Queuing Theory Jackson Networks

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# Queuing Theory Jackson Networks - PowerPoint PPT Presentation

Queuing Theory Jackson Networks. Network Model. Consider a simple 3 stage model where the output of 2 queues becomes the input process for a third. Station A. Station C. Station B. Network Model. Station A. Station C. Proposition 1: rate in = rate out, that is, if l A and l B

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Presentation Transcript
Queuing Theory

Jackson Networks

Network Model

Consider a simple 3 stage model where the output of 2 queues becomes the input process for a third

Station A

Station C

Station B

Network Model

Station A

Station C

Proposition 1: rate in = rate out, that is, if lA and lB

are the input rates for station A and B respectively, then the input rate for station C is lA+lB.

Proposition 2: exponential inter-arrivals at A and B provide exponential inter-arrivals at station C.

Station B

Network Model

Station A

Station C

Instructor Derivation of the minimum of 2 exponentials.

Station B

Jackson Network

Station 1

1

2

s1

1

2

sv

1

2

sj

Station j

.

.

Station v

Jackson Network
• All external arrivals to each station must follow a Poisson process (exponential inter-arrivals)
• All service times must be exponentially distributed
• All queues must have unlimited capacity
• When a job leaves a station, the probability that it will go to another station is independent of its past history and of the location of any other job
• Calculate the input arrival rate to each station
• Treat each station independently as an M/M/s queue
Example

Jobs submitted to a computer center must first pass through an input processor before arriving at the central processor. 80% of jobs get passed on the central processor and 20% of jobs are rejected. Of the jobs that pass through the central processor, 60% are returned to the customer and 40% are passed to the printer. Jobs arrive at the station at a rate of 10 per minute. Calculate the arrival rate to each station.

Example

.6

.2

10

.8

.4

Input

Central

Printer

Example (cont.)
• For the computer center the processing times are 10 seconds for the input processor, 5 seconds for the central processor, and 70 seconds for the printer. Our task is to determine the number of parallel stations (multiple servers) to have at each station to balance workload.
Example

10

8

3.2

Input

Central

Printer

For our initial try, we will solve for s1 = 2, s2 = 1, s3 = 4

10

8

3.2

Input

Central

Printer

If these numbers are correct, clearly Lq and Wq indicate the bottleneck is at the printer station. We may wish to add a printer if speedy return of printouts is required. Secondarily, overall processing may be increased by adding another input processor.

Job Shop Example
• An electronics firm has 3 different products in a job shop environment. The job shop has six different machines with multiple machines at 5 of the 6 stations.

Product Order rate Flow

1 30/month ABDF

2 10/month ABEF

3 20/month ACEF

Summary information is on the network below.

Job Shop Example

Machine D

s = 3

m = 11

Machine B

s = 2

m = 22

Machine C

s = 1

m = 29

Machine A

s = 3

m = 25

Machine E

s = 2

m = 23

Machine F

s = 4

m = 20

Job Shop Example

Machine D

s = 3

m = 11

Machine B

s = 2

m = 22

Machine C

s = 1

m = 29

Machine A

s = 3

m = 25

Machine E

s = 2

m = 23

Machine F

s = 4

m = 20

Job Shop Example

Machine D

s = 3

m = 11

Machine B

s = 2

m = 22

Machine C

s = 1

m = 29

Machine A

s = 3

m = 25

Machine E

s = 2

m = 23

Machine F

s = 4

m = 20

Job Shop Example

Machine D

s = 3

m = 11

Machine B

s = 2

m = 22

Machine C

s = 1

m = 29

Machine A

s = 3

m = 25

Machine E

s = 2

m = 23

Machine F

s = 4

m = 20

Lets calculate metrics for Machine B

M/M/2 model with

Use formulas in Fig. 16.5 p. 564 or use charts

Repeat for Machines A, C, D, E, F

Machine D

s = 3

m = 11

Machine B

s = 2

m = 22

Machine C

s = 1

m = 29

Machine A

s = 3

m = 25

Machine E

s = 2

m = 23

Machine F

s = 4

m = 20

Interesting Application to Manufacturing

Note that lead time is just the time in the system which for product 1 which has sequence ABDF is

W = total time in system = lead time

= WA + WB + WD + WF = .789

WIP = work in process = parts per month x lead time

= 30(.789) = 23.67

Repeat for Machines A, C, D, E, F

Machine D

s = 3

m = 11

Machine B

s = 2

m = 22

Machine C

s = 1

m = 29

Machine A

s = 3

m = 25

Machine E

s = 2

m = 23

Machine F

s = 4

m = 20

Some Recommendations

If lead time is too slow (WIP too high), one possibility is to add additional machining to the bottleneck area. This occurs at stations D and B. Note that this assumes the cost of the machines is not a critical part of the decision. This may or may not need to be included in a final recommendation.