCapacity();
}

contents[size] = x;
++size;
}


Note that the running time of this method is $$O(1)$$ if size != capacity (i.e, the array does not need to be resized), but is only $$O(n)$$ when a resize occurs.

Consider now the running time of pushing $$n$$ elements to a stack. For example we might have

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SimpleStack<Integer> stack;
...
for (int i = 0; i < n; ++i) {
stack.push(someValue);
someValue = nextValue();
}


A straightforward application of asymptotic analysis shows that for both implementations of SimpleStack above, the running time of the code snippet above is $O(n^2)$ (assuming nextValue() has running time $$O(n)$$): the stack.push(...) method runs in time $$O(n)$$, and this method is called $$n$$ times in the loop above. Yet running the same program with the two implementations of increaseCapacity() can give drastically different running times. Here are some example running times of the same program building a stack of size $$n$$ using the two different SimpleStack implementations:

Here are the running times just for ArraySimpleStackTwo to get a better picture of what is happening there:

Why is ArraySimpleStackTwo so much more efficient when both implementations’ add methods have the same worst case running time?

The answer lies in how frequently the increaseCapacity() method (which dominates the running time of add) is called. In ArraySimpleStackOne, this method only increases the capacity by one. Thus, after we’ve increased the capacity for one push, we might need to do it again on the next push. On the other hand, ArraySimpleStackTwo doubles the capacity of the array on each resize. Therefore, if a call to push invokes increaseCapacity when the size of the stack is $$n$$, the next call to increaseCapacity will not occur until the size reaches $$2 n$$. That is, we will have to perform another (at least) $$n$$ calls to push before another capacity increase.

In what follows, we will define the amortized cost of a method call. The basic idea of the method is to come up with an “accounting scheme” for the running times of method calls such that the running time of expensive (i.e., time-consuming) operations—such as increaseCapacity()—can be averaged out over less expensive method calls. Using amortized analysis, we will show that the amortized running time of the push method for ArraySimpleStackTwo is $$O(1)$$. Thus, when averaged over any sequence of method calls, the average push runs in time $$O(1)$$, whereas ArraySimpleStackOne still runs only in $$O(n)$$ time. Thus, we can explain the enormous difference in the running times depicted above.

The Banker’s View

In this section, we provide a definition of the amortized running time of a method via an accounting scheme. This view of amortized analysis is sometimes referred to as the “banker’s view” of amortized analysis. To this end, we associate a cost to each operation $$\mathrm{op}$$ that represents the operation’s running time. To perform an operation, the cost can be paid in two ways: either by paying the cost upfront, or by deducting (a portion of) the cost from an account $$A$$.

Each time an operation is performed, we can choose either to pay the cost (in part) from $$A$$ thereby decreasing its balance, or pay some extra cost upfront to increase the balance of $$A$$. In either case amortized cost of performing the operation $$\mathrm{op}$$ is

$\mathrm{ac}(\mathrm{op}) = \mathrm{cost}(\mathrm{op}) + \mathrm{bal}(A') - \mathrm{bal}(A).$

Here $$\mathrm{ac}$$ denotes amortized cost, while $$\mathrm{bal}(A')$$ and $$\mathrm{bal}(A)$$ denote, respectively, the balance of $$A$$ before and after performing the operation. Thus $$\mathrm{bal}(A') - \mathrm{bal}(A)$$ is positive if we pay additional funds into $$A$$ (in addition to paying $$\mathrm{cost}(\mathrm{op})$$ upfront), and this value is negative if we use funds from $$A$$ in order to pay (part of) the cost of $$\mathrm{op}$$.

Proposition. Suppose there is an accounting scheme as above such that for any sequence of operations $$\mathrm{op}_1, \mathrm{op}_2, \ldots, \mathrm{op}_m$$ the following hold:

1. for all $$i$$, we have $$\mathrm{ac}(\mathrm{op}_i) \leq C$$, and
2. the balance satisfies $$\mathrm{bal}(A) \geq 0$$ before and after every operation.

Then the total running time of the sequence of operations is at most $$m \cdot C$$. Thus, the average running time per operation for any sequence of operations is at most $$C$$.

Amortizing ArraySimpleStackTwo

Here, we show how to apply amortized analysis to the push method of ArraySimpleStackTwo. Suppose a call to increaseCapacity() occurs during a push operation where the previous capacity was $$N$$. That is, during that call to push, the capacity was increased to $$2 N$$, and the size of the stack increased to $$N + 1$$. Thus, the next call to increaseCapacity() (if any) will occur during a call to push after the size of the stack reaches $$2 N$$. Thus, a total of at least $$N$$ calls to $$\mathrm{push}$$ must be made before the next capacity increase.

Denote running time (cost) of the next call to increaseCapacity() by $$C_{2N}$$. By our worst-case analysis of the increaseCapacity() method, we have $$C_{2N} = O(N)$$. The idea of our accounting scheme is to do the following: for each push operation before the size reaches $$2 N$$, pay $$\frac{1}{N} C_{2N} = O(1)$$ additional into the account $$A$$ to increase its balance. This way, when the stack reaches size $$2 N$$ (after at least $$N$$ calls to push), the account balance is at least $$C_{2N}$$. Thus, the cost of increaseCapacity() can be paid out of the balance.

Using this accounting scheme, the for subsequent calls to push in which the stack size is less than $$2 N$$ incur an amortized cost of

$\mathrm{ac}(\texttt{push}) = \mathrm{cost}(\texttt{push}) + \frac{1}{N} C_{2N} = O(1).$

Here, $\mathrm{cost}(\texttt{push}) = O(1)$ because these calls do not call increaseSize(). For the next call to push (if any) in which the stack size is $$2 N$$, we have

\begin{align*} \mathrm{ac}(\texttt{push}) &= \mathrm{cost}(\texttt{push}) - (N - 1) \frac{1}{N} C_{2N}\\ &= C_{2N} - (N - 1) \frac{1}{N} C_{2N}\\ &= \frac{1}{N} C_{2N}\\ &= O(1). \end{align*}

Again, the account $$A$$ satisfies $$\mathrm{bal}(A) \geq (N - 1) \frac{1}{N} C_{2N}$$ before the operation because at least $$N - 1$$ previous calls to push were made. Thus, in all cases, the amortized cost of push is $$O(1)$$.