Task 2 A trout farmer and a pesticide maker are each located near a lake. The pe
ID: 1159825 • Letter: T
Question
Task 2 A trout farmer and a pesticide maker are each located near a lake. The pesticide maker can dump its waste in the lake at no cost or put in a safe place at a cost of $600 per tonne. The pesticide company produces 5 tgnnes of waste per year and will dispose of it in a cost minimizing manner. The trout farmer's profit depends on how much waste the pesticide maker dumps in the lake and is given as follows: Quantity of waste in lake (tonnesper year) Profits of the trout farmer tper yea $10 000 $9 000 $9 500 $9 o50 $8 000 6 800 If transactions costs are prohibitively high so that the trout farmer and the pesticide maker won't strike an agreement, then how much waste will be dumped in the lake? If the trout farmer is given the right to the lake and that right is enforceable, and there are no transaction costs to writing contracts, then how much waste will be dumped in the lake? a) b) Explain your answers to part (bExplanation / Answer
Solution:
a) If there is no agreement between the trout farmer (TF) and pesticide maker (PM), in order to minimize it's cost, PM will dump all it's waste in the lake, i.e., 5 tonnes of waste in the lake, which will cost him $0, as there in nothing to stop him.
b) If TF has the right to lake, PM cannot dump the waste in the lake without permission of TF. TF will only allow the PM to dump the waste in lake if he is compensated for the loss of his profits due to waste dumping.
If PM dumps 1 tonne of waste, profits for TF are reduced to $9,000. So, the loss in profits for TF = $10,000-$9,000 = $1000. If PM wants to dump 1 tonne of waste in lake, he will have to pay $1,000 (as compensation) so that TF agrees to it. Also, for all tonnes of waste not dumped in the lake, PM will have to keep it in a safe place, costing him $600 per tonne. Considering all this explanation, we can write the cost function of dumping to PM as follows:
Cost of dumping (beared by PM), C = (10000 - y(x)) + 600*(5 - x)
where, x = tonnes of waste dumped in lake
y(x) = profit to TF, when 'x' tonnes of waste is dumped in lake
(we have written x in the brackets alongside y, because value of y depends on value of x)
We derived this cost function in this way: (breaking the cost function)
1) Without any waste disposal, TF can earn profits of $10,000. So, this is the maximum he can earn, without any loss due to disposal. So, loss in profits/compensation amount = (10000 - y(x)) for x tonne of disposal.
2) Maximum wastage to be dumped is given as 5 tonnes and only choice of disposing that PM has is either lake or safe place. So, the amount of waste not dumped in lake, is kept in safe place. Thus, Tonnes of waste in safe place = (5 - x)
3) For every tonne in safe place, PM bears cost of $600 per tonne. So, for (5-x) tonnes of waste in safe place, cost = 600*(5-x)
PM's aim is to minimize the cost function, our aim is to find that value of x that minimizes it (no! you simply can't derivate it since it's not a continuous function, x is an integer)
Substituting different values of x in this cost function:
When x = 0, C = (10000 - 10000) + 600*(5-0) = $3,000
When x = 1, C = (10000 - 9000) + 600*(5-1) = $3,400
When x = 2, C = (10000 - 9500) + 600*(5-2) = $2,300
When x = 3, C = (10000 - 9050) + 600*(5-3) = $2,150
When x = 4, C = (10000 - 8000) + 600*(5-4) = $2,600
When x = 5, C = (10000 - 6800) + 600*(5-5) = $3,200
So, minimum cost is $2,150, when x= 3.
This means 3 tonnes of waste will be dumped in the lake.
P.S. - Profits to TF should decrease as the waste disposal in lake increases. But as per the table above, at waste of 1 tonne, profit = $9,000, while at waste of 2 tonnes, profit is higher ($9,500). The procedure in the answer is correct, but please do verify if the question is correct; final answer might change accordingly.
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