the current algorithm maximize the number of sockets opened, by re-using the first stacked free socket using freesockets.shift()
@see https://github.com/node-modules/agentkeepalive/blob/master/lib/_http_agent.js#L170
if (freeLen) {
// we have a free socket, so use that.
var socket = this.freeSockets[name].shift();
debug('have free socket');freesockets = [1][2][3][4][5]
# a socket is freed
freesockets = [1][2][3][4][5][6]
# new request arrive, current algorithm use "shift"
freesockets = [2][3][4][5][6]
# there might rarely be a timeout of any sockets.
an alternative could be to use freesockets.pop() to minimize the number of sockets opened
freesockets = [1][2][3][4][5]
# a socket is freed
freesockets = [1][2][3][4][5][6]
# new request arrive, using pop
freesockets = [1][2][3][4][5]
# first socket timeout
freesockets = [2][3][4][5]
implementation:
var algo = (this.options.algo === 'pop') ? 'pop' : 'shift';
var socket = this.freeSockets[name][algo]();
I tested this for 1K request/s, it divided the pool size by 5.
the current algorithm maximize the number of sockets opened, by re-using the first stacked free socket using freesockets.shift()
@see https://github.com/node-modules/agentkeepalive/blob/master/lib/_http_agent.js#L170
an alternative could be to use freesockets.pop() to minimize the number of sockets opened
implementation:
I tested this for 1K request/s, it divided the pool size by 5.