1 Module 4 Some Common applications Table o co.docx

1
Module 4: Some Common applications
Table o content
1- Inventory System Simulation
2- The M-N Inventory System
3- Machine reliability study
4- Evaluation of integral
5 - Simulation of hitting a Target
Case I: Inventory system simulation.
Introduction.
The Inventory management is one of the crucial aspects for any
manufacturing firm and well known
topic in both corporate and academic world. Inventory
management involves a set of decisions that aim
at matching existing demand with the supply of products and
materials over space and time in order to
achieve profitable operations. An inventory is considered as
one of the major assets of a business and it
represents an investment that is tied up until the item is sold or

used in the production of an item. It costs
money to store, track and insure inventory. Inventories that are
not well managed can create significant
financial problems for a business, whether the problem results
in an inventory glut or an inventory
shortage. Proper management of inventories would help to
utilize capital more effectively.
Why Is Inventory Control Important?
If your business requires maintaining an inventory, you might
sometimes feel like you're walking a
tightrope. Not having enough inventory means you run the risk
of losing sales, while having too much
inventory is costly in more ways than one. That's why having an
efficient inventory control system is so
important.
Avoiding Stock-outs.
One of the worst things you can do in business is to turn away
customers -- people who are ready to give
you their money -- because you've run out of the item they
want. "Stock outs" not only cost you money
from missed sales, they can also make you lose customers for

good, as people resolve to take their
business somewhere that can satisfy their needs. An efficient
inventory control system tracks how much
product you have in stock and forecasts how long your supplies
will last based on sales activity. This
allows you to place orders far enough ahead of time to prevent
stock-outs.
Overstock Hazards
When inventory isn't managed well, you can also wind up with
overstock -- too much of certain items.
Overstock comes with its own set of problems. The longer an
item sits unsold in inventory, the greater
the chance it will never sell at all, meaning you'll have to write
it off, or at least discount it deeply.
Products go out of style or become obsolete. Perishable items
spoil. Items that linger in storage get
damaged or stolen. And excessive inventory has to be stored,
counted and handled, which can add
ongoing costs.
2

Working Capital Issues
Inventory is expensive to acquire. When you pay, say, $15 for
an item from a supplier, you do so with
the expectation that you will soon sell the item for a higher
price, allowing you to recoup the cost plus
some profit. As long as the item sits on the shelf, though, its
value is locked up in inventory. That's $15
you can't use elsewhere in your business. So inventory control
isn't just about managing the "stuff"
going in and out of your company; it's also about managing your
working capital, keeping you from
having too much precious cash tied up in operations.
Manufacturer's Angle
Inventory control isn't just a concern for companies that deal in
finished goods, such as retailers and
wholesalers. It's also critical for manufacturers, who maintain
three types of inventory: raw materials,
works in process and finished goods. If you run out of an
essential ingredient or component, production
will halt, which can be extremely costly. If you don't have a
supply of finished goods on hand to fill
orders at they come in, you risk losing customers. Staying on
top of inventory is essential if you're to

keep the line running and keep products moving out the door.
I-2: Analysis, Modeling and Simulation
Careful inventory management is critical to the financial health
of businesses whose primary
venture is manufacturing or retailing. In retail and
manufacturing companies, huge amounts
of time and $'s are expended in keeping and managing
inventory. A simple graphical
representation of inventory system is shown in Figure 1 below
Basic concepts involved in inventory management. we will
build an inventory model to
answer the following two questions:
• How much do we order? (see figure 1)
• When to order? - with the goal of minimizing the total
inventory costs.
X: Flow rate in (order)

Inventory fluctuation in the Tank
is a function of (X-Y)
Y:
Flow rate out (Demand)
I
X: flow rate (i.e., gallons /hour)
Y = Demand rate ( used by customers)
Figure 1. A
simple inventory system
3
In most basic inventory models we are going to make several
important assumptions in order
to keep the model simple.
Assumptions:
• Only one item is considered.
• An entire order arrives at once.
• Shortages may or may not be allowed.

• The demand is probabilistic and its probability distribution is
known
• The time value of money is zero.
• Price for items is not a function of order quantity (no Quantity
discount)
• Lead-time is known and constant.
There are three basic types of inventory - raw materials,
work-in-process, finished
goods. Our analysis may apply to any of the three with minor
variations. The principle
remains the same. In inventory analysis, we work with two
variables
a) Independent demand which in most cases is represented as a
random variable since
we do not have any control on number of customers and their
demand level.
Independent demand is most frequently associated with finished
goods where the
demand is more or less unknown.
b) Dependent demand refers to those items (i.e., inventory level,
demand for labor hours
and so on) which are determined as a function of the

independent demand.
The costs associated with inventory fall into two broad
categories or components. All
the costs associated with keeping stock in inventory is “lumped
together as carrying cost (we
may also call it the holding cost component). All the costs
associated with ordering and
delivering the stock is "lumped" together, and called the
ordering cost component. Some
costs which may be included in the holding cost component are:
• opportunity cost
• taxes
• insurance
• storage
• shrinkage
Some costs which may be included in the ordering cost
component are:
• ordering costs
• set-up costs
• transportation costs

• small lot costs
• stock-outs and backorders
4
To simulate a simple inventory system, we first build a
mathematical (analytical) model.
Following typical conventions, we need some descriptions and
shorthand notation for
variables used in the model (These notations are those used in
different textbook in modeling
a general inventory system. The analysis may use alternative
notations).
Order quantity (Q) - Number of units ordered, also called the lot
or batch size.
Demand (D) - Usually the annual demand. You may need to
convert available information
to annualized data.
Item Cost (C) - Purchase price of raw materials or value of

finished goods or WIP.
Carrying charge rate (i) - Composite % or decimal fraction of
the item's cost that reflects
the cost of keeping one unit in inventory for one year. Usually,
but not always, expressed in
% per unit per year.
Holding Cost per unit (H) - Cost, in dollars, to keep one unit in
inventory for one year, H =
i x C. Make sure you don't confuse this with the holding cost
component.
Ordering Cost (Co) - Cost to place one order, not to be confused
with the Ordering Cost
Component. Frequently, this is designated S, the set-up cost,
when dealing with WIP or
finished goods inventory.
Lead Time (LT) - The time that elapses between placing an
order and receipt of that order.

Re-Order Point (ROP) - The on-hand inventory level at which
we should place the order for
the next batch.
Stock-out Cost (S)- The cost incurred when there is insufficient
inventory.
1-3. M-N Inventory System simulation &&Assignment
An M-N inventory system, generally used by small companies,
is a system that has an inventory review
every N time periods (i.e., every 20 days), and when we
replenish, we always bring the inventory level
to A maximum of M units. At each N time units, the inventory
level is checked and an order placed to
bring the inventory up to level M. The operation of the system
(shown in Figure 2 below) is based on
the following assumptions
1)- The lead time (for ordering and receiving) is assumed zero
2)- There are two customers, Company A and B. Demand by
these customers (called X and Y) are
random. Note that since X and Y are random, then total

demand Z= X+Y is also random
3)- Shortages are backordered. This means, if we do not have it
now, we will supply the customer
next period. 4)- At the end of time period “0” (initial
inventory for period 1) = I0.
5
Time
Inventory level Figure 2. An M-N Inventory
System
M
Io
Let: BINi = Beginning Inventory, period i

EINi = Ending Inventory, period i
Xi & Yi = Demand for the product by internal customers (Y),
and external customers (X) in period
ORDi = Amount to order @ the end of period I (This is shown
as Qi in the graph)
EXCi = Excess inventory, end of period i
SHTi = Shortage end of period i
n = # of replications
N = Time between orders ( we order every N units of
time) M = Max Inventory Level allowed
J = Number of time shortage happened
What you need to do:
a). Develop the equations (Models) of the system
b)- Use spreadsheet and simulation the system 100 tomes.
c)- From the output, for a given “M”, determine what % of time
the end of period inventory will be
negative (need backordering). Develop a histogram of the
variable “ENIi”
d)- If we want s safety stock of 50 units, what the optimum
value of “M” should be?

HINT: How to use Spreadsheet To simulate this type of
systems?
Step 1; We start in period 0 (this is for setting initial condition
and getting ready to start period
( 1, 2, 3,…). Now let:
EIN0 = I0, then ORD0 = M-Io (this is because we are
assuming that at the end of each period we
will order enough material to bring the total inventory
to “M” for the start of next day). Enter
in top row of excel the following information for period
“0”.
BIN0 =0, Z0 = 0, EIN0=I0, ORD0 =M -I0,
EXC0 =0, and SHT0 =0
X1+Y1
X2+Y2
X3+Y3
X4+Y4
Q1

Q2 Q3
Q4 Q5
0 1
2
3 4
N N
N N
6
Row A B
Period Starting
1 Inventory

C D E F
Order Demand Ending Shortage
Quantity Inventory Units
Excess
Units
2
3
0
1
0
= E2 =M-E2
0
=Z
Io
=M-D3
0 0
=IF(E3>0, 0,E3) =IF(E3<0, 0,E3)

copy
Step 2; for periods 1, 2, 3,…n, find the value of different
elements as follows;
BINi = EIN i --1 +ORD i –1 [i.e., beginning inventory on
period 10 equals ending inventory of
period 9 plus what we receive
(ordered before) beginning of period 10]
Xi is generated randomly This is generated randomly from the
probability distribution of Z
EINi = BINi – Zi [i.e., ending inventory of say
period 10 is equal to it beginning – demand
In period 10]
ORDi = M – EINi [i.e., The amount we order (say at
the beginning of period 10) must be
enough to bring inventory
level to “M” units
If EINi > 0, then EXCi = EINi, and SHTi = 0 and vice-vera
n excel worksheet, the simulation will proceed as in Table
1below.
,

Table 1
Note that in simulating an inventory system, the objective may
be to determine response to many (other
than what is discussed here) managerial questions, including;
calculating average or an
appropriate value for M, or testing validity of assumptions (on
which this simulation is based),
and so on. In those cases, we have to modify some or all of the
above equations (models)
Step 3; Run the simulation n times (n.>50). And
Step 4: Calculate system metrics/performance measures,
including, A). Percent of time there was
shortage
B). Maximum or average shortage quantity
C). Probability distribution of shortage quantity D).
Probability distribution of order quantity
E) If we desire to have “K” units in inventory as safety stock,
what the optimum value of “M”

should be
F). Others …………………………………
7
Problem 4A-1 Basic N_N Inventory
Simulate the M-N inventory system, discussed above, with the
following input data:
• I0 =120 units (initial Inventory)
• M=480 units
• N= 1 week (5 working days)
• X= Uniform distribution with Min =270 units/week and
Max=350
Y(units/week) 160 175 195
Pr(Y) 0.40 0.32 0.28
What you need to do:
a). Develop the equations (Models) of the system

b)- Use spreadsheet and simulation the system 100 tomes.
c)- From the output, determine what % of time the end of period
inventory will
be negative (need backordering). Develop a histogram of
the variable “ENIi”
d)- If we want a safety stock of 50 units, what the optimum
value of “M” should be?
Problem 4A-2: The newspaper seller's problem
A classical inventory problem concerns the purchase and sale of
newspapers. The paper seller
buys the papers for 33 cents each and sells them for 50 cents
each. Newspapers not sold at the
end of the day are sold as scrap for 5 cents each. Newspapers
can be purchased in bundles of
10. Thus, the paper seller can buy 50, 60, and so on. There are
three types of news days,
“good,” “fair,” and “poor,” with probabilities of 0.35, 0.45, and
0.20, respectively. The
distribution of papers demanded on each of these days is given
in Table 1 below. The problem
is to determine the optimal number of papers the newspaper
seller should purchase. This

will be accomplished by simulating demands for 40 days and
recording profits from sales
each day.
Table 1. Distribution of Newspapers Demanded
Demand
Demand Probability Distribution
Good Fair Poor
40 0.03 0.10 0.44
50 0.05 0.18 0.22
60 0.15 0.40 0.16
70 0.20 0.20 0.12
80 0.35 0.08 0.06
90 0.15 0.04 0.00
100 0.07 0.00 0.00
Problem 4A-3: Simulation of a Special (M, N) inventory
system.
Consider a system where the inventory level follows the pattern
of the probabilistic order-level
inventory system shown in Figure 2. Suppose that the maximum

inventory level, M =11 units
ASSIGNMENT 4A
8
and the review period, N, is 5 days. The problem is to estimate,
by simulation, the average
ending units in inventory and the number of days when a
shortage condition occurs. The
distribution of the number of units demanded per day is shown
in Table 2. In this problem, lead
time is a random variable, as shown in Table 2. Assume that
orders are placed at the close of
business day and are received for inventory at the beginning of
business day as determined by
the lead time
Figure 2. Probabilistic
Order level Inventory System
Amount in inventory
Time

Table 2: Probability distribution of
demand and lead time
Demand/day Probability Lead Time
(Days) Probability
0
0.1 1 0.60
1 0.25 2
0.30
2 0.35 3
0.10
3 0.21
4 0.09
Part I: Estimate, by simulation, the average ending units in
inventory and the number of days
when a shortage condition occurs.
Part II: a)- Extend the case for 15 more cycles and draw
conclusions.
b)- Rework the example for 10 cycles with M = 10,
and N=6
====================================
2. Simulating machine reliability

II-1)- What is Reliability?
Reliability refers to the % of time a machine is up and working
(A machine may be down for a
number of reasons). For a brief discussion of reliability and
related subjects, see he Appendix 1
at the end of this section.is presented below.
Q3
Time
Q2 Q1
9
EXAMPLE: Machine A is (on average) down 40 minutes per
day (day of work = 8 hours).
The reliability factor = (8*60 -40)/(8*60) = 91.67
%
This says that, the machine is only 91.67%
reliable and about
8.33% of time will be Idle
II-2) how to use simulation to determine reliability of a

machine?
EXAMPLE: A machine has two different bearing that fail in
service. The distribution of
life (time between breakdowns) are as follows
life in minutes (ti)➔ 40 100
120
Probability ➔ 0.3 0.45
0.25
When a bearing fails, maintenance department is called to
install a new bearing. Maintenance
jobs are started immediately after a breakdown. The time it
takes to fix the problem is as
follows
Time to install a bearing (RTi)➔ 5 Min. 10Min.
15Min.
Probability ➔ 0.35 0.45
0.20
Note: When a Bearing fails, the company incurs two types of
costs:
1) The machine becomes idle. This cost the company $20 an
hour
2) We pay for a new bearing. The cost for bearing = $45

What is required.
1) Determine the reliability of the machine
2) Number of stoppage per day
3) Average cost of maintenance per day
4) What % of time the maintenance operator is busy working for
this machine.
SOLUTION
First let us show the operation of the system on a time line
Let:
ti = time between arrival of successive failures on bearing
#1, and
tĩ = time between arrival of successive failures on bearing
#2
10
Figure 1 T=End of simulation
Repair Time

t1 t2 t3

Clock
0 Fail(A) Fail(B) Fail (A)
Fail (A) Time
t1̃ t2̃
t3̃
B Fails
B Fails Clock Time
Fail B
Legend:
: This Symbol is used for Repair Time
: The (length of) inter-arrival s (Brown color for Bearing
type A, and Green for B)
: Continuous Clock Time
Now we can generate ti and tĩ and on each case convert the
numbers to clock hours. As shown
in Figure 1, above. This may be done as follows
Repair time

Number of Generate ti Clock# 1 generate tĩ
Clock # 2 Bearing Bearing
Trials (Min) (Min) (min)
(min) #1 (min) #2 (min)
------------- ---------------- ----------- ------------- ---
---------------- -------------------------
1 40 40 100
100 10 1 5
2 120 160 120
220 5 5
3 100 260 100
320 15 10
4 … …. ….
… … ….
Totals
30 30
Conclusions:
a- T= Clock time = 320 +10 =330 min (See Figure 1). # of
breakdowns = 6
b- Downtime = 30+30 = 60 minutes
c- Cost= $20*60min/60 = $20. Machine cost = (6) *$45 = $270
(for T min).
Total cost = $270+$20 = $290

d- Operator busy time = (330 – 60) =270 ➔ Availability =
270/330 = 81,8%
A Spreadsheet simulation format
A B C D E F
Number of
breakdown
ti (Time between
Breakdowns
Repair time
(RT)
Clock time
After repair
Clock time
Before repair
Cumulative
Down time
11

Run the simulation for 15 hours per day and answer the
questions a, b, c, and d
(above) with the following additional assumptions:
1)- When there is a breakdown, in addition to repair time we
spend X minutes for
testing and resetting up the system (X has a normal distribution
with mean 10 and
standard deviation of 5 minutes)
2)- After every 4 hours of operations, we need to change a filter
on the machine
which takes 20 minutes each time
===============================================
===========
APPENDIX 1
A brief Review of Reliability and Availability
Maintainability:
It is the effort and cost of performing maintenance. It is

affected by factors such as, the ease of
access to equipment for maintenance, availability of spare parts
and the skill level
required doing the maintenance. One measure of maintainability
is Mean Time To Repair
(MTTR). A high MTTR is an indication of low maintainability
MTTR= (Downtime for repair)/( Number of repairs)
Here,
Downtime for repair includes
0 0 0 480 0 0
1
2
3
………
n
Totals ➔
Assignment 4B
12

a) Time waiting for repair TW
b) Time spent doing the repairs TR
c) Time spent for testing and getting the equipment ready to
resume production (of good
parts)
Note that in some organizations repair time is defined as the
downtime for repair only.
Reliability:
It is the probability that the equipment will perform properly
under normal operating
conditions for a given period of time. In some cases reliability
is not defined over time but
over another measurement such as miles traveled etc.
One measure of reliability (R) is the probability of successive
performance or
R = ( Number of successes) / (Number of repetitions)
Example 1: A machine produces 500 parts of which 480 are
good, and then the machine
is 480/500 = 96% reliable
Example 2: A machine used to test circuit boards for defects
work 99% of the times
(it misses 1% of all defective boards), then the

machine is 99% reliable.
Availability:
Availability is the proportion of time equipment is actually
available to perform work out of the
time it should be available one measure of availability
(A) =MTBF / (MTBF-MTTR)
It is obvious that availability is increased through a
combination of increasing MTBF,
decreasing MTTR or both. This relationship, while taking into
account the repair-related
sources of downtime(MTTR), it ignores non-repair sources of
downtime as a result, it
overstates the actual equipment availability. A better measure of
availability is
A=( Actual running time)/ ( Planned running time)
Where, Planned Running Time= Total Plant Time- Planned
Down Time Actual
Running Time= Planned Running Time- All
Downtimes
Example: Suppose a plant 2 shifts (16 hrs) per weekday. During
each shift, the plant has

two hours of planned downtime
Planned running time= 16-2(2)= 12 hrs
Suppose the machine is stopped each day, an average of 110
minutes for setups and 75
minutes for breakdowns and repairs, then: Actual
running time= 12(60)-(110+75=535 mins
13
The availability of equipment is thus A= 535/12(60)= 0.743 or
74.3%
Quality:
Quality is defined as the degree by which product/services
satisfy the user’s requirements.
Quality of a product/service and reliability are dependent on
each other. A high quality product
will have a high reliability and vice versa.
Failure:
Failure here simply means that equipment/component’s
performance is not satisfactory. It can
also mean that equipment /component malfunctioning in some
aspects or is completely broken.
A failure may be treated as random or deterministic by studying
the physics of the failure

process.
The more reliable equipment/component is, the less likely it
will fail. The likelihood of
failure is shown with failure pattern (also called bath-tub-curve)
which will be discussed later.
Mean Time Before Failure:
A measure reliability and equipment failure is the mean time
between failure (MTBF).
For equipment that can be repaired, MTBF represents the
average time between failures. For
equipments that cannot be repaired, it is average time to the
first failure. If we assume a constant
failure rate, then
MTBF=( Total running time)/ (Number of failure)
MTBF is usually determined on the basis of historical data on
product/equipment downtime as
shown below
| tw1 | tD1 | tw2 | tD2 | tw3 | tD3 | | tDn | Time line
Note: tW = System working, tD = System idle
MTBF = { ∑ Twi }/ n
Example:
Twenty machines are operated for 100 hours. One machine fails

after 60 hours and another after
70 hours. What is MTBF?
MTBF={ (20)(100)-[(100-70) +(100-60)]} /2 = 1930/2= =965
hours/failure
Alternatively MTBF= (18)(100)+70+60/2= 1930/2 = 965
hours/failure
Failure Distribution
Reliability function
It is generally assumed that the probability distribution of tW is
Exponential. And item will
not fail before t. Then
R(t)= e-λt (Here R(t) is the reliability function)
14
Where, 0 < = R(t) < =1
e = natural logarithmic base
t = specified time
λ=failure rate= 1/MTBF

Example:
If MTBF for a machine is 965 hours/failure, what is the
reliability of the machine at 500
hours, 900 hours?
λ=1/965= 0.0010362
R(500)= e-(500)(0.0010362) = 0.596
R(900)=e-(900)(0.0010362) = 0.39
===============================================
============
Case 3: Evaluation of Integral, Method 1
Consider the following polynomial. It is desired to evaluate
the area under the curve from
x=0 to x =3. Use the method we discussed in class on
Thursday. To assist you in this process,
below, I have provided a briefly explanation of the method and
how you should apply it to this
(or any) function.
Objective:
Determining area under the above curve (integral) between
Xmin and Xmax. This is the area

under the curve between point C and Point D. Suppose the
equation of the curve [Y = f(X) ] is
very complex and could not be integrated, using simple
integration rules. Therefore, we have
decided to use digital simulation to answer the problem.
Y
Ymax.: A
B
YE
E
Figure 1.
YR
Ymin C
X E D X
Xmin
Xmax
Procedure: In Fig 1, p = a ratio defined as:
a)-p = (Area under the curve) (Total area of
rectangle ABDC). (1)

15
b)-The area of the rectangle = (Xmax - Xmin)*( Ymax
– Ymin).
c)- If p is known, then from equation (1),,
Area under the curve= (p)*(Area of
rectangle ) =(p)*( CD x CA)
= (p)*[ (Xmax – Xmin )(Ymax - Ymin) ]
How to estimate p ?
Step 1: Generate a random number between Xmax and Xmin.
Call it XE
Step 2: Substitute XE in Y = f(X) and find the value of Y.
This is called YE
Step 3: Generate a random number between Ymax and Ymin.
Call it YR (See Figure 2)
Step 4: Compare YR .with YE. . If YR ≤ YE, then, the
point (XE, YR) is inside the curve,
(or under the curve) otherwise it is outside the curve .
See Figure 2.
Example: In Fig. 2, point (XE, YR) is point F which
is under the curve.
Step 5: Do steps 1 to 4 above, m times and count n = Number
of times that, the

point was under the curve. After repeating m times,
you can find p = n/m
Y
Ymax A
B
E
YE
Figure 2:
YR F
Ymin C
X E D X
Xmin
Xmax
How to determine the area of the rectangle ?
Area of the rectangle ABDC = (Xmax - Xmin)*( Ymax – Ymin).
Note that in this equation;
• Xmax and Xmin are given quantities..
• Ymin = 0
• Ymax must be determined as follows:

To determine the value of Ymax, we can use one of the
following methods:
A). Find the derivative of the function Y = f(X) and set the
derivative equal to zero. Assuming
it is possible to solve the equation, the solution will
enable us to find all max and min
points of the function.
16
B). Use enumeration method. Determine the value of Y for
different/all possible
values of X (see below) . Pick the largest value of Y and
call it Ymax . For
instance we may change the values of X as follows
Values assigned to X : Xmin, 0.10 +Xmin , 0.20 +Xmin,
0.30 +Xmin,,……. Xmax
Note that since we are assuming it is not possible to use
mathematical methods (i.e., using derivative of
the function) to determine the coordinates of the point where
the function takes its maximum value, we
have to use an alternative method such as the one discussed in

the previous paragraph. It should be noted
that, when we use this method, the Ymax obtained is an
approximation and may not necessarily be equal
to the real maximum. However, it is possible to get as close to
the real Maximum as possible if we
minimize the incremental increase from one (assigned) X value
to the next one in the process.
Example
Suppose the function we want to use simulation and determine
the integral of that function is given as:
Y = 2X – X ² where 0 ≤ X ≤2. This implies that Xmax =2 and
Xmin = 0 We can find Ymax by
assigning a series of possible values to X and calculating the
value of Y for each X as follows:
Assign X = 0 0.1 0.2 0.3,………… 1, 1.1
1.2……………2
Calculate Y = 0 0.19 0.36 0.51 ………...1,
0.99 0.96…………..0
Ymax
Y
Therefore:
Xmax =2,

Xmin = 0 ,
Ymax =1, 1.0 Ymax
Ymin =0.
Figure 3
X
0 1 2
(Xmax )
Therefore for this example, Area of rectangle = (2-0)*( 1-0) =
2
Use either method 1 or method 2 (discussed below) to estimate
area under each curve.
ASSIGNMENTS 4C

17
A)-Use Monte Carlo simulation to approximate the integral
B)- Use Monte Carlo simulation to approximate the area under
the curve 4/(1+x2 )
Between x = 0.5 and x =2
III-2:Evaluation on Integral, Method 2(1)
An Overview
In this part we introduce an alternative method for estimating
the area of a shape using the Monte
Carlo technique. The principle of a basic Monte Carlo
estimation is this: imagine that we want to
integrate a one-dimensional function f(x) from aa to bb such as:
b
F= ∫ f(x)dx.
a
As you may remember, the integral of a function f(x) can be
interpreted as calculating the area
below the function's curve. This idea is illustrated in Figure 1.
Now imagine that we just pick up
a random value, say x in the range [a,b], evaluate the function
f(x) at x and multiply the result by
(b-a). Figure 2 shows what the result looks like: it's another
rectangle (where f(x) is the height of

that rectangle and (b-a) its width), which in a way you can also
look at a very crude
approximation of the area under the curve. If we evaluate the
function at x1 (figure 3) we quite
drastically underestimate this area. If we evaluate the function
at x2, we over-estimate the area.
But as we keep evaluating the function at different
Figure 1: the integral over the domain [a,b] can be seen as the
area under the curve.
18
Figure 2: the curve can be evaluated at x and the result can be
multiplied by (b - a).
If we pick random points between a and b, adding up the area of
the rectangles and averaging
the sum, the resulting number gets closer and closer to the
actual result of the integral. It's not
surprising in a way as the rectangles which are too large
compensate for the rectangles which are
too small. And in fact, we can prove that summing them up and
averaging their areas actually

converges to the integral "area" under the curve, as the number
of samples used in the calculation
increases. This idea is illustrated in the following figure. The
function was evaluated in four
different locations (In the simulation process, xi values must be
generated randomly). The result
of the function as these four values of x randomly chosen, are
then multiplied by (b-a), summed
up and averaged (we divide the sum by 4). The result can be
considered as an approximation of
the actual integral.
19
Of course, as usual with Monte Carlo methods, this
approximation converges to the integral
result as the number of rectangles or samples used increases.
We can formalize this idea with the following formula:
N-1
FN = (1/N) (b−a) ∑ f(Xi) (1)

0
N in equation (1), is the number of samples used (or number of
trials in simulation) in the study
to estimate the area under the curve. FN is an approximation of
area using N samples. As N
increases, the error of estimate approaches zero. This equation
is called a basic Monte Carlo
estimator
Case 4. Simulation of hitting a target
In this case study we simulate a bombing mission where the
actual location a bomb hits varies
from its target by a random amount defined by Normal
distribution. The normal distribution is
used to estimate the random location the bomber hits each time.
Recall that the Normal
distribution is symmetric around its mean with likely range
determined by its standard deviation.
This is a classic use of normal distribution where the deviation
from the mean (target) represent
error/ missing the target.
---------------------------------------------------------------------------
---

(1). https://www.scratchapixel.com/lessons/mathematics-
physics-for-computer-graphics/monte-
carlo-methods-in-practice/monte-carlo-integration
A Military Bomber must hit a target shown as a blue box in
Figure 1. We want to use Digital
Simulation to estimate % of time the bomber will hit the target.
Assumptions:
1)- Bomber flies on Y access (as shown below)
2)- Real target is point “O” with X=Y=0
a). Deviation in X direction follows a Normal distribution with:
Mean = 0. & Standard Deviation = 150 meter
b). Similarly, deviations in Y direction follows a Normal
Distribution with:
Mean = 0. & Standard deviation = 200 meter
Case study : Assignment: 4D-1
https://www.scratchapixel.com/lessons/mathematics-physics-
for-computer-graphics/monte-

20
Figure 1
Y Target inside this Square
X
a). Deviation in X direction follows a Normal distribution with:
Mean = 0. & Standard Deviation = 150 meter
b). Similarly, deviations in Y direction follows a Normal
Distribution with:
Mean = 0. & Standard deviation = 200 meter
General Procedure to simulate the system.
1)- generate X and Y randomly (2)
2)- find out whether the point (X,Y) is inside the box or not,
3)- If it is inside the bob, add to count (numbers inside)

4)- Repeat steps 1, 2,, and 3, above “N” times. Suppose only M
time your point was inside the box.
➔ Probability of hitting the target = M/N
Case Study 2.
Case study 2 . Assignment 4D-2
21
In general, How to determine a randomly generated point is
inside the target area or not.
Consider the following Example.
EXAMPLE: Suppose You have generated X and Y Randomly
(Point A ). And were planning to
hit the Target, which is inside the triangle OBC
Y
B
Y1

A
Yc
X
0 X1
C
Point is INSIDE or OUTSIDE the Target Area?
To determine answer to this question, we assume that the
equation of Border lines ( like Line
BC) Are given. For example, in this case (above). Equation
are:
• Line OC: Y=0
• Line OB: X=0
• Line BC: Y= 7 -1.75X ➔ Coordinated of B and C are:
B:(0, 7), C: (4, 0)
Now you can use the following procedure to determine whether
a point hit by the Bomber
(generated randomly) is inside or outside the OBC target area:

a)- Check to see if X1 (randomly generated) > XMax, if yes,
point A is outside the triangular
area If it is not go to step b
b)- Substitute X1 in the equation of the line BC. This will
give generate Yc ( Calculated
the value of “Y”)
c)- If y1 >Yc, then the point is outside the triangular area,
if not it is inside
Assignment 4-4
Problem #1: Simulate the system (Case study 1, Above) 50
times and estimate % of time the
Bomber will hit the target. Based on your simulation, which
direction (Y or X) is the least
reliable/causes more Miss than the other
ASSIGNMENT 4-4:
22
Problem # 2: Solve problem #12 on page 81 of the textbook.
Note that you have to run the

model40 times (not 5 time as the problem statement indicates).

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