I am trying to work out a queuing theory calculation for a Markov M/M/n+M queue, but cannot get my numbers to match the correct output.
I am trying to calculate the probability that a call abandons.
The formula is given on the last page of this link.
http://iew3.technion.ac.il/serveng/References/Erlang_A_Avi.pdf
The paper provides an answer to the problem (Image of section 4.3) where.
- Number of Agents handling calls = 10
- Average Time to Handle 1 call = 2 minutes
- 300 Calls per 60 minutes
- Average Caller Patience = 2 minutes
This gives a Probability to Abandon of 12.5%
I have been using this method to work out the calculation
• λ – arrival rate (calls per unit of time); = 300/ 3600 = 0.083333333 calls per second
• µ – service rate (1/µ is the average duration of service); = 1/ 120 = 0.008333333 calls per second
• n – number of servers/agents; n = 10
• θ – individual abandonment rate (1/θ is the average patience); = 1/120 = 0.0083333333 calls per second
I then try to work out A(x,y) as given in Equation 3.5
$$A(x,y)=1+\sum_{j=1}^\infty \frac{(y^j)}{\prod_{k=1}^j(x+k)}$$
Where $$ x = \frac{n \mu} { \theta } $$ = (10 x 0.083333333)/ 0.0083333333 = 10
And $$ y = \lambda / \theta $$ = 0.083333333 / 0.0083333333 = 10
I then work out that A(x,y) = 1.718282
Offered load per agent = $$ \rho = \lambda / n \mu $$ = 0.0083333333 / (10 * 0.008333333 ) = 1
I then try to work out the probability to Abandon
$$P(Ab)=\frac{1} {\rho A(\frac{n \mu} { \theta },\frac{\lambda } { \theta })} + 1 - \frac{1} {\rho} $$
P(abandon) = 1/( 1 x 4.332748 ) + 1 - (1/1) = 23.08%
I cannot figure out where I am going wrong.
Where am I going wrong?
$\theta = 0.008333$ gives x = 10 and y = 10. Please rework A(x,y) and I hope you get the right answer
$$E_{1,n} = \frac{2755}{12842} = .214582$$
$$P(W>0) = \dfrac{.4332748\times0.214582}{1+(.4332748-1)*(.214582)} = .54207$$
$$P\text{{Ab}} =.23\times0.54207 \approx .125$$