Solution to Problem 179

Solution to Problem 180

No correct solutions were received for this problem.

The sketch above summarizes the situation. The leaves begins at B. Let A_k, B_k,..., H_k, denote the percent of leaves in the various yards at time k. The following relations then hold:

A_k= A_k-1+ .3B_k-1B_k = .3D_k-1C_k = .1B_k-1 + C_k-1
D_k = .6B_k-1+ .3F_k-1E_k = .1D_k_-1+ E_k_-1F_k = .6D_k_-1G_k = .1F_k_-1+ G_k_-1F_k = .6F_k_-1+ H_k_-1

We also have the starting conditions B₀= 1, A₀= C₀= D₀= E₀= F₀= G₀= H₀= 0, that is, 100% of the leaves start in yard A. We are looking for the "limit" of these equations as k ® ¥. Let M and v_k be the following matrices.

M =	1	.3	0	0	0	0	0	0	,	v_{k =}	A_k	.
	0	0	0	.3	0	0	0	0			B_k
	0	.1	1	0	0	0	0	0			C_k
	0	.6	0	0	0	.3	0	0			D_k
	0	0	0	.1	1	0	0	0			E_k
	0	0	0	.6	0	0	0	0			F_k
	0	0	0	0	0	.1	1	0			G_k
	0	0	0	0	0	.6	0	1			H_k

Then M v_k-1 = v_k and inductively M^k v₀ = v_k. We need to find lim_{k ®
¥} M^k. Using any reasonably powerful calculator, you can compute that

M¹⁰⁰v₀₌	0.384375	,
	0
	.128125
	0
	.09375
	0
	.05625
	.3375

and the numbers are stationary at that point. (There are more rigorous ways of doing this computation, but they all yield the same result.) The percent that doesn't turn to mulch is therefore 0.384375 + .3375 = 0.721875.

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ã2006 Alberto L. Delgado