mintuhouse

While going through https://www.speedshop.co/2020/05/11/the-ruby-gvl-and-scaling.html, I read

Amdahl's Law is simply 1 / (1 - p + p/s), where p is the percentage of the task that could be done in parallel, and s is the speedup factor from the part of the task that gained improved resources (the parallel part).

So, in our example, let's say that half of SatelliteDataProcessorJob is GVL-bound and half is IO-bound. In this case, p is 0.5 and s is 10, because we can wait for IO in parallel and there are 10 threads. In this case, Amdahl's Law shows that a Sidekiq process would go through our jobs up to 1.81x faster than a single-threaded Resque or DelayedJob process.

Question: In this case, shouldn’t speed up be 2x as IO-bound portion takes practically zero CPU time rather than p/s i.e., s = ♾️. Wondering what I am missing here? 🤔

mintuhouse

// ==UserScript==
// @name        GitHub PR Add Merge Description
// @version     1.0.0
// @description A userscript that adds merge commit description
// @license     MIT
// @author      Hasan Kumar
// @namespace   https://github.com/mintuhouse
// @include     https://github.com/*
// @run-at      document-idle
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mintuhouse

Ckeditor.setup do |config|
    config.current_user_method  do
        current_user
    end
    config.authorize_with :chronus
 end

mintuhouse

class Net::LDAP

  def initialize(args = {})
    @host = args[:host] || DefaultHost
    @port = args[:port] || DefaultPort
    @verbose = false # Make this configurable with a switch on the class.
    @auth = args[:auth] || DefaultAuth
    @base = args[:base] || DefaultTreebase
    encryption args[:encryption] # may be nil