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Monte Carlo Simulation

This document provides an overview of Monte Carlo simulation and examples of how to apply it. Monte Carlo simulation uses random numbers to estimate outcomes of complex problems that are difficult to solve analytically. Example problems include simulating daily demand at a grocery store, coin flips, dice rolls, and traffic light colors. The document also provides a newsboy problem simulation and car accident penalty simulation to demonstrate how to set up and run Monte Carlo simulations.

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Modeling & Simulation(IS331) Tutorial Session 11 Ayman S. Abdelaziz
MONTE CARLO SIMULATION
What is Monte Carlo ? ▪From the previous description we can define Monte Carlo methods generally as an experimental mathematical tool that is used to experiment with random numbers to solve or estimate complex problems. ▪Monte Carlo simulation are usually associated with problems with theoretical interests or optimizing certain values with large amount of variables and restrictions which is otherwise almost impossible to optimize. ▪Unlike discrete event simulation Monte Carlo simulation are not concerned with the passage of time. ▪The interesting property of Monte Carlo simulation is that, with the appropriate set up of the problem, it can be used to solve (or estimate) both deterministic and stochastic problems as will be explained in the following examples.
Problem 1 Get the average daily demand for a small grocery store selling a fresh bread according to the following table: Note that: the proportion of balls of a specific color corresponds exactly to the probability of a specific level of daily demand.
Solutions To simulate the daily demand(5 days) Draw one ball at a time, notice its color and then place it back in the bowl. Then translate the outcomes into unique values of demand. Expected value of simulated demand = 560/5 = 112 units / day Analytical solution: Expected daily demand =100(0.2) + 110(0.5) +120(0.3) = 111 units / day
Problem 2: 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. Solution 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) R. Observations (T, H, T, H, T)
Solutions 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) R. Observations (6, 3, 5, reject, reject) 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) R. Observations (Red, Green, Red ,Green, Red)
Problem 3: News boy 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 Demand Types of new days Good Typical Poor 40 0.05 50 0.20 0.03 60 0.41 0.07 70 0.26 0.28 80 0.08 0.30 0.02 90 0.25 0.15 100 0.07 0.28 110 0.24 120 0.12 130 0.09 140 0.06 150 0.04
Solutions
Solutions 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.
Solutions 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
Problem 4: 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:
Problem 4: 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
Solutions The probability accidents occur is 30%  00 --> 29 The probability accidents don’t occur is 70%  30 --> 99
Solutions 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
Solutins: ▪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
Solutins: 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
References ▪ “Simulation modeling and analysis, Fourth edition” Averill M. Law, McGraw-Hill.
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Thanks! ANY QUESTIONS? You can find me at: Ayman_shamel@hotmail.com Fb.com/ayman.shamel