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Problem 4. Draw a decision tree for the following problem, and then solve it. A

ID: 3277232 • Letter: P

Question

Problem 4. Draw a decision tree for the following problem, and then solve it. A spreadsheet will help. You have a water supply reservoir that can hold up to 100 units of water. You are considering how much R(t) to release in each year t for the next two years (t-1,2) There are two objectives: get as close as possible to the release target, and get as close as possible to the storage target Presently (at the start of year 1). you have S(1)-40 units in the reservoir, where S(t) is the amount of water at the start of year t. For year 1, you know that the total inflow Q1)-90 units. In each year after year 1, water flows into the reservoir, but the amount of flow Q(t) is uncertain: it might be L (30 units), M (90 units), or H (140 units). The probabilities of each flow are P(L) 0.3, P(M) 0.4, and P(H 0.3. The probability of a level of flow in year 2 (0(2)) is statistically independent of the flo in year 1 Note that you have to satisfy mass balance, so S(t+ S(t)+Q)-R(t) (ignore evaporation and seepage!). You'l have to calculate S(t+1), given your decision R(t), the random inflow Q(t), and the previous year's storage S(). Let's assume that you can choose from 3 releases in each year: release R(t 30, 80, or 130. You can only choose feasible releases (that is, if you choose a release R so that S(t+1) zero or above 100, then that release is infeasible). Assume that at the time you choose R(t). you will already know what Q(t will be (you have a good forecast), but you don't know Q(t+1), which follows the above probability distribution. (E.g.. you can choose R(1) knowing what Q) is; then in the next year t-2, you can choose R(2) after you forecast what Q(2) is, etc.)) would be below There is a cost associated with each R (you need to consider R(1) and R(2)); a medium R is best because if R is too low, there are water shortages, and if it is too high, there is flooding damage Meanwhile there is also a cost associated with each S (you need to consider S(2) and S(3)); too low and boaters complain they can't use their boats, but too high you get erosion and flooding along the shoreline, so a medium level is best. The cost functions are (R- 70)2 for each R, and (S-50)2 for each S (except if S is less than 0 or more than 100, the cost is infinite, because those values are infeasible. Use 1,000,000 for the cost in those cases). Use a weight of 50 for the R cost function, and a weight of 30 for the storage cost function (weight each cost by those weights) The decision tree will have decisions for R(1) and R(2), and the consequence will be the total cost associated with R(1), S(2), R(3), and S(3). a. What is the optimal R(1)? b. What is the optimal R(2), given that Q(2)-L? Given that Q(2) M? Given that Q(2) H? What is the expected cost of the optimal strategy? (A strategy is the optimal here-and-now decision R(1), and the optimal "wait-and-see" R(2) given each possible Q(2)) storage S(1 or S(2) that is either negative or greater than 100)? (Please describe) e highest amount 130 is optimal? (Please describe under what conditions you'd do that d. Are there any combinations of R(I), Q(2), and R(2) that result in infeasible storages (some e. Is there any situation in which releasing the lowest amount 30 is optimal? In which releasing

Explanation / Answer

The cost function is 50*(R-70)^2 + 30*(S-50)^2

S(1) = 40, Q(1) = 90

S(2) = S(1) + Q(1) - R(1) = 40+90-R(1) = 130-R(1)

R could be 30, 80 or 130

Optimal R(1) is 80, S(2) is 50

b.

If Q(2) = 30, S(3) = 50+30-R(2) =80-R2

So, R(2) could be 30 or 80

If Q(2)=90, S(3) = 50 +90-R(2) = 140-R(2)

So, optimal R(2) is 80

If Q(2)=140, S(3) = 190-R(2)

So, R(2) must be 130

c. Expected cost = P(Q=30).Cost when Q=30 + P(Q=90).Cost when Q=90 + P(Q=140).Cost when Q=130

=0.3*80000+0.4*8000+0.3*183000 = 82100

d. For R(1)=50, infeasible storage is when Q(2)=30,R(2)=130

Q(2)=90,R(2)=30

Q(2)=140,R(2)=30,80

R(1) S(2) Cost 30 100 50*1600+30*2500=155000 80 50 5000+0=5000 130 0 50*3600+30*2500=255000
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