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 March 8th, 2018, 01:13 AM #1 Newbie   Joined: Mar 2018 From: Malaysia Posts: 5 Thanks: 0 Systems of Linear Equations Hi all, need help in this. I am so confused. Suppose that in a fixed population of food consumers, there are just three brands X Y Z that share the market. Suppose that each year, 10% of brand X buyers shift to brand Y and 20% to brand Z and 20% of brand Y buyers shift to brand X and 20% to brand Z 10% of brand Z buyers shift to brand X and none to brand Y Suppose that it is found that despite all this shifting, the market shares of these brands are nevertheless unchanged, what are these market shares? TIA Last edited by skipjack; March 8th, 2018 at 02:09 AM.
 March 8th, 2018, 02:53 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,004 Thanks: 1862 You need to assume that each consumer uses just one brand at a time. During each year, for every 5 consumers of brand Y, 1 moves to Brand X and 1 moves to brand Z, so there must have been 20 consumers of brand X originally, 2 of which shift to brand Y. I'll leave you to show by similar means that the corresponding number of consumers of brand Z was 50. Hence brand Z has 2/3 of the market, brand X has 4/15 of the market, and brand Y has the remaining 1/15 of the market. Thanks from speedzone
 April 5th, 2018, 12:19 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Let $x_n$ be the number of buyers of brand X in the nth year, $y_n$ be the number of buyers of brand Y, and $z_n$ be the number of buyers of brand Z. "10% of brand X buyers shift to brand Y and 20% to brand Z" "20% of brand Y buyers shift to brand X and 20% to brand Z" "10% of brand Z buyers shift to brand X and none to brand Y" So $X_{n+1}= .7X_n+ .2Y_n+ .1Z_n$ $Y_{n+1}= .1X_n+ .6Y_n$ $Z_{n+1}= .2X_n+ .2Y_n+ .9Z_n$ The "market share change" for each is $X_{n+1}- X_n= .3X_n- .2Y_n- .1Z_n$ $Y_{n+1}- Y_n= -.1X_n+ .4Y_n$ $Z_{n+1}- Z_n= -.2X_n- 2Y_n+ .1Z_n$ Saying that "the market shares are unchanged" means that those changes are all 0. Solve $.3X_n- .2Y_n- .1Z_n= 0$ $-.1X_n+ .4Y_n= 0$ and $-.2X_n- .2Y_n+ .1Z_n= 0$. An obvious solution is the "trivial solution" $X_n= Y_n= Z_n= 0$. The question is whether there at e other, non-trivial, solutions and, if so, what are they?

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