Simulation and modeling lab mnit jaipur.pptx

Modelling and Simulation Lab
(MEP-328P)
Paras Garg, PhD Scholar, Department of Mechanical
Engineering, MNIT Jaipur

List of
Experiments
Lab No. Name of experiment Research
Scholar
WEEK
1 Monte Carlo Simulation(Manually) Rohit WEEK-1
2 Simulation of single server queuing system
(Manually)
Rohit WEEK-2
3 Introduction to WITNESS software Vaibhav WEEK-3
4 Simulation of multiple sever queuing system with
WITNESS
Vaibhav WEEK-4
5 To find out existing operational bottlenecks in a
manufacturing system & carry possible what-if
analysis with WITNESS
Vaibhav WEEK-5
6 Forecasting for a Manufacturing System using
WITNESS
Shobha WEEK-6
7 Analysis of Manufacturing Plant layout for
Capacity Planning using WITNESS simulation
software
Shobha WEEK-7
8 Developing machine learning models for sound
classification using the Nano Edge AI toolkit.
Paras WEEK-8
9 Developing machine learning algorithms for
Motion Classification using Nano edge AI tool kit
Paras WEEK-9
10 Life Cycle Assessment of a PIN using Gabi
Software
Shobha/Rohit WEEK-10
11 Statistical Modelling using SPSS-1 Ved Prabha WEEK-11
12 Statistical Modelling using SPSS-2 Ved Prabha WEEK-12
Paras

Lab File
• AIM
• THEORY
• SOFTWARE USED
• PROCEDURE
• SIMULATION
• RESULTS
• CONCLUSION
• REFRENCES

EC
No.
Evaluation
Component
Weightage
1 MTE 20%
2 Lab file 20%
3 Viva 20%
4 Attendance
20%
5 ETE 20%
Evaluation Scheme

What is a Model?
This is not an apple just its graphical
representation
This graphical representation is enough if we want to know
how apple looks, but it’s not an abstracting of the complete
reality.

Why need Model?
• Describe, classify
• Understand
• Predict
• Control a phenomenon

What is Simulation?
Simulation includes developing a model of a real system and experimental
manipulation of the model to predict real system behavior.
The real system are often complex, dynamic and interactive.
A model is an abstraction of a real system representing the objects within
the system and their interactions.

Why Use Simulation?
• Ability to experiment with model system rather than the real system.
• Ability to obtain results in less time.
• Ability to control sources of variations.
• Ability to avoid errors in measurement.
• Ability to generate alternate scenarios for decision making.
• Ability to obtain replications.

Types of system
- Discrete System:- State Variables Changes instantaneously at separated
points in time.
• Bank Model: state changes occur only when a customer arrives
or departs.
- Continuous system:- State variables change continuously as a function
of time.
• Airplane flight:- State variables like position, velocity change
continuously
Many Systems are partly discrete, partly continuous

Discrete Random Variable
• Random Variable X is a function whose value describes the outcome
of a random experiment. If the number of the possible outcomes is
countable in a set of positive integers, the random variable is called a
discrete random variable. The associated probability mass
function(pmf) gives the probability that this random variable takes
one of the discrete value of k = 0,1,2,….infinity.

Continuous Random Variable
• A continuous Random Variable X can take any real value x from an
interval [a,b]. The cumulative distribution function F(x)(cdf) gives the
probability that the random variable is smaller to equal to x.
• A continuous random variable is one which takes an infinite number
of possible values. Continuous random variables are usually
measurements. Examples include height, weight, the amount of sugar
in an orange, the time required to run a mile

Steps-by-Step Procedure
• Establishing Probability Distribution
• Cumulative Probability Distribution
• Setting Random Intervals
• Generating Random Numbers
• To find the answer of question asked using the above 4 steps

Example: Shelly’s Supermarket
y’s Inventory Example: Shelly’s Supermarket
Inventory Supermarket
Shelly's makes a Rs.0.04 profit per postage stamp. The vending
machine holds 230 stamps and it costs Shelly’s Rs.4.00 in labor
to fill the machine. Shelly's will fill the machine at the beginning of
every 5th day. Conduct a 20-day simulation and determine the
expected profit per day. Assume the machine must be filled on
the first day.
Inventory Example: Shelly’s Supermarket
Case Study-1 Shelly’s Super Market

The set of random numbers corresponding to each sales level is:
Number Sold Random
Per Day Probability Numbers
20 .10 .00 but less than .10
30 .15 .10 but less than .25
40 .20 .25 but less than .45
50 .25 .45 but less than .70
60 .20 .70 but less than .90
70 .10 .90 but less than 1.0
Use the following stream of random numbers:
.71, .95, .83, .44, .34, .49, .88, .56, .05, .39,
.75, .12, .03, .59, .29, .77, .76, .57, .15, .53

Number Profit
of Stamps From Cost of
Random Left In Sale of Refilling Daily
Day Number Demand Machine Stamps Machine Profit
1 .71 60 170 2.40 4.00 -1.60
2 .95 70 100 2.80 -- 2.80
3 .83 60 40 2.40 -- 2.40
4 .44 40 0 1.60 -- 1.60
5 .34 40 0 0 -- 0
6 .49 50 180 2.00 4.00 -2.00
7 .88 60 120 2.40 -- 2.40
8 .56 50 70 2.00 -- 2.00
9 .05 20 50 .80 -- .80
10 .39 40 10 1.60 -- 1.60

Number Profit
11 .75 60 170 2.40 4.00 -1.60
12 .12 30 140 1.20 -- 1.20
13 .03 20 120 .80 -- .80
14 .59 50 70 2.00 -- 2.00
15 .29 40 30 1.60 -- 1.60
16 .77 60 170 2.40 4.00 -1.60
17 .76 60 110 2.40 -- 2.40
18 .57 50 60 2.00 -- 2.00
19 .15 30 30 1.20 -- 1.20
20 .53 50 0 1.20 -- 1.20

 Summary: Filling Every 5th Day
Total profit for 20 days = $19.20
Expected profit per day = 19.20/20 = $0.96
Would filling every 4th day be more profitable?

Number Profit
1 .71 60 170 2.40 4.00 -1.60
2 .95 70 100 2.80 -- 2.80
3 .83 60 40 2.40 -- 2.40
4 .44 40 0 1.60 -- 1.60
5 .34 40 190 1.60 4.00 -2.40
6 .49 50 140 2.00 -- 2.00
7 .88 60 80 2.40 -- 2.40
8 .56 50 30 2.00 -- 2.00
9 .05 20 210 .80 4.00 -3.20
10 .39 40 170 1.60 -- 1.60

Number Profit
11 .75 60 110 2.40 -- 2.40
12 .12 30 80 1.20 -- 1.20
13 .03 20 210 .80 4.00 -3.20
14 .59 50 160 2.00 -- 2.00
15 .29 40 120 1.60 -- 1.60
16 .77 60 60 2.40 -- 2.40
17 .76 60 170 2.40 4.00 -1.60
18 .57 50 120 2.00 -- 2.00
19 .15 30 90 1.20 -- 1.20
20 .53 50 40 2.00 -- 2.00

 Summary: Filling Every 4th Day
Total profit for 20 days = $17.60
Expected profit per day = 17.60/20 = $0.88
The simulation results suggest that Shelly
should fill the stamp machine very 5th day, rather
than every 4th day.

Waiting Line Example: Wayne Airport
Wayne International Airport primarily serves domestic air traffic. Occasionally,
however, a chartered plane from abroad will arrive with
passengers bound for Wayne's two great amusement parks, Algorithmland
and Giffith's Cherry Preserve.
Whenever an international plane arrives at the airport the two customs
inspectors on duty set up operations to process the passengers.
Incoming passengers must first have their passports and visas checked. This is
handled by one inspector. The time required to check a passenger's
passports and visas can be described by the probability distribution on the
next slide.

Time Required to
Check a Passenger's
Passport and Visa Probability
20 seconds .20
40 seconds .40
60 seconds .30
80 seconds .10

After having their passports and visas checked, the passengers next proceed
to the second customs official who does baggage inspections. Passengers
form a single waiting line with the official inspecting baggage on a first come,
first served basis. The time required for baggage inspection follows the
probability distribution shown below.
Time Required For
Baggage Inspection Probability
No Time .25
1 minute .60
2 minutes .10
3 minutes .05

• Random Number Mapping
Time Required to
Check a Passenger's Random
Passport and Visa Probability Numbers
20 seconds .20 .00 < .20
40 seconds .40 .20 < .60
60 seconds .30 .60 < .90
80 seconds .10 .90 < 1.0

• Random Number Mapping
Time Required For Random
Baggage Inspection Probability Numbers
No Time .25 .00 < .25
1 minute .60 .25 < .85
2 minutes .10 .85 < .95
3 minutes .05 .95 < 1.0

• Next-Event Simulation Records
For each passenger the following information must be
recorded:
• When his service begins at the passport control inspection
• The length of time for this service
• When his service begins at the baggage inspection
• The length of time for this service

• Time Relationships
Time a passenger begins service
by the passport inspector
= (Time the previous passenger started passport service)
+ (Time of previous passenger's passport service)

Time a passenger begins service
by the baggage inspector
( If passenger does not wait in line for baggage inspection)
= (Time passenger completes service
with the passport control inspector)
(If the passenger does wait in line for baggage inspection)
= (Time previous passenger completes
service with the baggage inspector)

Time a customer completes service
at the baggage inspector
= (Time customer begins service with baggage inspector) + (Time
required for baggage inspection)

A chartered plane from abroad lands at Wayne
Airport with 80 passengers. Simulate the processing
of the first 10 passengers through customs.
Use the following random numbers:
For passport control:
.93, .63, .26, .16, .21, .26, .70, .55, .72, .89
For baggage inspection:
.13, .08, .60, .13, .68, .40, .40, .27, .23, .64

• Simulation Worksheet (partial)
Passport Control Baggage Inspections
Pass. Time Rand. Serv. Time Time Rand. Serv. Time
Num. Begin Num. Time End Begin Num. Time End
1 0:00 .93 1:20 1:20 1:20 .13 0:00 1:20
2 1:20 .63 1:00 2:20 2:20 .08 0:00 2:20
3 2:20 .26 :40 3:00 3:00 .60 1:00 4:00
4 3:00 .16 :20 3:20 4:00 .13 0:00 4:00
5 3:20 .21 :40 4:00 4:00 .68 1:00 5:00

• Simulation Worksheet (continued)
Passport Control Baggage Inspections
Pass. Time Rand. Serv. Time Time Rand. Serv. Time
Num. Begin Num. Time End Begin Num. Time End
6 4:00 .26 :40 4:40 5:00 .40 1:00 6:00
7 4:40 .70 1:00 5:40 6:00 .40 1:00 7:00
8 5:40 .55 :40 6:20 7:00 .27 1:00 8:00
9 6:20 .72 1:00 7:20 8:00 .23 0:00 8:00
10 7:20 .89 1:00 8:20 8:20 .64 1:00 9:20

• Explanation
For example, passenger 1 begins being served by the passport
control inspector immediately. His service time is 1:20 (80 seconds) at
which time he goes immediately to the baggage inspector who waves
him through without inspection.
Passenger 2 begins service with passport inspector 1:20 minutes
(80 seconds) after arriving there (as this is when passenger 1 is
finished) and requires 1:00 minute (60 seconds) for passport
inspection. He is waved through baggage inspection as well.
The process continues in this manner.

• Question
How long will it take for the first 10 passengers to clear
customs?
• Answer
Passenger 10 clears customs after 9 minutes and 20 seconds.

• Question
What is the average length of time a customer waits before having
his bags inspected after he clears passport control? How is this
estimate biased?

• Answer
For each passenger calculate his waiting time:
(Baggage Inspection Begins) - (Passport Control Ends)
= 0+0+0+40+0+20+20+40+40+0 = 120 seconds.
120/10 = 12 seconds per passenger
This is a biased estimate because we assume that the
simulation began with the system empty. Thus, the results tend to
underestimate the average waiting time.

• =RAND()
• =IF(C5<0.2, 20, IF(C5<0.6, 40, IF(C5<0.9, 60, 80)))
• =MAX()
• =SUM()
Excel Formulas used in Case-2

Simulation and modeling lab mnit jaipur.pptx

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