Graduate Program in Engineering and Technology Management Simulation of Discrete Event Systems Aslı...

Graduate Program in Engineering and Technology Management

Simulation of Discrete Event Systems

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2

Dynamic Simulation:Queueing System

Arrivals Departures

Service

is identified by:

•Arrival rate, interarrival time distribution•Service rate, service time distribution•# servers•# queues•Queue capacities•Queue disciplines, FIFO, LIFO, etc.

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M/M/1 Queueing System

Arrivals Departures

Service

M: interarrival time is exponentially distributedM: service time is exponentially distributed1: There is a single server

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Ex3: Model Specifics Initially (time 0) empty and idle Base time units: minutes Input data (assume given for now …), in minutes:

Part Number Arrival Time Interarrival Time Service Time1 0.00 1.73 2.902 1.73 1.35 1.763 3.08 0.71 3.394 3.79 0.62 4.525 4.41 14.28 4.466 18.69 0.70 4.367 19.39 15.52 2.078 34.91 3.15 3.369 38.06 1.76 2.37

10 39.82 1.00 5.3811 40.82 . .

. . . .

. . . . Stop when 20 minutes of (simulated) time have

passed

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Queuing Simulation Random variables:

Time between arrivals Service time represented by probability distributions.

Events: Arrival of a customer to the system Departure from the system.

State variables: # customers in the queue Worker status {busy, idle}

Output measures: Average waiting time in the queue % utilization of the server Average time spent in the system

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Output Performance Measures

Total production of parts over the run (P) Average waiting time of parts in queue:

Maximum waiting time of parts in queue:

N = no. of parts completing queue waitWQi = waiting time in queue of ith partKnow: WQ1 = 0 (why?)

N > 1 (why?)N

WQN

ii

1

iNi

WQmax,...,1

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Output Performance Measures (cont’d.)

Time-average number of parts in queue:

Maximum number of parts in queue:

Average and maximum total time in system of parts:

Q(t) = number of parts in queue at time t

20

)(200 dttQ

)(max200

tQt

iPi

P

ii

TSP

TS

max,...,1

1 ,

TSi = time in system of part i

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Output Performance Measures (cont’d.)

Utilization of the machine (proportion of time busy)

Many others possible (information overload?)

t

ttB

dttB

timeat idle is machine the if0

timeat busy is machine the if1)(,

20

)(200

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Simulation by Hand

Manually track state variables, statistical accumulators

Use “given” interarrival, service times Keep track of event calendar “Lurch” clock from one event to the next Will omit times in system, “max” computations

here (see text for complete details)

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System

Clock

B(t)

Q(t)

Arrival times of custs. in queue

Event calendar

Number of completed waiting times in queue

Total of waiting times in queue

Area under Q(t)

Area under B(t)

Q(t) graph B(t) graph

Time (Minutes) Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...

Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...

Simulation by Hand: Setup

0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

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System

Clock 0.00

B(t) 0

Q(t) 0


<empty>

Event calendar [1, 0.00, Arr] [–, 20.00, End]

Number of completed waiting times in queue 0

Total of waiting times in queue 0.00

Area under Q(t) 0.00

Area under B(t) 0.00



Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...

Simulation by Hand: t = 0.00, Initialize

0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

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System

Clock 0.00

B(t) 1

Q(t) 0


<empty>

Event calendar [2, 1.73, Arr] [1, 2.90, Dep] [–, 20.00, End]







Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...

Simulation by Hand: t = 0.00, Arrival of Part 1

0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

1

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System

Clock 1.73

B(t) 1

Q(t) 1


(1.73)

Event calendar [1, 2.90, Dep] [3, 3.08, Arr] [–, 20.00, End]







Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

12

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System

Clock 2.90

B(t) 1

Q(t) 0


<empty>








Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...

Simulation by Hand: t = 2.90, Departure of Part 1

0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

2

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System

Clock 3.08

B(t) 1

Q(t) 1


(3.08)








Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

23

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System

Clock 3.79

B(t) 1

Q(t) 2


(3.79, 3.08)








Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

234

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System

Clock 4.41

B(t) 1

Q(t) 3


(4.41, 3.79, 3.08)







Time (Minutes)

Interarrival times 1.73, 1.35, 0.71, 0.62, 14.28, 0.70, 15.52, 3.15, 1.76, 1.00, ...

Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

2345

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System

Clock 4.66

B(t) 1

Q(t) 2


(4.41, 3.79)







Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

345

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System

Clock 8.05

B(t) 1

Q(t) 1


(4.41)







Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

45

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System

Clock 12.57

B(t) 1

Q(t) 0


()







Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

5

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System

Clock 17.03

B(t) 0

Q(t) 0

Arrival times of custs. in queue ()

Event calendar [6, 18.69, Arr] [–, 20.00, End]






Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

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System

Clock 18.69

B(t) 1

Q(t) 0

Arrival times of custs. in queue ()

Event calendar [7, 19.39, Arr] [–, 20.00, End] [6, 23.05, Dep]






Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

6

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System

Clock 19.39

B(t) 1

Q(t) 1


(19.39)

Event calendar [–, 20.00, End] [6, 23.05, Dep] [8, 34.91, Arr]






Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...


0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

67

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Simulation by Hand: t = 20.00, The End

0

1

2

3

4

0 5 10 15 20

012

0 5 10 15 20

67

System

Clock 20.00

B(t) 1

Q(t) 1


(19.39)

Event calendar [6, 23.05, Dep] [8, 34.91, Arr]






Time (Minutes)


Service times 2.90, 1.76, 3.39, 4.52, 4.46, 4.36, 2.07, 3.36, 2.37, 5.38, ...

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Ex3:Complete Record of the Hand Simulation

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Ex3: Simulation by Hand:Finishing Up Average waiting time in queue:

Time-average number in queue:

Utilization of drill press:

part per minutes 53261715

queue in times of No.queue in times of Total

..

part 79.020

78.15

eclock valu Final

curve )(under Area

tQ

less)(dimension 92.020

34.18

eclock valu Final

curve )(under Area

tB

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Entity Based-Simulation

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ITEM #INTERARRIVAL TIME

ARRIVAL TIME

SERVICE TIME

START SERVICE TIME

END SERVICE TIME WAIT TIME

TIME IN SYSTEM

1 1,73 0,00 2,90 0,00 2,90 0,00 2,90

2 1,35 1,73 1,76 2,90 4,66 1,17 2,93

3 0,71 3,08 3,39 4,66 8,05 1,58 4,97

4 0,62 3,79 4,52 8,05 12,57 4,26 8,78

5 14,28 4,41 4,46 12,57 17,03 8,16 12,62

6 0,70 18,69 4,36 18,69 23,05 0,00 4,36

7 15,52 19,39 2,07 23,05

8 3,15 34,91

Average wait time=(0+1.17+1.58+4.26+8.16+0)/6=2.53min

Average time in system=(2.9+2.93+4.97+8.78+12.62+4.36)/6=36.56/6=6.09min

Since simulation ends at 20th minute, 6th item’s process

will not be completed!

The last event that occurs in a simulation will be the arrival of 7th item!

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Randomness in Simulation The above was just one “replication” — a sample of size

one (not worth much) Made a total of five replications:

Confidence intervals for expected values: In general, For expected total production,

nstX n //, 211 )/.)(.(. 56417762803 042803 ..

Notesubstantialvariabilityacrossreplications

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Comparing Alternatives

Usually, simulation is used for more than just a single model “configuration”

Often want to compare alternatives, select or search for the best (via some criterion)

Simple processing system: What would happen if the arrival rate were to double? Cut interarrival times in half Rerun the model for double-time arrivals Make five replications

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Results: Original vs. Double-Time Arrivals

Original – circles Double-time – triangles Replication 1 – filled in Replications 2-5 – hollow Note variability Danger of making

decisions based on one (first) replication

Hard to see if there are really differences

Need: Statistical analysis of simulation output data

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Graduate Program in Engineering and Technology Management Simulation of Discrete Event Systems Aslı...

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