Transcript How_Swine_Flu_Spreads

Swine Flu
How Diseases Spread
Imagine you told your secret the two of your best friends
and your friends passed on your secret to two of their best
friends and this happened every hour
How many people would know your secret after 24 hours?
1
2
3  2 1
1 2  3
3
7  2 1
1 2  4  7
4
15  2  1
1  2  4  8  15
2  1  16,777,215
24
That’s about a quarter of the population of the UK who
know your secret!!!!
Draw a graph of ‘Number of Hours’ (plot ‘n’ on the x-axis)
versus ‘Total Number of People who Know your Secret’ (plot
‘P’ on the y-axis)
Take care when scaling your axes – make sure you know
what numbers you need to plot
Fully label your graph (axes, titles, etc)
What does the graph look like – what shape is it?
P  2 1
n
Imagine you were slightly less discrete and told your secret
the three of your best friends who then passed on your
secret to three of their best friends and this happened
every hour
How many people would know your secret after 24 hours?
1
1 3  4
1  3  9  13
1  3  9  27  40
3 1
Total 
2
24
3 1
9
 14110 (3SF )
2
n
141 BILLION!!! That’s more than the population of the
world (approx 7 billion) – AND YOU ONLY TOLD 3
PEOPLE!!!!!
How would the graph of ‘Number of Hours’ (x-axis) versus
‘Total Number of People who Know your Secret’ (y-axis)
look different from the one you drew earlier?
Draw this graph
Take care when scaling your axes – make sure you know
what numbers you need to plot
n
Fully label your graph (axes, titles, etc)
3 1
P
2
What formula would you use to calculate the total number
of people knowing your secret if you tell 4 people who then
each go on to tell another 4 people etc?
4n  1
P
3
What about 5 people, 6 people etc?
5n  1
P
4
6n  1
P
5
You have now worked out that the total number of people
who get to know your secret (we have been calling this ‘P’)
after a number of hours (we have been calling this ‘n’) when
each person tells your secret to 2 people, can be calculated
using the formula:
P  2 1
n
What formula would you use to calculate the total number
of people who get to know your secret if each person tells 3
people instead of 2?
3n  1
P
2
4n  1
What about 4 people?
P
n
3
5
1
What about 5 people?
P
4
xn 1
P
What about ‘x’ people?
x 1
2 People
Hours New
(n)
People
3 People
Total
(P)
Hours New
(n)
People
4 People
Total
(P)
Hours New
(n)
People
1
1
1
1
1
2
2
3
2
2
3
4
7
3
3
4
8
15
4
4
5
16
31
5
5
6
32
63
6
6
P  2 1
n
P
P
Total
(P)
Clearly this model is not perfect for our gossip spreading
investigation but it does serve as a good demonstration of
how numbers can theoretically grow very big very quickly
The reality is that people get bored with the secret or they
are not interested or they cannot find someone who doesn’t
already know the secret so they either don’t pass on your
secret or they tell fewer than 3 people
The spread of your secret then starts to slow down
For the purposes of our investigation we will refer to the
measure of the ability of a rumour to spread as the
SPREAD FACTOR
Imagine that so far 64 people know your secret, by now it’s
no longer a secret, people are becoming bored and there
are fewer people to tell because many people already know
the secret
You are going to investigate what happens when some
people don’t spread your precious secret – Assume our 64
people only tell 48 new people about your secret
48
The spread factor will be
 0.75 and we will assume this
64
remains the same
Calculate how many people know your secret every hour
64
64  64  0.75  112
64  64  0.75  64  0.75 0.75  148
2
3
64  64  0.75  64  0.75  64  0.75  175
Look at how many MORE people get to know your secret
each hour…
112
64
48
148
36
175
27
Each hour there are fewer new people learning your secret
and we can therefore predict that eventually nobody new
will learn your secret
Draw a graph of ‘Hours’ (x-axis) versus ‘Number of People
who Know your Secret’ (y-axis)
Take care when scaling your axes – make sure you know
what numbers you need to plot
Fully label your graph (axes, titles, etc)
What does the graph look like – what shape is it?
Assuming the spread factor is less than 1, in other words
the spread of your secret is reducing, there is a formula
that we can use to calculate how many people in total will
learn your secret…
If we call the number of people who already know your
secret A and the spread factor is S, then
A
Total who hear your secret 
1 S
How many people in total would learn your secret if 64
people already know your secret when the spread factor
becomes steady at 0.75?
64
 256
Total who hear your secret 
1  0.75
Use the formula…
A
Total who hear your secret 
1 S
to calculate the number of people who will get to know your
secret if, out of the 200 people, one person in every ten
tells another person
1
Spread Factor 
10
A  200
200
Total who hear your secret 
 222 people
1  0.1
Clearly, when the spread factor is small, fewer new people
get to know your secret
Diseases Spread Like Secrets
If the spread factor is less than 1, the disease will die out
If the spread factor is greater than 1, the disease will grow
and spread
Typical Infectious
Period
Spread Factor
HIV
4 years
3
Smallpox
25 days
4
Flu
5 days
4
Measles
14 days
17
In other words, at the start of an outbreak, a person with
flu may be infectious for about 5 days during which time
they will infect about 4 people
Investigation:
Imagine you have £1 and you put it in a bank account that
pays 100% interest every year (wow!)
How much do you have at the end of one year?
Now imagine a different bank that pays 50% interest every
6 months - how much would your £1 become at the end of
one year?
If a bank pays 25% interest every 3 months how much
would your £1 become after one year?
If a bank pays 12.5% interest every 1½ months how much
would your £1 become after one year?
Continue with this process… what do you notice is happening
to your end of year total?
Number
of
Payments
per Year
Initial
Amount
(£)
%age
Interest
£1
100%
£1
50%
£1
25%
4
£1
12.5%
8
£1
6.25%
16
£1
3.125%
32
£1
1.5625%
64
£1
0.78125%
128
£1
0.390625%
256
£1
0.1953125%
512
Multiplier
1
1.5
2
Equation
End of
Year
Value (£)
Multiplier
Number
of
Payments
per Year
Equation
End of
Year
Value (£)
100%
2
1
1x21
£2
£1
50%
1.5
2
1x1.52
£2.25
£1
25%
1.25
4
1x1.254
£2.44
£1
12.5%
1.125
8
1x1.1258
£2.57
£1
6.25%
1.0625
16
1x1.062516
£2.64
£1
3.125%
1.03125
32
1x1.0312532
£2.68
£1
1.5625%
1.015625
64
1x1.01562564
£2.70
£1
0.78125%
1.0078125
128
1x1.0078125128
£2.71
£1
0.390625%
1.00390625
256
1x1.00390625256
£2.71
£1
0.1953125% 1.001953125
512
1x1.001953125512
£2.72
Initial
Amount
(£)
%age
Interest
£1
Investigation - Summary:
Although it initially appears that your £1 can be turned into
an ever increasing amount by reducing the interest but
paying more often it becomes clear that there is a limit to
how much your £1 can be worth after one year
This value is £2.72 (rounded to the nearest penny)
The actual value is 2.71828 (5 decimal places)
This number (Euler’s number) occurs whenever you have
natural growths in populations and can be calculated using
the formula…
 1
e  1  
 n
n
Euler’s Number (2.718):
If the number of people infected at the start of an
outbreak of a disease such as Swine Flu is A and the spread
factor is S we could create a very simple formula…
New Infections  A S
But newly infected people start infecting other people thus
creating their own set of ‘New Infections’
Euler’s number is instrumental in predicting the total
number of disease carriers during an outbreak of disease
such as Swine Flu
The following formula has been found to work as a good
model for predicting early stage infections…
Number of Carriers after 1 Infectious Period  Ae S 1
If 10 people have a variety of flu which is infectious for 5
days with a spread factor of 4, how many carriers of the
disease will there be after the first 5 days?
Number of Carriers after 1 Infectious Period  Ae
S 1
4 1
 10  2.718
 10 2.7183
 201 (rounded to nearest wh ole person)
How many carriers of the disease will there be after the
first 10 days (i.e.) after 2 infectious periods?
 201 2.71841
3
 201 2.718
 4036 (rounded to nearest wh ole person)
How many carriers of the disease will there be after the
first 15 days (i.e.) after 3 infectious periods?
Number of Carriers after 1 Infectious Period  Ae S 1
4 1
 4036  2.718
 4036 2.7183
 81040 (rounded to nearest wh ole person)
How many carriers of the disease will there be after the
first 20 days (i.e.) after 4 infectious periods?
 81040  2.71841
3
 81040 2.718
 1,627,226 (rounded to nearest wh ole person)
If the formula for calculating the number of carriers after
1 period is…
Number of Carriers after 1 Infectious Period  Ae S 1
What would be the formula for calculating the number of
carriers after T infection periods?
Number of Carriers after T Infectious Period  Ae

S 1 T
Use this formula to calculate the number of carriers after
the first 6 infectious periods (i.e. about 1 month)
Number of Carriers after T Infectious Period  Ae
 10  2.718 
 655,000,000 (3 SF)
41 6

S 1 T
Clearly this last figure is nonsense, this is because the
formula works as a good model during the early stages of
the spread of the disease but becomes inaccurate over
longer periods
What factors prevent the numbers getting as high as 655
million?
…Many people infected so fewer people left to become
infected
…Those remaining uninfected may already be immune to the
disease
…Medication and drugs may have had enough time to start
being effective
…People may be quarantining themselves as a form of
protection
Typical Infectious
Period
Spread Factor
HIV
4 years
3
Smallpox
25 days
4
Flu
5 days
4
Measles
14 days
17
Number of Carriers after T Infectious Period  Ae

S 1 T
Teacher’s Notes:
a(1  r )
Sum Geometric Series with n terms Sn 
(1  r )
n
a(r  1)
Sum Geometric Series with n terms Sn 
(r  1)
n
A( S n  1)
Sum Geometric Series with n terms Pn 
(S  1)