MGTSC 352

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MGTSC 352 Lecture 25: Congestion Management MEC example Manufacturing example

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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 - PowerPoint PPT Presentation

Transcript of MGTSC 352

Page 1: MGTSC 352

MGTSC 352

Lecture 25: Congestion Management

MEC example

Manufacturing example

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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.

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Modeling Approaches

• Waiting line analysis template

• Simulation

• We’ll use both for this exampleTo Excel …

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The Laws of Queueing

• eff = (1 – PrBalk) Effective service rate = entering rate * % sticking around

• = eff /(s) utilization = demand / capacity

• W = Wq + 1/ Time in the system = time in the Q + time in service

• L = = Lq + S* # people in the system = # in the line + # in service # in service = # servers * probability of server being busy

• Little’s Law: L = effW• Little’s Law for the queue: Lq = effWq• Little’s Law for the servers: s* = eff (1/)

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Manufacturing Example (p. 184)

Machine

(1.2 or 1.8/minute)1/minute

Poisson arrivals

Exponential service times

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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 …

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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.

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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 Law L = × W

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Manufacturing Example 2(p. 184)

Machine

(1.2 or 1.8/minute)1.1/minute

Poisson arrivals

Exponential service times

Changed from 1 to 1.1

Reminder:

Cost of idle jobs

= holding cost

= $2.50 / minute / job

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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