Chapter 2 Machine Interference Model Long Run Analysis Deterministic Model Markov Model

IE 512Chapter 22 Problem Description Group of m automatic machines Operator must change tools or perform minor repairs How many machines should be assigned to one operator? Performance measures –Operator utilization:  = fraction of time the operator is busy –Production rate: TH = # finished items per unit time –Machine availability:  = TH/G, where G is the gross production rate, or the production rate that would be achieved if each machine were always available Note: In this queuing system, the machines are the customers!

IE 512Chapter 23 Long Run Analysis Each machine has gross production rate h P n is the proportion of time that exactly n machines are down: Then, given P n,

IE 512Chapter 24 Eliminate some unknowns Suppose the mean time to repair a machine is 1/ , and the mean time between failures for a single machine is 1/. = avg. # of repairs in (0,t] =  t = avg. # of failures in (0,t] = In the long run, assuming the system is stable,

IE 512Chapter 25 Queuing Measures of Performance = average # of machines waiting for service = average number of machines down = average downtime duration of a machine = average duration of waiting time for repair

IE 512Chapter 26 Little’s Formula Observe from the previous equations: where is the total average number of failures per unit time = the arrival rate of customers to the queuing system Little’s formula relates mean # of customers in system to mean time a customer spends in the system.

IE 512Chapter 27 A Deterministic Model Suppose each machine spends exactly 1/ time units working followed by exactly 1/  time units in repair. Then if and we could stagger the failure times, we would have no more than one machine unavailable at any time, so that (Otherwise,

IE 512Chapter 28 A Markov Model Let be the time between the (n-1) st repair and the n th failure of machine j, and be the time duration of the n th repair (indep.) The time until the first failure is N(t) = # of machines down at time t follows a CTMC with S = {0, 1, …, m} and

IE 512Chapter 29 Steady-State Probabilities satisfy the balance equationsor level-crossing equations

IE 512Chapter 210 Solution

IE 512Chapter 211 Erlang Distribution If failure and/or repair times are not exponential, can fit an Erlang distribution by matching moments: Big advantage: Can still model as a CTMC. Consider time to machine failure (each machine) as Erlang k. Can think in terms of k phases in the time to failure, where the time the m/c spends in each phase is exponential (k ): Mean time spent in each phase = Mean total time to failure =

IE 512Chapter 212 Expanded State Definition M i (t) = # of machines operating in phase i at time t For example, if k = 2, then a single machine without interference follows the CTMC (1 = failed state): 1 0;1 0;2   

IE 512Chapter 213 Transitions among States (k=2) Steady state probabilities: Rate into state  2 (l 1 +1) 2 (l 2 +1)

IE 512Chapter 214 Balance Equations This system of equations (for any k) has the solution:

IE 512Chapter 215 SS Number of Machines Working From the previous equation and get Find probabilities by normalizing to 1. This distribution is independent of k or any other characteristics of the failure time distribution. It can be shown that the same state distribution holds for any failure time distribution!