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Consider a decision faced by a cattle breeder. The breeder must decide how many cattle he should sell in the market each year and how many he should retain for breeding purposes. Suppose the breeder starts with a herd of 200 cattle. If he breeds cattle, he obtains 1.4 times as many cattle per year as he started with. The cost of breeding is $30 for each head of cattle not sold. Breeding takes one year, and the $30 cost includes all expenses of maintaining an animal and its offspring. Alternatively, the breeder may sell cattle in the market, at a price which depends on how many cattle he sells. If Y represents the number of cattle which are sold in a given year, then the Price, P, is given by the following equation:
P = 200 - 0.2Y, 0 ≤Y ≤ 1,000
Start at the end of the actual decision process and represent the size of the herd in year 10 as x10. The herd is to be sold at $150 per head at the beginning of year 10 and consider f10(x10) = 150 x10.
Assume that the breeder plans to sell his entire herd at the beginning of 10 years from now and retire. Determine the optimal strategy (of breeding and selling in the market place) for the breeder over the next ten years using dynamic programming, so that his profits are maximised at the time of retirement.
5. Packages from a machine a normally distributed with a mean 200g and its standard deviation 2grams. Find the probability that a package from the machine weighs a) Less than
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