Successive matrix multiplication

Click For Summary
The discussion focuses on modeling the spread of a flu epidemic using matrix multiplication to predict the percentage of ill individuals over time. The initial conditions include a percentage Y of the population being ill and probabilities a and B for remaining healthy or sick, respectively. The equations for healthy and sick populations are derived, leading to a matrix-vector representation. Participants seek clarification on forming the appropriate coefficients for the transition matrix A, specifically A(21) and A(22), while expressing uncertainty about the equations for X[n+1]. The conversation emphasizes the need for a MATLAB function to compute the percentage of ill individuals after n days based on these parameters.
sara_87
Messages
748
Reaction score
0
On a given day of a flu epidemic, a given percentage Y of the population is ill and (1-Y) is healthy. The probability of remaining healthy on the next day is a, and that of remaining sick is B. The question is, what percentage will be ill after a given number of days as a function of B,a and Y.

Write down the problem as a matrix multiplication of the vector of healthy and ill people. Define a MATLAB function that gives the percentage of ill people after n days for arbitrary B a,and Y.

any hints will be appreciated v much
thank you
 
Computer science news on Phys.org
sara_87 said:
On a given day of a flu epidemic, a given percentage Y of the population is ill and (1-Y) is healthy. The probability of remaining healthy on the next day is a, and that of remaining sick is B. The question is, what percentage will be ill after a given number of days as a function of B,a and Y.

Write down the problem as a matrix multiplication of the vector of healthy and ill people. Define a MATLAB function that gives the percentage of ill people after n days for arbitrary B a,and Y.

any hints will be appreciated v much
thank you

# healthy people = # people staying healthy + # new healthy people

The odds of staying healthy are a, so
# people staying healthy = a * # healthy people

New healthy people are sick people who did not stay sick. The odds of staying sick are B, so the odds of NOT remaining sick, i.e. new healthy people is (1-B), so
# new healthy people = (1-B) * # sick people

So if Y is the number of healthy people and X is the number of sick people at a given time,
Y[n+1] = a*Y[n] + (1-B)*X[n]

Write down a similar equation for X[n+1]. Do you see how this can be written as a matrix-vector product?
 
would the equation for X[n+1] be:
X[n+1]=B*X[n]+(1-B)*X[n]

i understood everything you did but I'm struggling to write it as a matrix vector product.
would it be something like:
(X,Y)=(somthing with a)+(something with b)
?
thank you
 
You want to write

<br /> \begin{bmatrix} A_{11} &amp; A_{12} \\ A_{21} &amp; A_{22} \end{bmatrix}<br /> \begin{bmatrix} X_n \\ Y_n \end{bmatrix} =<br /> \begin{bmatrix} X_{n+1} \\ Y_{n+1} \end{bmatrix} <br />

Multiplying this out gives

<br /> \begin{array}{c} X_{n+1} \\ Y_{n+1} \end{array} =<br /> \begin{array}{cc} A_{11}X[n] + A_{12}Y[n] \\ A_{21}X[n] + A_{22}Y[n] \end{array}<br />

so find the coefficients for the A matrix that make this match your equations.
 
oh ok so A(21) would be=(1-B)
and A(22) would be=a
is that right?
and how do i find A(11) and A(12) because I'm not so convinced about my equation for X[n+1]??
 
I’ve spent nearly my entire life online, and witnessed AI become integrated into our lives. It’s clear that AI is apart of us now whether we like it or not, unless your a anti tech cabin lover. AI has some form of control over your life. But what I’ve seen very recently is that people are loosing their ingenuity and deciding to use AI. I feel as if it’ll bleed into STEM which is kinda has already and, every idea or thought could become fully reliant on AI. Yeah AI makes life easier but at a...

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K