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Erlang B determines the probability that a call is blocked and is defined as:

$$Pr[blocking]= {{A^C \over C!} \over {\sum_{k = 0}^{C} {A^k \over k!}}}$$

while Erlang C determines the probability that a call is delayed and is defined as:

$$Pr[delay>0]= {{A^C \over C!} \over {A^C+C!(1-{A\over C})\sum_{k = 0}^{C-1} {A^k \over k!}}}$$

But, the only difference lies in the fact that we use Erlang B when a user is denied access due to the unavailability of a channel. Whereas, we use Erlang C when a user is denied access due to the unavailability of a channel and further, put in a queue.

So my question is why do both formulas need to be different? In both cases, we are determining the probability if a channel would be available or not regardless of the fact that the user is put in a queue. This sounds intuitively right, right?

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The basic, qualitative reason for you to have a higher probability of delay is that there are more potential scenarios for delay than there are for simple blocking.

For example, suppose that all channels are used, and I make a call. 1 minute later you make a call. In the B scenario I am blocked. Perhaps a channel frees up in the next minute for you (probability p), perhaps not (probability 1-p).

In the C scenario if I hold for the minute of delay you will be delayed even if a channel comes through, since it would go to me (probability of not delayed p minus something instead of p). Obviously that's pretty specific, which is why (especially with alot of channels) the difference is not overly large.

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  • \$\begingroup\$ Yes that came in my mind after I posted this question but left it open just to see if that was the right reason. Other thing that troubles me is that why do Erlang C has values greater than one being a probability? \$\endgroup\$ Commented Oct 11, 2018 at 19:12

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