MGTSC 352

Lecture 23: Congestion Management

Introduction: Asgard Bank example

Simulating a queue

Types of congested systems, queueing template

Ride’n’Collide example

MEC example

Manufacturing example

Analyzing a Congested System (pg. 174)

System Description

Measures of Quality of Service

Measures important to Servers

Model of the System

Inputs

Outputs

Asgard Bank: Times Between Arrivals

(pg. 173)

0

10

20

30

40

50

60

70

0.00 - 0.250.25 - 0.500.50 - 0.750.75 - 1.001.00 - 1.251.25 - 1.501.50 - 1.751.75 - 2.002.00 - 2.252.25 - 2.502.50 - 2.752.75 - 3.003.00 - 3.253.25 - 3.503.50 - 3.753.75 - 4.00

> 4.00

Time between Arrivals (min.)

Frequency

Average = 1.00 min.St. dev. = 0.92 min.Arrival rate = λ=?

pg. 168

Asgard Bank: Service Times

0

10

20

30

40

50

60

70

80

0.00 - 0.100.20 - 0.300.40 - 0.500.60 - 0.700.80 - 0.901.00 - 1.101.20 - 1.301.40 - 1.501.60 - 1.701.80 - 1.902.00 - 2.102.20 - 2.302.40 - 2.502.60 - 2.702.80 - 2.903.00 - 3.103.20 - 3.303.40 - 3.503.60 - 3.703.80 - 3.90Duration of Service (min.)

Frequency

Average = 0.95 min. (57 sec.)St. dev. = 0.17 min. (10 sec.)

service rate = μ=?

Including Randomness: Simulation• Service times:

Normal distribution, mean = 57/3600 hrs, stdev = 10/3600 hrs.

MAX(NORMINV(RAND(),57/3600,10/3600),0)

• Inter-arrival times:Exponential distribution, mean = 1/ 60 hrs.

– (1/60)*LN(RAND())To Excel …

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

1

11

21

31

41

51

61

Customer number

Time (hours)

Waiting time

Service time

Press F9 to recalculate

Simulated Lunch Hour 1:

71 arrivals

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

1

11

21

31

41

51

61

Customer number

Time (hours)

Waiting time

Service time

Press F9 to recalculate

Simulated Lunch Hour 2:

50 arrivals

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

1

11

21

31

41

51

61

Customer number

Time (hours)

Waiting time

Service time

Press F9 to recalculate

Simulated Lunch Hour 3:

Unused capacity

Causes of Congestion

• Higher than average number of arrivals

• Lower than average service capacity

• Lost capacity due to timing

Lesson: For a service where customers arrive randomly, it is not a good idea to operate the system close to its average capacity

Template.xls

• Does calculations for– M/M/s– M/M/s/s+C– M/M/s//M– M/G/1

• Want to know more? Go to http://www.bus.ualberta.ca/aingolfsson/qtp/

• Asgard Bank Data– Model: M/G/1– Arrival rate: 1 per minute– Average service time: 57/60 min.– St. dev of service time: 10/60 min.

Asgard Conclusions

• The ATM is busy 95% of the time.

• Average queue length = 9.3 people

• Average no. in the system = 10.25 (waiting, or using the ATM)

• Average wait = 9.3 minutes

• What if the service rate changes to …– 1.05 / min.?– 1.06 / min.?

Ride’n’Collide (pg. 178)

• Repair personnel cost: $10 per hour

• Average repair duration: 30 minutes

• Lost income: $50 per hour per car

• Number of cars: 20 cars

• A car will function for 10 hours on average from the time it has been fixed until the next time it needs to be repaired.

• How many repair-people should be hired?

Ride’n’Collide

• Customers =• Servers =• Average number in system =

• Lost revenue per hour =• Arrival rate =• Service rate =• Model to use:

Waiting Line Analysis Template:Which Model to Use?

• Who are the customers?

• Who are the servers?

• Where is the queue?

… not always obvious

• If you are told how many customers there are

… then you should consider using the “finite population” template

waiting room = queue

potential customers parallel servers

Number is small enough to worry about

• If you are told the maximum number of customers that can wait (the size of the waiting room)

… then you should consider using the “finite Q” template

waiting room = queue

potential customers parallel servers

Capacity is small enough to worry about

• If you are told the standard deviation of the service times, and there is 1 server

… then you should consider using the “MG1” template

waiting room = queue

potential customers

one server, non-exponential service

time distribution

• If you are told nothing about the size of the pool of potential customers, or the maximum number that will wait, or the standard deviation of the service times,

… then you should probably use the “MMs” template

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.

Modeling Approaches

• Simulation

• Waiting line analysis template

• We’ll use both for this exampleTo Excel …

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