Estimating bandwidth requirements and productivity gain from traffic data

Question

My small office has a 30Mbit/s internet connection, on which two or three people work. We could get twice the bandwidth at a substantial cost.

I imagine a bandwidth upgrade would slightly improve latency for all internet-related tasks and thus provide a small increase in productivity.

I'd like to estimate the productivity gain that could be achieved by this upgrade, based on per-second traffic data. Are there any rules-of-thumb that can be used for such a calculation?

For example, one could count how often the bandwidth maxes out. If the bandwidth maxes out for 10 minutes every day, then doubling the bandwidth may reduce waiting times by 5 minutes (in the best case). Obviously the relationship between latency and bandwidth is more complex than that. Is there any way to do this estimation more accurately?

Answer 1

I've learned a rule of thumb by a Cisco CCIE and he said, that if a link is utilized more than 50% on average over a period of a workday, you should consider upgrading the connection.

Here's an official way of calculating Link Utilization and Efficiency:

Link utilization is simply the average traffic over a particular link expressed as a percentage of the total link capacity. Link efficiency is a less commonly used term that is defined as the ratio of the time taken to transmit a frame (or frames) of data to the total time it takes to transmit and acknowledge the frame or frames:

Utilization (U) or Efficiency=(Time taken to transmit frame)/(Total transmission time)

The efficiency of a link with stop-and-wait ARQ can be determined as follows: If the time taken to transmit a frame or block of data is tf, the propagation delay for both frame and acknowledgement is td, the time taken to transmit an acknowledgement is ta and the total processing time is tp, then:

U = (tf) / (tf + ta + tp + 2td)

In many situations the acknowledgement transmission time and processing times can be ignored, giving:

U = (tf) / (tf + 2td) = (1) / (1 + 2a) ; where a = td / tf

Example:

A point-to-point satellite transmission link connecting two computers uses a stop-and waits ARQ strategy and has the following characteristics:

Data transmission rate = 64 kbps
Frame size, n = 2048 bytes
Information bytes per frame, k = 2043 bytes
Propagation delay, td = 180 ms
Acknowledgement size, ta = 10 bytes
Round-trip processing delay, tp = 50 ms

Determine the throughput and link efficiency.

Solution:

Frame transmission time tf = (2048 x 8) / (64000) = 0.256 sec.
Acknowledgement transmission time ta = (10 x 8) / (64000) = 1.25 msec.
Total time to transmit frame and receive an acknowledgement is:
tf + ta + tp + 2td = 0.256 + 0.00125 + 0.05 + 0.36 = 0.66725 sec.
Throughput k = (2043 x 8) / (0.66725) = 24.494 Kbps.

Note that the resulting throughput is considerably less than the transmission rate of 64 kbps.

The link efficiency (Utilization) can now be calculated, neglecting ta and tp, as follows:

a = (td) / (tf) = (0.18) / (0.256) = 0.7
U = (1) / (1 + 2a) = (1) / (1 + 1.4) = 41.67 %.

Source: https://uomustansiriyah.edu.iq/media/lectures/5/5_2017_03_18!09_07_27_PM.pdf