Imagine a relatively new device, of which there are two competing versions, A and B. (We assume that no-one owns both.) The potential market is 100 million, but at present, 20 million own A, while 12 million own B. Currently, however, B outsells A 2:1. Combined sales total 600 000 per month.<\/p>\n
Assuming only new sales (as opposed to replacement), and only one per customer, when will ownership of B equal A? How many of each will have sold by then? What will be the total market penetration?<\/p>\n
Solution:<\/p>\n
Let<\/p>\n
x=new sales of A (which means 2x=new sales of B)
\nt=time<\/p>\n
We want the owned units of A to equal the owned units of B<\/p>\n
20 000 000 + x = 12 000 000 +2x<\/p>\n
Subtracting x and also 12 000 000 from both sides gives<\/p>\n
8 000 000 = x<\/p>\n
Right now, 20 000 000 people own A. So when 8 000 000 more units of A have sold, 28 000 000 will own one. During that time, 16 million units of B will sell; 12 million people own B right now. 16 million + 12 million = 28 million, the same as A will be.<\/p>\n
At market equalization, 28 000 000 customers will own each. Therefore, 56 000 000 will own one or the other: total market penetration will be 56%. <\/p>\n
How long will market equalization take? Of the 600 000 units selling each month, 200 000 are A (while 400 000 are B). At 200 000 units per month, the time for 8 000 000 units to sell is 8 000 000 \u00f7 200 000 = 40 months.<\/p>\n
HTH:)<\/p>\n
Source:<\/p>\n
Tan, S.T. Applied Finite Mathematics<\/u>. Boston: PWS-KENT, 1990.<\/p>\n