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MGTSC 352

MGTSC 352. Lecture 25: Congestion Management MEC example Manufacturing example. MEC (p. 181). One operator, two lines to take orders Average call duration: 4 minutes exp Average call rate: 10 calls per hour exp Average profit from call: $24.76 Third call gets busy signal

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MGTSC 352

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  1. MGTSC 352 Lecture 25: Congestion Management MEC example Manufacturing example

  2. MEC (p. 181) • One operator, two lines to take orders • Average call duration: 4 minutes exp • Average call rate: 10 calls per hour exp • Average profit from call: $24.76 • Third call gets busy signal • How many lines/agents? • Line cost: $4.00/ hr • Agent cost: $12.00/hr • Avg. time on hold < 1 min.

  3. Modeling Approaches • Waiting line analysis template • Simulation • We’ll use both for this example To Excel …

  4. The Laws of Queueing • leff = l (1 – PrBalk) • Effective service rate = entering rate * % sticking around • r = leff /(sm) • utilization = demand / capacity • W = Wq + 1/m • Time in the system = time in the Q + time in service • L = = Lq + S*r • # people in the system = # in the line + # in service • # in service = # servers * probability of server being busy • Little’s Law: L = leffW • Little’s Law for the queue: Lq = leffWq • Little’s Law for the servers: s*r = leff (1/m)

  5. Manufacturing Example (p. 184) Machine (1.2 or 1.8/minute) 1/minute Poisson arrivals Exponential service times

  6. Manufacturing Example • Arrival rate for jobs: 1 per minute • Machine 1: • Processing rate: 1.20/minute • Cost: $1.20/minute • Machine 2: • Processing rate: 1.80/minute • Cost: $2.00/minute • Cost of idle jobs: $2.50/minute • Which machine should be chosen? To Excel …

  7. Manufacturing Example • Cost of machine 1 = $1.20 / min. + ($2.50 / min. / job)  (5.00 jobs) = $13.70 / min. • Cost of machine 2 =$2.00 / min. + ($2.50 / min. / job)  (1.25 jobs) = $5.13 / min.  Switching to machine 2 saves money – reduction in lost revenue outweighs higher operating cost.

  8. Cost of waiting (Mach. 1) • Method 1: • Unit cost × L = ($2.50 / min. job)  (5.00 jobs) = $13.70 / min • Method 2: • Unit cost ×× W = = ($2.50 / min. job)  (5.00 min)  (1 job/min) = $13.70 / min • Little’s LawL = × W

  9. Manufacturing Example 2(p. 184) Changed from 1 to 1.1 Machine (1.2 or 1.8/minute) 1.1/minute Reminder: Cost of idle jobs = holding cost = $2.50 / minute / job Poisson arrivals Exponential service times

  10. Manufacturing Example 3(p. 184) Two machines, each taking twice as long Machine (1.2 or 2 at .6/min) 1/minute Reminder: Cost of idle jobs = holding cost = $2.50 / minute / job Poisson arrivals Exponential service times

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