# Queuing Theory: Delay of a Message on a Link

I'm researching on Load Balancing in networks and using queueing theory for modeling, but I cannot understand the reason for some calculations. Some papers and books mention that on the condition that packets arrive at a link at a Poisson rate and the packet lengths are exponentially distributed, the mean delay of a message on a link can be calculated by using M/M/1 queue formulas which means:

$$Delay =\frac {\text{MessageLength (bits)}}{\text{LinkCapacity(bits/sec)}} * \frac{1}{1-ρ}$$

while

$$\rho = \frac {\text{PacketArrivalRate (packet/sec)}}{\frac{\text{LinkCapacity(bits/sec)}}{\text{MessageLength (bits)}}}$$

My first question is why should we use the above way to calculate the delay instead of using well-known $$\frac{1}{μ-λ}$$ which in this case is

$$\frac{1}{\text{LinkCapacity(bits/sec)}-\text{PacketArrivalRate (packet/sec)}\cdot \text{MessageLength (bits)}}?$$

My second question is why service time or $$\frac{1}{\mu}$$ is

$$\frac{\text{MessageLength (bits)}} {\text{LinkCapacity(bits/sec)}}?$$

• I can't answer the first question, but for the second question, this seems like a consequence of Little's Law? – LarrySnyder610 May 16 at 1:35
• @LarrySnyder610 How can I apply Little's Law to service rate? It seems to be related to the arrival rate because Little's Law says the average number of customers, etc. equals the arrival rate multiplied by average time that a customer spends in the system. – Benyamin T May 16 at 6:59
• Actually it's simpler than Little's Law. I forgot that the processing time per bit is deterministic. So the service time for a given message is just # bits in the message divided by link capacity in bits/sec. Then the mean service time is expected message length divided by link capacity, no? – LarrySnyder610 May 16 at 11:46
• @LarrySnyder610 You mean fixed by deterministic, right? Thanks, you are right, however, I still don't know how to relate it to my first question. Actually, there are two ways to calculate delay as I wrote above but I can't understand the reason that the second way isn't true. – Benyamin T May 16 at 12:26
• Yes, by "deterministic" I mean "fixed," i.e., the same every time. Unfortunately I don't have an answer to your first question. – LarrySnyder610 May 16 at 13:55