Modelling and Simulation IS 331. Lec (3) презентация

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Monte Carlo simulation Estimating π Craps (dice game) Time-stepped simulation

Monte Carlo simulation
Estimating π
Craps (dice game)
Time-stepped simulation
Mortgage scenarios
Trace-driven simulation
Single-server queue (ssq1.c)
Discrete-event

simulation
Witchcraft hair salon

Simulation Examples

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Monte Carlo simulation Estimating π Craps (dice game) Simulation Examples

Monte Carlo simulation
Estimating π
Craps (dice game)

Simulation Examples

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Example 3 Use the one-digit Random Number (6, 3, 5,

Example 3

Use the one-digit Random Number (6, 3, 5, 0, 8)

to generate random observations for :
Throwing an unbiased coin,
Throwing a die,
The color of a traffic found by a randomly arriving car when it is green 40% of the time, yellow 10% of the time, and red 50% of the time.
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To simulate a coin using one digit R.N.(0-9), let 0

To simulate a coin using one digit R.N.(0-9),
let 0

- 4 Represent a Head and
5 – 9 Represent a Tail;
so the solution will be:

(6, 3, 5, 0, 8) (T, H, T, H, T)

R. Observations

Example 3: Solution

(6, 3, 5, 0, 8)

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Example 3: Solution b) To simulate a die using number

Example 3: Solution

b) To simulate a die using number 0

– 9.
Let: 1 – 6 represents the faces of the die; then
0, 7, 8 and 9 are rejected.
So the solution will be

(6, 3, 5, 0, 8) (6, 3, 5, reject, reject)

R. Observations

(6, 3, 5, 0, 8)

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Example 3: Solution c) Divide 0 – 9 into 3

Example 3: Solution

c) Divide 0 – 9 into 3 classes

according to the given probability:
Green 40% 0, 1, 2, 3
Yellow 10% 4
Red 50% 5, 6, 7, 8, 9

So (6, 3, 5, 0, 8) (Red, Green, Red ,Green, Red)

R. Observations

(6, 3, 5, 0, 8)

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A news boy buys paper for 15 P.T. each and

A news boy buys paper for 15 P.T. each and

sells them for 25 P.T. each. Unsold papers have no value in analyzing past sells, it is found that three demand distribution exist, depending on news events in the papers. A good interesting news day yield to many potential customer, typical and poor days yield to less interest. Historically 20% of days have been “good”, 50% “typical”, 30% “Poor”, The demand distributions are summarized :
Simulate 10 days period and determine main daily profit, max profit, min profit if he decides to buy 80 papers each day. Use the following R.N :-
09 65 69 66 57 64 84 24 92 22 49 50 61 01 45 48 48 45 52 35

Example 4: A news boy

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Example 4: Solution (A news boy)

Example 4: Solution (A news boy)

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06/10/2019 1st Day: RN 09 --> Good RN 65 -->

06/10/2019

1st Day:
RN 09 --> Good
RN 65 --> demand = 60
Profit =

60*25 – 80*15 = 300 P.T
 2nd Day:
RN 69 --> Typical
RN 66 --> demand = 80
Profit = 80*25 – 80*15 = 800 P.T
 3rd Day:
RN 57 --> Typical
RN 64 --> demand = 80
Profit = 80*25 – 80*15 = 800 P.T
4th Day:
RN 84 --> poor
RN 24 --> demand = 100
Profit = 80*25 – 80*15 = 800 P.T ; we calculate with demand 80 NOT 100 as the newsboy buy only 80 news paper each day.

Example 4: Solution (A news boy)

Use the following R.N :-
09 65 69 66 57 64 84 24 92 22 49 50 61 01 45 48 48 45 52 35

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06/10/2019 5th Day: RN 92 --> poor RN 22 -->

06/10/2019

5th Day:
RN 92 --> poor
RN 22 --> demand = 100
Profit =

80*25 – 80*15 = 800 P.T
6th Day:
RN 49 --> Typical
RN 50 --> demand = 80
Profit = 80*25 – 80*15 = 800 P.T 
7th Day:
RN 61 --> Typical
RN 01 --> demand = 50
Profit = 50*25 – 80*15 = 50 P.T
8th Day:
RN 45 --> Typical
RN 48 --> demand = 80
Profit = 80*25 – 80*15 = 800 P.T

9th Day:
RN 48 --> Typical
RN 45 --> demand = 80
Profit = 80*25 – 80*15 = 800 P.T
10th Day:
RN 52 --> Typical
RN 35 --> demand = 70
Profit = 70*25 – 80*15 = 550 P.T
MIN profit: in 7th day which is 50 P.T
MAX profit: is 800 P.T
Main daily profit:
300+800+800+800+800+800+50+800+800+550/10=650 PT

Use the following R.N :-
09 65 69 66 57 64 84 24 92 22 49 50 61 01 45 48 48 45 52 35

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06/10/2019 Example 5: The owner of car renting company, pays

06/10/2019

Example 5: The owner of car renting company, pays a penalty

for
each accident according to the following distribution:

The accidents occur with the probability of 30% (number of accidents per day is up to 6). Number of accidents per day has the following distribution:

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Simulate 12 days. Based on your simulation what is the

Simulate 12 days. Based on your simulation what is the minimum

money the owner must have to cover the penalties occurred per day.
Use the following R.N:
44 99 15 97 21 47 80 28 87 13 33 42 84 27 64
59 33 84 00 10 50 51 09 31 12 94 96 97 77

Example 5: The owner of car renting company

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Solution 5: The owner of car renting company The probability

Solution 5: The owner of car renting company

The probability accidents occur

is 30% ? 00 --> 29
The probability accidents don’t occur is 70% ? 30 --> 99
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Solution 5: The owner of car renting company 1st Day:

Solution 5: The owner of car renting company

1st Day:
RN: 92 -->

NO ACCIDENT
2ND Day:
RN: 44 --> NO ACCIDENT
3RD Day:
RN: 99 --> NO ACCIDENT
4TH Day:
RN: 15 --> ACCIDENT
RN: 97 --> Number of accidents = 6 accidents.
? 1th Accident:
RN: 21 --> Amount of money paid= 50 L.E
? 2ND Accident:
RN: 47 --> Amount of money paid= 100 L.E
?3rd Accident:
RN: 80 --> Amount of money paid= 150 L.E
 ?4th Accident:
RN: 28 --> Amount of money paid = 50 L.E
?5th Accident:
RN: 87 --> Amount of money paid = 200 L.E
?6th Accident:
RN: 13 --> Amount of money paid= 50 L.E
Then the total paid in this day = 50 + 100 + 150 + 50 + 200 + 50 = 600 L.E

Use the following R.N:
44 99 15 97 21 47 80 28 87 13 33 42 84 27 64
59 33 84 00 10 50 51 09 31 12 94 96 97 77

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5th Day: RN: 33 --> NO ACCIDENT 6TH Day: RN:

5th Day:
RN: 33 --> NO ACCIDENT
6TH Day:
RN: 42 --> NO

ACCIDENT
7th Day:
RN: 84 --> NO ACCIDENT
8th Day:
RN: 27 --> ACCIDENT.
RN: 64 --> Number of accidents = 5 accidents.
1st Accident:
RN: 59 --> Amount of money paid = 100 L.E
2nd Accident:
RN: 33 --> Amount of money paid = 100 L.E
3rd Accident:
RN: 84 --> Amount of money paid = 150 L.E
4th Accident:
RN: 00 --> Amount of money paid = 50 L.E
5th Accident:
RN: 10 --> Amount of money paid = 50 L.E
Then the total paid in this day = 100 + 100 + 150 + 50 + 50 = 450 L.E

Solution 5: The owner of car renting company

Use the following R.N:
44 99 15 97 21 47 80 28 87 13 33 42 84 27 64
59 33 84 00 10 50 51 09 31 12 94 96 97 77

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9TH Day: RN: 50 --> NO ACCIDENT. 10TH Day: RN:

9TH Day:
RN: 50 --> NO ACCIDENT.
10TH Day:
RN: 51 --> NO

ACCIDENT.
11TH Day:
RN: 09 --> ACCIDENT.
RN: 31 --> Number of accidents = 4 accidents
?1st Accident:
RN: 12 --> Amount of money paid= 50 L.E
?2nd Accident:
RN: 94 --> Amount of money paid= 200 L.E
?3rd Accident:
RN: 96 --> Amount of money paid= 200 L.E
?4th Accident:
RN: 97 --> Amount of money paid= 200 L.E
Then the total paid in this day = 50 + 200 + 200 + 200 = 650 L.E
12th Day:
RN: 77 --> NO ACCIDENT
Based on the above simulation :- The minimum money the owner must have to cover the penalties occurred per day (The maximum paid money in all days) = 650 L.E

Solution 5: The owner of car renting company

Use the following R.N:
44 99 15 97 21 47 80 28 87 13 33 42 84 27 64
59 33 84 00 10 50 51 09 31 12 94 96 97 77

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