# Parallel Cost Models

A cost model is a framework used to describe the cost of a program or algorithm, often measured in the number of instructions executed or steps taken. Often we describe the cost in using big-O notation. For example, we may say that merge sort requires O(n log(n)) comparisons to sort a sequence. Most programmers have some sort of informal cost model that they use for reasoning about the performance of their program. Usually this is just counting the “steps” taken. Sometimes we acknowledge that some steps are more expensive than others (e.g. a memory access compared to an integer addition), but often we do not. Unfortunately, counting the number of steps is not a good way to describe a parallel algorithm. This is because we are not just interested in the total number of steps, but also how well the algorithm might take advantage of a parallel computer. To describe such algorithms, we use a parallel cost model.

Consider adding together the integers from 1 to 9, both inclusive. If we simply add together the numbers in order starting from 1, the dependency graph of the intermediates and final result looks like this:

``````1 2 3 5 6 7 8 9
\| | | | | | |
3 | | | | | |
\| | | | | |
6 | | | | |
\| | | | |
11 | | | |
\| | | |
17 | | |
\| | |
24 | |
\| |
32 |
\|
41``````

This involves a total of 7 additions. If we counted from 1 to n, we would need n-1 additions. We say that this algorithm has a total work of O(n). The work is just the count of all steps, exactly as we are used to quantify the performance of algorithms. However, in a parallel cost model we also measure the span (sometimes called depth). The span is the length of the longest path in the dependency tree from the final result to an input value. In this case, the span is 7. Or, generalising, the span is O(n).

But now consider if instead of adding the numbers from left to right, we repeatedly added them pairwise, making our dependency graph a balanced tree structure:

``````1 2 3 5 6 7 8 9
\ / \ / \ / \ /
3   8   13  17
\  /    \  /
11      30
\    /
\  /
41``````

There are still 7 additions, and the work is still O(n). But now the longest path has only 3 edges, and generalising to summing n numbers, we can show that the span is O(log2(n)). Imagine that we were executing this on a parallel computer with an unlimited amount of processors. At each step of the parallel computer, we could evaluate every node in the dependency graph for which its predecessors have already been evaluated. In the tree above, it would correspond to evaluating an entire “level” of the summation tree for every step. On such a computer, evaluation would require only 3 steps - or for adding n numbers, O(log2(n)) steps. The span tells us how many steps the algorithm would need on an infinitely parallel computer! (Asymptotically.)

Of course, infinitely parallel computers do not exist. However, Brent’s Lemma (see references) tells us that we can simulate a machine with x processors on a machine with only y processors at a cost proportional to x/y. Essentially, the cost is proportional to the amount of “missing” parallelism, relative to what the algorithm could exploit. This means that the span is a useful way of quantifying the parallel scalability of an algorithm, even though we will only ever execute it on finitely parallel computers.

## Work Effiency

If a parallel algorithm has the same asymptotic work as the best known sequential algorithm, then we say that it is work-efficient. For example, the Hillis-Steele algorithm for parallel prefix sum requires O(n log(n)) work, while a sequential algorithm requires O(n) work. For prefix sums, a work-efficient algorithm is known, but there are problems for which no work-efficient parallel algorithm is known.

A parallel algorithm that is not work efficient can still be practically useful. Perhaps the ability to exploit a parallel computer outweighs the wasted work. But eventually, for sufficiently large input, the best sequential algorithm will outperform a parallel algorithm that is not work-efficient.