Answers the Chapter 5, Question 26 in Peterson and Davie's "Networking: A Systems Approach"

peterson_davie_ch5q26.ml

(* OCaml program to answer this question *)

let delta = 0.125
let mu = 1.0
let phi = 4.0

let timeout (estimate_rtt, deviation) = mu *. estimate_rtt +. phi *. deviation;;

let calculate_estimate (old_estimate_rtt, old_deviation, sample_rtt) =
    let difference = sample_rtt -. old_estimate_rtt in
    let new_estimate_rtt = old_estimate_rtt +. (delta *. difference) in
    let new_deviation = old_deviation +. ((delta *. (abs_float difference)) -. old_deviation) in
    
    (new_estimate_rtt, new_deviation);;

let time_till_target (start_estimate_rtt, start_deviation, rtt, target_timeout) =
    let rec iter estimate_rtt deviation time =
        let current_time = time +. rtt in
        let (new_estimate_rtt, new_deviation) = calculate_estimate (estimate_rtt, deviation, rtt) in
        if timeout (new_estimate_rtt, new_deviation) < target_timeout then
            current_time
        else
            iter new_estimate_rtt new_deviation current_time
    in

    iter start_estimate_rtt start_deviation 0.0;;

(* Find out the time required to reach when the timeout < 4 seconds *)
let () = Printf.printf "%F" (time_till_target ((*estimate_rtt*) 4.0, (*deviation*) 1.0, (*rtt*) 1.0, (*target*) 4.0));;