Auto insurance company classifies it's customer in three

Question

An auto insurance company classifies its customers in three​ categories: poor,​ satisfactory, and preferred. Each​ year, 30​% of those in the poor category are moved to satisfactory and 5​% of those in the satisfactory category are moved to preferred.​ Also, 5​% in the preferred category are moved to the satisfactory​ category, and 5​% of those in the satisfactory category are moved to the poor category. Customers are never moved from poor to​ preferred, or​ conversely, in a single year. Assuming these percentages remain valid over a long period of​ time, how many customers can the company expect to have in each category in the long​ run? Poor=enter your response here​%

Answer 1

Let in t years

a(t) is a proportion of customers who have poor category;

b(t) is a proportion of customers who have satisfactory category;

c(t) is a proportion of customers who have preferred category.

a(t) + b(t) + c(t) = 1

What is happened in one year?

0.3a(t) move from poor to satisfactory category;

0.05b(t) move from satisfactory to preferred category;

0.05c(t) move from preferred to satisfactory category;

0.05b(t) move from satisfactory to poor category.

So,

a(t+1) = a(t) - 0.3a(t) + 0.05b(t) = 0.7a(t) + 0.05b(t)

b(t+1) = b(t) - 0.05b(t) - 0.05b(t) + 0.3a(t) + 0.05c(t) = 0.3a(t) + 0.9b(t) + 0.05c(t)

c(t+1) = c(t) - 0.05c(t) + 0.05b(t) = 0.05b(t) + 0.95c(t)

a(t+1) + b(t+1) + c(t+1) = 1

This is a Markov chain (https://en.wikipedia.org/wiki/Markov_chain) that approaches a stationary distribution: in the long run a(t) → a, b(t) → b, c(t) → c.

Values a, b, and c satisfy the equations:

(1) a = 0.7a + 0.05b

(2) b = 0.3a + 0.9b + 0.05c

(3) c = 0.05b + 0.95c

(4) a + b + c = 1

From (3) we get c = b, from (1) we get b = 6a. Substitute these expressions into (4):

a + 6a + 6a = 1

a = 1/13, b = 6/13, c = 6/13

Convert these fractions to a percent form and round to the whole percentage: in a long run the company expects to have about 8% of customers in poor category and about 46% of customers in both satisfactory and preferred categories.

Answer 2

Hi Madhav B,.,

You could set this up as a customer matrix * a ratechange matrix problem, but that's overkill!

Just reason it through. #Poor x rate(0.30) -> .delta.#Satisfactory, and #Satisfactory x rate(0.05) -> .delta.#Poor. since these must balance for stability, then #Poor = (1/6)#Satisfactory. Similarly, #Satisfactory = #Preferred. So ratios #(Poor:Satisfactory:Preferred) = 1:6:6 . I think you can set that ratio into the 100%CustormerBase yourself?

If you want a more general problem along these lines, that you cannot reason so simply (because 3 populations are all feeding into each other), email me back at: [email protected] and I'll send you one. Of course, you COULD always solve these sorts of problems by trial-and-error, aka successive approximations -- but it might profit you to be able to set the solution up elegantly? A little insurance for the future, you might say?

-- Cheers, --Mr. d.