Exponential Distribution Mean of X Exponential (k): E [X] = 1/k Variance of X Exponential (k):2/X =1/k2 Probability X is less than t: P (X < t) = 1 ekt Probability X is greater than t:P (X > t) = ekt

Geometric Distribution Number in system:L Geometric (1 ) Mean number in system:E [L] = /1 Variance of number in system: 2/L =/(1 )2 Probability L is greater than n:P (L > n) = n+1

M/M/1 (Single-server) Queue Interarrival time: A Exponential () Service time:T Exponential () Time spent in system: W Exponential ( ) Utilization factor: = / Mean number in line: E [Lq] = 2/1 Mean time in line: E [Wq] = E [Lq]/ Mean time in system:E [W] = E [L]/ Probability of empty system:P0 = 1

3 Manipulating the E [Lq] Equation An interesting problem is deconstructing the familiar M/M/1 diagram to solve for unknown parameters. Given all of the statistical manipulation has been done to get steady-state solutions, moving things around is now in the realm of algebra, and familiar to most of you. At this point,all weve done is basic addition, subtraction, multiplication, and division, without all the need for more complex operators. One simple piece we have not covered so far is that, given E [Lq], we could solve for the input parameter, , via the quadratic equation: =E [Lq] (E [Lq] + 4) E [Lq]/2

Think about the quadratic equation in the following questions. 1. Given = 10 per hour and E [Lq] = 0.9 people, what is the distribution of the interarrival time, A? (Hint: make sure to first derive the input parameter .)

2. Using the above distribution, what is the probability P (A > E [T])? (Hint: first consider the distribution of the service time T, then find its associated mean E [T].)

3. What is the average time spent in the system (in minutes)? (a) 6 minutes (b) 9 minutes (c) 15 minutes (d) 20 minutes

Please solve all the questions, show all the steps and provide explaination. Also, please refer to the formula provided above.