An oil spill has fouled 200 miles of Pacific shoreline The o

An oil spill has fouled 200 miles of Pacific shoreline. The oil company responsible has been given 14 days to clean up the shoreline, after which a fine will be levied in the amount of $10,000/day. The local clean up crew can scrub five miles of beach per week at a cost of $500/day. Additional crews can be brought in at a cost of $18,000 plus $800/day for each crew.

(a) How many additional crews should be brought in to maximize the total cost to the company? (You may use the five step method). How much will the clean-up cost?

(b) Examine the sensitivity to the amount of the fine. Consider the number of days the company will take to clean up the spill and the total cost to the company.

(c) Examine the sensitivity to the amount of the fine. Consider the number of days the company will take to clean up the spill and the total cost to the company.

(d) The company has filed an appeal on the grounds that the amount of the fine is excessive. Assuming that the only purpose of the fine is to motivate the company to clean up the oil spill in a timely manner, is the fine excessive?

Solution

Step 1: Ask the question.

Variables:       c = number of additional crews t = time to clean up spill (days) T = total cost of clean-up ($)

F         = fine ($)

Assumptions: T

= 500 t + (18000 + 800 t) c + F

200

= (5 / 7) (c + 1) t

F

= 0

if

t <= 14

F

= 10,000 (t - 14)

if

t > 14

c is a nonnegative integer, and t >= 0

Objective:       Minimize T.

Step 2: Select the modeling approach.

We will model this problem as a one variable optimization problem. See text p. 6.

Step 3: Formulate the model.

Let x = c and y = T, and write

y = f(x) = 500 (280 / (x+1)) +(18,000 + 800 (280 / (x+1))) x if x >= 19 or

y = f(x) = 500 (280 / (x+1)) +(18,000 + 800 (280 / (x+1))) x + 10,000 (280 / (x + 1) - 14)

if x < 19. Our goal is to maximize f(x) over the set of nonnegative integers x.

Step 4: Solve the model.

One way to solve is to minimize over the nonnegative reals, and then specialize to the integers. On 0 <= x < 19 we have

f \'(x) = 2000*(9*x^2+18*x-1349)/(x+1)^2

which is negative on [0, 11.28) and positive on (11.28, 19), so x = 11.28 is the minimum. Then the minimum over the integers occurs at either x = 11 or at x = 12, and we can check that x = 11, f (x) = 508333 is smaller. On x >= 19 we have

f \'(x) = 6000*(3*x^2+6*x+17)/(x+1)^2

which is always positive, and so on this interval the minumum occurs at x = 19, f (x) = 561800. Then the global minimum occurs at x = 11.

Step 5: Answer the question.

According to this model, the optimal policy is to bring in 11 additional crews, resulting in a total clean-up cost of around $510,000. Clean-up will take around 23.3 days and the resulting fine will be around $93,000.

(b) Examine the sensitivity to the rate at which a crew can clean up the shoreline.

Consider both the optimal number of crews and the total cost to the company.

Generalize the model in part (a) to obtain the assumption 200 = r (c + 1) t

where currently r = (5 / 7) miles per day per crew. Then we have

y = f(x) = 500 (200 / (r (x+1))) +(18,000 + 800 * 200 / (r (x+1))) x if x >= 19 or

y = f(x) = 500 (200 / (r (x+1))) +(18,000 + 800 * 200 / (r (x+1))) x + 10,000 (200 / (r (x+1)) - 14)

if x < 19. For values of r near (5 / 7) the minimum should still occur on the interval (0, 19). Solving

0 = f \'(x) = 2000*(9*r*x^2+18*r*x+9*r-970)/(r*(x+1)^2)

yields

x = (SQRT(970)-3*SQRT(r))/(3*SQRT(r))

and so

dx / dr = -SQRT(970)/(6*r^(3/2)). Substituting r = (5 / 7) we obtain

S(x, r) = (dx / dr) (r / x) = -0.54

so that if the cleanup crews are 10% faster than expected, the optimal number of crews decreases by about 5.4%. If we substitute the formula for the optimal x in terms of r into the formula for y = f(x) in the case x < 19, we obtain

y = -2000*(79*r-6*SQRT(970)*SQRT(r)-80)/r and then we can also calculate

dy / dr = -2000*SQRT(10)*(3*SQRT(97)*SQRT(r)+8*SQRT(10))/r^2 Substituting r = (5 / 7) we obtain

S(y, r) = (dy / dr) (r / y) = -0.88

so that if the cleanup crews are 10% faster than expected, the total cost of clean-up decreases by about 8.8%.

(c) Examine the sensitivity to the amount of the fine. Consider the number of days the company will take to clean up the spill and the total cost to the company.

Generalize the model in part (a) to obtain the assumption F = a (t - 14) if t > 14

where currently a = 10,000 dollars per day. Then for values of a near 10,000 we have y = f(x) = 500 (280 / (x+1)) +(18,000 + 800 (280 / (x+1))) x

+ a (280 / (x + 1) - 14)

if x < 19. The optimal number of crews is x = SQRT(14)*SQRT(a-300)/30-1 and so the time to finish the clean-up is

t = 280 / (x+1) = 600*SQRT(14)/SQRT(a-300) and then we can compute that at a=10,000 we have

S(t,a) = (dt/da)(a/t) = (-300*SQRT(14)/(a-300)^(3/2))*(a/t) = -0.52

so that if the fine is raised by 2% then the cleanup time should decrease by about 1%. Substituting the optimal formula for x in terms of a into the equation for y above, we obtain

y        = 2*(600*SQRT(14)*SQRT(a-300)-7*a+103000)

dy/da = 2*SQRT(7)*(300*SQRT(2)-SQRT(7)*SQRT(a-300))/SQRT(a-300) so that at a = 10,000 we have

S(y,a) = (dy/da)(a/y) = 0.17

which means that if the fine is increased then the total cleanup cost to the company will go up by about 1.7% for each additional $1,000 per day of fine.

(d) The company has filed an appeal on the grounds that the amount of the fine is excessive. Assuming that the only purpose of the fine is to motivate the company to clean up the oil spill in a timely manner, is the fine excessive?

Reasonable answers will differ on this question. On the one hand, the fine is only 19% of the total cost, and if the fine were reduced by 50% then the number of days to clean up the spill would increase by about 25% and it would only save the company about 8.5% of the total cost. So the amount of the fine does not seem excessive. On the other hand, if the 14 day limit were extended to 21 days, cleanup would proceed exactly as before, only the company would save $70,000. So in this case the 14 day limit does seem excessive.

Assumptions: T	= 500 t + (18000 + 800 t) c + F
200	= (5 / 7) (c + 1) t
F	= 0	if	t <= 14
F	= 10,000 (t - 14)	if	t > 14