EEL6507sp09L14

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EEL6507 Spring 2009, Lecture 14, Monday 2009/02/09 (Notes created by Adam Flynn)

Covered in this lecutre:

  • Continuation of discouraged arrivals from lecture 13
  • Little's Law and M/M/\infty
  • Solution to problem 2.12
  • M / M / m

Discouraged arrivals (cont'd from Lecture 13)

For discouraged arrivals, the arrival rate decreases as the number of customers in the queue decreases. If the decreasing arrival rate is directly inversely proportional the the number of customers in the queue, the following relationship is true:

\lambda_k = \frac{\lambda}{k}

Little's Law and M/M/\infty

From previous lectures, we recall that:

\bar{N} = \lambda \bar{T}

and

\rho = 1 - \rho_0 = \lambda \bar{x}

, where \bar{x} = \frac{1}{\mu} and ρ0 = e − α / μ for a Poisson process.

From this, we see that:

\lambda = \frac{1-\rho_0}{\bar{x}} = \mu(1-e^{-\alpha / \mu})

For low-intensity systems, \lambda \approx \alpha .

For high-intensity systems, \lambda \approx \mu .

Thus, combining above equations:

\bar{T} = \frac{\bar{N}}{1-\rho_0}\bar{x}

Alternatively:

(1-\rho_0)\bar{T} = \bar{N} \bar{x}

In layman's terms, the above equation says that: (probability of being busy)(average system time) = (average number in system)(average service time).


For a M/M/\infty system, λ is the arrival rate and kμ is the service rate.

Solution to problem 2.12

For a birth-death process, λk = λ and μk = kμ (same as M/M/\infty).

We need to show that P(z,t) = e^{\frac{\lambda}{\mu}(1-e^{-\mu t})(z-1)} .

The differential-difference equation for the system is:

\frac{\partial P(z,t)}{\partial t} = \lambda P_{k-1}(t) + \mu (k+1) P_{k+1}(t) - (\lambda + k \mu) P_{k}(t)

If the Z-transform is defined as:

P(z,t)=\sum_{k=0}^{\infin} P_{k}(t)z^k

We can derive the following relationships:

z P(z,t)=\sum_{k=0}^{\infin} P_{k-1}(t)z^k
\frac{\partial P(z,t)}{\partial z} = \sum_{k=1}^{\infin} (k+1) P_{k+1}(t)z^k
z \frac{\partial P(z,t)}{\partial z} = \sum_{k=1}^{\infin} k P_{k}(t)z^k

The partial derivatives of the original equation for P(z,t) with respect to t and z are:

\frac{\partial P(z,t)}{\partial t} = P(z,t) \frac{\lambda}{\mu}(z-1)e^{-\mu t} \mu
\frac{\partial P(z,t)}{\partial z} = P(z,t) \frac{\lambda}{\mu}(1-e^{-\mu t})

If we substitute everything into the original differential-difference equation, everything will cancel, proving that the given Z-transform is correct.

M / M / m System

Here, we develop a new "trick" to help us out:

p_k = p_0 \prod_{i = 0}^{k - 1} \frac{\lambda_i}{\mu_{i + 1}} = p_0 \prod_{i = n}^{k - 1} \frac{\lambda_i}{\mu_{i + 1}} \prod_{i = 0}^{n - 1} \frac{\lambda_i}{\mu_{i + 1}}

For a M / M / m system, we define the following:

p_k = p_0 \prod_{i = 0}^{k - 1} \frac{\lambda}{(i+1) \mu} = p_0 \left(\frac{\lambda}{\mu}\right)^k \frac{1}{k!}, k \le m

p_k = p_m \prod_{i = m}^{k - 1} \frac{\lambda}{m \mu} = p_m \left(\frac{\lambda}{m \mu}\right)^{k-m}, k > m

Lecture Voice Recording

lecture_recording

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