Congestion Management: Queuing Theory and Manufacturing Examples, Slides of Production and Operations Management

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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.

• 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.

The Laws of Queueing

• λeff = λ (1 – PrBalk)  Effective service rate = entering rate * % sticking around
• ρ = λeff /(sμ)  utilization = demand / capacity
• W = Wq + 1/μ  Time in the system = time in the Q + time in service
• L = = Lq + S*ρ  # people in the system = # in the line + # in service  # in service = # servers * probability of server being busy
• Little’s Law: L = λeff W
• Little’s Law for the queue: Lq = λeff Wq
• Little’s Law for the servers: s* ρ = λeff (1/μ)

Manufacturing Example (p. 184)

Machine (1.2 or 1.8/minute) 1/minute

Poisson arrivals

Exponential service times

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.

Cost of waiting (Mach. 1)

• Method 1:

• Unit cost × L = ($2.50 / min. job) × ( 5.00 jobs)

= $13.70 / min

• Method 2:

• Unit cost × λ × W =

= ($2.50 / min. job) × ( 5.00 min) × (1 job/min)

= $13.70 / min

• Little’s Law L = λ × W Docsity.com

Manufacturing Example 3 (p. 184)

Machine (1.2 or 2 at .6/min) 1/minute

Poisson arrivals

Exponential service times

Reminder : Cost of idle jobs = holding cost = $2.50 / minute / job

Two machines, each taking twice as long