Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

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

Examples of congestion management using queuing theory in manufacturing systems. It includes details on call center management and manufacturing examples with poisson arrivals and exponential service times. The document also covers the laws of queuing and cost calculations.

Typology: Slides

2012/2013

Uploaded on 01/01/2013

dipal
dipal 🇮🇳

4.5

(18)

106 documents

1 / 10

Toggle sidebar

Related documents


Partial preview of the text

Download Congestion Management: Queuing Theory and Manufacturing Examples and more Slides Production and Operations Management in PDF only on Docsity!

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.

Modeling Approaches

• Waiting line analysis template

• Simulation

• We’ll use both for this example

To Excel …

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

  • 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 …

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 2 (p. 184)

Machine (1.2 or 1.8/minute) 1.1/minute

Poisson arrivals

Exponential service times

Changed from 1 to 1.

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

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