JavaScript - test algorithm performance

In this article, we would like to show how to test algorithms' performance in JavaScript.

Usually, when we compare algorithms that execution times are short, we repeat computations multiple times to get precise results. The number of repeats is selected empirically. The source code that is not an algorithm part shouldn't be measured.

Main concept:

// ONLINE-RUNNER:browser;

const test = (repeats, description, func) => {
	const t1 = Date.now();
  	for (let i = 0; i < repeats; ++i) {
  		func();
    }
  	const t2 = Date.now();
  	const dt = t2 - t1;
  	console.log(`${description}: total time is ${dt} ms`);
};


// Usage example:

const repeats = 100000;  // set number of repeats depending on cases complexity

test(repeats, 'case 1', () => { /* source code case 1 */ });
test(repeats, 'case 2', () => { /* source code case 2 */ });
test(repeats, 'case 3', () => { /* source code case 3 */ });

// add more cases here ...

Hint: presented concept describes only a simple way how to test performance - in practice tests made by benchmarks are more complex.

Practical example

In the below example we compare algorithms that computes Fibonacci sequence values.

In the test, we made 100000 repeats on Fibonacci algorithms using Ryzen 9 x5900 processor and V8 JS Engine (Google Chome v110) under Windows 11.

Test result:

Iterative algorithm: total time is 69 ms
Binet's formula: total time is 2 ms

Conclusion: iterative algorithm is ~35 times slower in the tested cases.

Test source code:

// ONLINE-RUNNER:browser;

const test = (repeats, description, func) => {
	const t1 = Date.now();
  	for (let i = 0; i < repeats; ++i) {
  		func();
    }
  	const t2 = Date.now();
  	const dt = t2 - t1;
  	console.log(`${description}: total time is ${dt} ms`);
};


// Tested algorithms:

function fibonacci1(number) {
    if (number < 1) return 0;
    if (number < 2) return 1;
    return fibonacci1(number - 2) + fibonacci1(number - 1);
}

function fibonacci2(number) {
    var a = Math.pow(1 + 2.23606797749979, number);
    var b = Math.pow(1 - 2.23606797749979, number);
    var c = 2.23606797749979 * Math.pow(2, number);
    return Math.round((a - b) / c);
}


// Performance tests:

const repeats = 100000;  // set number of repeats depending on cases complexity

test(repeats, 'Iterative algorithm', () => {
    for (let i = 1; i < 10; ++i) {
        fibonacci1(i);
    }
});

test(repeats, 'Binet\'s formula', () => {
    for (let i = 1; i < 10; ++i) {
        fibonacci2(i);
    }
});