Mahmoud Abas, Full-Stack Software Engineer & AI

Modeling and Simulation - Lesson 5

Monte Carlo Simulation

Random numbers read as demand, treatments and service times: Exercise 5 solved step by step.

Objectives

Turn a frequency table into random number intervals
Compute the mean and standard deviation of demand and lead time, then EOQ and ROL
Build the inventory simulation table and decide when to order
Simulate a scheduled queue and find the average waiting time
Generate exponential variates with the inverse transform
Use a mixed congruential generator to simulate a factory

Part 1

Inventory control

Example 1: demand, lead time and when to order

Example 1

The problem

We know the demand for the past 100 days
We know the lead time in which the ordered quantity arrived
We need the demand in the coming days, the possible lead time and when to order

Example 1

The data

Lead time	Frequency
3 days	28%
4 days	44%
5 days	28%

Example 1

The costs

Symbol	Cost of	Value
`C0`	Placing an order	100 pound/order
`Ci`	Holding an item in stock	0.1 pound/unit/day
`Cs`	Shortage	5 pounds/unit

Example 1

Mean demand

μ_D = ∑ D_in

μ_D = 4523100

μ_D = 45.23

Example 1

Standard deviation of demand

σ_D = √∑ D_i² - n (μ_D)²n - 1

214077 - 100 (45.23)²

= 9501.71

σ_D = √9501.7199

σ_D = 9.8

Example 1

Mean lead time

28 × 3 + 44 × 4 + 28 × 5

= 84 + 176 + 140 = 400

μ_L = 400100 = 4

Example 1

Standard deviation of lead time

28 (3)² + 44 (4)² + 28 (5)²

- 100 (4)² = 56

σ_L = √5699

σ_L = 0.75

Mean lead time

A slip on the exercise sheet

Example 1

Histogram of the demand

Group the demand into intervals: 20 - 24, 25 - 29, ..., 70 - 74
Count the frequency and the cumulative frequency of each interval
Assign two-digit random numbers 00 to 99 according to the frequency
To predict a demand, take the average value of its interval

Example 1

Demand frequencies: part 1 of 3

Interval	Average	Freq.	Cum.
20 - 24	22	1	1
25 - 29	27	3	4
30 - 34	32	12	16
35 - 39	37	10	26

Example 1

Demand frequencies: part 2 of 3

Interval	Average	Freq.	Cum.
40 - 44	42	26	52
45 - 49	47	17	69
50 - 54	52	13	82
55 - 59	57	12	94

Example 1

Demand frequencies: part 3 of 3

Interval	Average	Freq.	Cum.
60 - 64	62	2	96
65 - 69	67	2	98
70 - 74	72	2	100

Example 1

Demand random numbers: part 1 of 3

Average	Cum.	Random numbers
22	1	00
27	4	01 - 03
32	16	04 - 15
37	26	16 - 25

Example 1

Demand random numbers: part 2 of 3

Average	Cum.	Random numbers
42	52	26 - 51
47	69	52 - 68
52	82	69 - 81
57	94	82 - 93

Example 1

Demand random numbers: part 3 of 3

Average	Cum.	Random numbers
62	96	94 - 95
67	98	96 - 97
72	100	98 - 99

Range of the demand

A slip on the exercise sheet

Example 1

Lead time random numbers

Lead time	Freq.	Cum.	Random numbers
3	28	28	00 - 27
4	44	72	28 - 71
5	28	100	72 - 99

Example 1

Economic order quantity (EOQ)

EOQ = √2 μ_D C₀C_i

EOQ = √2 × 45.23 × 1000.1

EOQ ≈ 300

Example 1

The reorder level (ROL)

ROL = μ_D μ_L

+ K √(μ_L σ_D² + μ_D² σ_L²)

K	Chance of shortage	Sure of no shortage
1.65	5%	95%
2.33	1%	99%

Example 1

ROL: the parts

For K = 1.65

μ_D μ_L = 45.23 × 4 = 180.92

4 (9.8)² + (45.23)² (0.75)²

= 1534.90

√1534.90 = 39.18

Example 1

ROL: the result

For K = 1.65

ROL = 180.92 + 1.65 × 39.18

ROL ≈ 245.6

Rounding of ROL

A slip on the exercise sheet

Example 1

Starting the simulation

We use EOQ = 300 and ROL = 245 units
The initial stock is 300 units on Friday evening
The simulation starts from Monday onwards
Random numbers 00 to 99 come from Excel or from another generator

Excel

A slip on the exercise sheet

Example 1

The random numbers

For	Random numbers
Demand	81, 78, 89, 59, 40, 33, 60, 67, 19, 21
Lead time	65, 67, 22

The named random numbers

A slip on the exercise sheet

Example 1

Rules of the table

Initial stock = previous final stock
Final stock = initial stock - daily demand + quantity received
Quantity in hand + order = final stock + the order on the way
If that quantity falls below ROL, order EOQ and draw a lead time

Example 1

Demand: week 1

Day	Initial	R. No.	Demand
Mon	300	81	52
Tue	248	78	52
Wed	196	89	57
Thu	139	59	47
Fri	92	40	42

Example 1

Demand: week 2

Day	Initial	R. No.	Demand
Mon	50	33	42
Tue	308	60	47
Wed	261	67	47
Thu	214	19	37
Fri	177	21	37

Example 1

Final stock: week 1

Day	Shortage	Received	Final
Mon	-	-	248
Tue	-	-	196
Wed	-	-	139
Thu	-	-	92
Fri	-	-	50

Example 1

Final stock: week 2

Day	Shortage	Received	Final
Mon	-	300	308
Tue	-	-	261
Wed	-	-	214
Thu	-	-	177
Fri	-	-	140

(50 - 42) + 300 = 308

Example 1

When do we order?

Example 1

Orders: week 1

Day	In hand + order	Ordered	Lead time (R. No.)
Mon	248	0	-
Tue	196	300	4 (65)
Wed	439	0	-
Thu	392	0	-
Fri	350	0	-

Example 1

Orders: week 2

Day	In hand + order	Ordered	Lead time (R. No.)
Mon	308	0	-
Tue	261	0	-
Wed	214	300	4 (67)
Thu	477	0	-
Fri	440	0	-

Quantity in hand + order

A slip on the exercise sheet

Example 1

When the order arrives

Lead time 4 working days from Tuesday: received on next Monday
Final stock on next Monday: 308 units
The second order, placed on Wednesday, is received on next Tuesday, after the 10 days of the table

(50 - 42) + 300 = 308

Example 1

Costs in a period

Part 2

A dentist clinic

Example 2: a queue with scheduled arrivals

Example 2

The problem

Find the average waiting time of the customers
The simulation starts at 8:00
Arrivals are scheduled: one customer every 30 minutes from 8:00
In other cases arrivals may be random, from a probability distribution
Each customer comes for one of five treatments

Example 2

The five services

Service	Time (min)	Probability
Filling	40	0.40
Crown	60	0.15
Cleaning	15	0.15
Extraction	45	0.10
Checkup	15	0.20

Example 2

Random number intervals

Service	Prob.	Cum.	Interval
Filling	0.40	0.40	00 - 39
Crown	0.15	0.55	40 - 54
Cleaning	0.15	0.70	55 - 69
Extraction	0.10	0.80	70 - 79
Checkup	0.20	1.00	80 - 99

Example 2

Which service: patients 1 to 4

Patient	Arrival	R. No.	Service (min)
1	8:00	40	Crown (60)
2	8:30	82	Checkup (15)
3	9:00	11	Filling (40)
4	9:30	34	Filling (40)

Example 2

Which service: patients 5 to 8

Patient	Arrival	R. No.	Service (min)
5	10:00	25	Filling (40)
6	10:30	66	Cleaning (15)
7	11:00	17	Filling (40)
8	11:30	79	Extraction (45)

Example 2

Rules of the table

Service starts at the later of the arrival and the previous service end
Service ends at start + service time
Waiting time = service start - arrival

Example 2

Waiting times: patients 1 to 4

Patient	Service (min)	Starts	Ends
1	60	8:00	9:00
2	15	9:00	9:15
3	40	9:15	9:55
4	40	9:55	10:35

Patient	Arrival	Starts	Wait (min)
1	8:00	8:00	0
2	8:30	9:00	30
3	9:00	9:15	15
4	9:30	9:55	25

Example 2

Waiting times: patients 5 to 8

Patient	Service (min)	Starts	Ends
5	40	10:35	11:15
6	15	11:15	11:30
7	40	11:30	12:10
8	45	12:10	12:55

Patient	Arrival	Starts	Wait (min)
5	10:00	10:35	35
6	10:30	11:15	45
7	11:00	11:30	30
8	11:30	12:10	40

Example 2

The average waiting time

0 + 30 + 15 + 25 + 35

+ 45 + 30 + 40 = 220

2208 = 27.5

Part 3

The inverse transform

Example 3: the exponential distribution

Example 3

The distribution function

F(x) = 1 - e^-x/β if x ≥ 0

F(x) = 0 otherwise

Example 3

Finding the inverse

u = 1 - e^-x/β

F^-1(u) = -β ln (1 - u)

Example 3

Generating the variate

Generate U ~ U(0, 1)
Let X = -β ln U
U replaces 1 - U, because both have the same U(0, 1) distribution

Part 4

An exponential queue

Example 4: random arrival and service times

Example 4

The problem

Inter-arrival time is exponential with mean β1 = 2
Service time is exponential with mean β2 = 1.33
Build the table for 8 customers and find the average waiting time

Example 4

The inverse transform equations

X_i = -β₁ ln U₁

Y_i = -β₂ ln U₂

Xi is the inter-arrival time and Yi is the service time
U1 and U2 lie between 0 and 1
Divide each two-digit number by 100: U1 = u1 / 100 and U2 = u2 / 100

Example 4

The first customer

Stream	Random numbers
u1	20, 23, 86, 09, 92, 35, 38, 01
u2	21, 44, 27, 70, 73, 36, 59, 42

X₁ = -2 ln (0.20) = 3.22

Y₁ = -1.33 ln (0.21) = 2.08

Example 4

Arrivals: customers 1 to 4

Customer	u1	Xi	Arrival
1	20	3.22	3.22
2	23	2.94	6.16
3	86	0.30	6.46
4	09	4.82	11.28

3.22 + 2.94 = 6.16

Example 4

Arrivals: customers 5 to 8

Customer	u1	Xi	Arrival
5	92	0.17	11.45
6	35	2.10	13.55
7	38	1.94	15.49
8	01	9.21	24.70

Example 4

Service times: customers 1 to 4

Customer	u2	U2	Yi
1	21	0.21	2.08
2	44	0.44	1.09
3	27	0.27	1.74
4	70	0.70	0.47

Example 4

Service times: customers 5 to 8

Customer	u2	U2	Yi
5	73	0.73	0.42
6	36	0.36	1.36
7	59	0.59	0.70
8	42	0.42	1.15

Service times

A slip on the exercise sheet

Example 4

Rules of the table

Service starts at the later of the arrival and the previous departure
Departure = service start + Yi
Waiting time = service start - arrival

7.25 - 6.46 = 0.79

Example 4

Departures: customers 1 to 4

Customer	Arrival	Departure	Wait
1	3.22	5.30	0
2	6.16	7.25	0
3	6.46	8.99	0.79
4	11.28	11.75	0

Example 4

Departures: customers 5 to 8

Customer	Arrival	Departure	Wait
5	11.45	12.17	0.30
6	13.55	14.91	0
7	15.49	16.19	0
8	24.70	25.85	0

Example 4

The average waiting time

0.79 + 0.308 = 1.098 ≈ 0.14

Part 5

Manufacturing

Example 5: units produced and vehicles available

Example 5

Units produced

Units Xi	P(x = Xi)	Interval u1
500	0.06	00 - 05
550	0.14	06 - 19
600	0.20	20 - 39
650	0.40	40 - 79
700	0.20	80 - 99

Example 5

Vehicles available

Vehicles Yi	P(y = Yi)	Interval u2
5	0.16	00 - 15
6	0.36	16 - 51
7	0.20	52 - 71
8	0.16	72 - 87
9	0.12	88 - 99

Example 5

The mixed congruential generator

X_i = (a X_i-1 + c) mod m

Stream	a	c	Seed X0
u1	21	3	20
u2	41	7	1

The formula

A slip on the exercise sheet

X_i = a X_i-1 + c mod (m)

X_i = (a X_i-1 + c) mod m

Example 5

Stream u1: X1 to X5

i	21 × X + 3	Xi
X1	21 × 20 + 3 = 423	23
X2	21 × 23 + 3 = 486	86
X3	21 × 86 + 3 = 1809	09
X4	21 × 9 + 3 = 192	92
X5	21 × 92 + 3 = 1935	35

Example 5

Stream u1: X6 to X9

i	21 × X + 3	Xi
X6	21 × 35 + 3 = 738	38
X7	21 × 38 + 3 = 801	01
X8	21 × 1 + 3 = 24	24
X9	21 × 24 + 3 = 507	07

Example 5

Stream u2: X1 to X5

i	41 × X + 7	Xi
X1	41 × 1 + 7 = 48	48
X2	41 × 48 + 7 = 1975	75
X3	41 × 75 + 7 = 3082	82
X4	41 × 82 + 7 = 3369	69
X5	41 × 69 + 7 = 2836	36

Example 5

Stream u2: X6 to X9

i	41 × X + 7	Xi
X6	41 × 36 + 7 = 1483	83
X7	41 × 83 + 7 = 3410	10
X8	41 × 10 + 7 = 417	17
X9	41 × 17 + 7 = 704	04

Example 5

Units: days 1 to 5

Day	R. No.	Units
1	20	600
2	23	600
3	86	700
4	09	550
5	92	700

Example 5

Units: days 6 to 10

Day	R. No.	Units
6	35	600
7	38	600
8	01	500
9	24	600
10	07	550

Example 5

Vehicles: days 1 to 5

Day	R. No.	Vehicles
1	01	5
2	48	6
3	75	8
4	82	8
5	69	7

Example 5

Vehicles: days 6 to 10

Day	R. No.	Vehicles
6	36	6
7	83	8
8	10	5
9	17	6
10	04	5

Example 5

The averages

600010 = 600 units/day

6410 = 6.4 vehicles/day

Part 6

Common mistakes

and practice

Common mistakes

A demand drawn from an interval is its average: 81 gives 52, not 50 or 54
Boundaries: 40 is a crown, because filling ends at 39
Test ROL against in hand + order, not the final stock alone
Count lead time in working days: 4 days from Tuesday is next Monday
Service starts at the later of the arrival and the previous end
Divide u by 100 before ln, and keep the minus sign in X = -β ln U

Practice

Example 1 tables: demand for 94, 51, 37, and lead time for 20, 71, 72
Holding and ordering cost of the 10-day inventory table
Dentist, 4 patients from 8:00: random numbers 39, 55, 80, 70
Exponential queue: u1 = 50, 90, 40 and u2 = 30, 60, 80
Stream u1 with seed X0 = 5: 4 days of units produced

Key Takeaways

An interval takes as many two-digit numbers as its frequency; it ends at the cumulative frequency minus 1
EOQ sets how much to order, and ROL sets when
An order counts from the day after it is placed, and it enters the stock on the day it arrives
Waiting time = service start - arrival
Exponential variate: X = -β ln U, from the inverse transform
Mixed congruential generator: the next X is (aX + c) mod m

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Mahmoud AbasMonte Carlo Simulation