Samson Hydro operates a pumped hydro storage facility.  When power costs are low water can be          pumped up into a raised reservoir and when power costs are high the same water can be released to     generate power. The total usable capacity of the reservoir is 9,000,000 litres (which we write as 9,000 kl or kilolitres)

There are only three states that the plant at Samson Hydro can be in: (a) pumping water up into the        reservoir using power, (b) idle, (c) allowing water to flow out of the reservoir to generate power. When  pumping water Samson Hydro requires 290 MW and during this phase a total of 14 kl per minute can be pumped. When water is flowing out of the dam, the rate of flow is 16 kl per minute and this will               generate 260 MW of power.

There is a loss of efficiency in storing power in this way (it costs more power to pump than Samson         Hydro generates through letting water out, even though the flows are greater when Samson Hydro is     generating).  This will mean that it makes sense for there to be significant periods when the plant is idle.

The spreadsheet gives 6 weeks of price data (42 days, or 1008 hours). These are electricity prices in € per MWh and are taken from the prices in Denmark for a period starting on 9 September 2021. (Note that prices can occasionally be slightly negative)

Part 1 (16 Marks)

Suppose that a single daily schedule of operation will be determined that will be used throughout the  six-week period. It is desired to maximize average daily profit with the reservoir at a level of 4,000 kl at the beginning and end of every day (12 midnight) and allowing for the maximum capacity of the reservoir. Formulate this as a linear program.

Use the notation  pn,t for the price in hour t of day n.  You may wish to use, as variables for each of 24    hours, the proportion of the hour pumping, and the proportion of the hour generating, 48 numbers in    total - since the proportion idle can then be deduced.  You may also want to define a variable for the       reservoir level at the end of hour t . You need to carefully write down expressions (using Sigma notation) for the objective and each of the constraints, as well as specifying which variables are non-negative.

Then solve the optimization problem. Find the maximum average daily profit and the daily schedule of operation that achieves this.

Part 2 (28 marks)

Now consider a scenario in which a decision needs to be made without knowledge of future prices. Suppose that Samson Hydro needs to decide on what to do for the next hour (proportions of time  pumping; or generating; or idle) on the basis of the current price and reservoir level. We consider a simple linear threshold policy of the following form:

(A)  At the end of hour t calculate a number wt  = αyt  + βpt where yt  is the reservoir level at the end of hour  t and pt  is the price in hour t . Here a and F are parameters to be chosen.

(B)  The policy is to pump in the following hour (hour t + 1) if wt  ≤ C, generate in hour t + 1 if Wt  ≥ D, and do nothing if wt  is between these two values. Here C and D are parameters to be chosen.

We allow a, F, C and D to depend on the time of day. The version of this policy that you should use has two different values for these parameters: one for the morning (first 12 hours before midday) and one   for the afternoon/evening (second 12 hours, after midday). For decisions on whether to pump for the    hour from 12 noon to 1 pm you need to use the second set of parameters (applying these to the prices  in the period from 11 to noon and the reservoir level at noon). Please be careful about the timing of this switch over.  Because of the linearity we can normalise so that a = 1 in both cases. For the first hour of the day please use a fixed policy based on your answer to Part 1, rather than looking at the last hour of  the previous day.

In fact, the policy cannot be applied in exactly this manner – so your first task is to determine a version of this policy that will ensure that the reservoir level stays positive and below 9,000 kl and returns to a 4,000 kl at the end of the day (12 midnight).   Except where forced to do otherwise by these constraints the policy will, as before, pump as much as possible if Wt  ≤ C, generate as much as possible if   wt  ≥ D, and be idle if wt  is between these two values.  Note that, for example, the constraint of needing to be at 5,000 kl at midnight may force Samson Hydro to pump, even if wt  > C .  Please describe your modified policy carefully.

(Hint: Consider adding a restriction that the reservoir level (in 1000s of litres) at the end of hour t lies in

the region between the values  min(9000,  4000 + (24 − t)60 × 16) and max(0,  4000 − (24 − t)60 × 14) . )

Now find values for the parameters F, C and D  with both am and pm values for each (so 6 parameters in total) that optimize the average profit made over the 42 days of data. Note that this is a nonlinear optimization problem. Because there is more than one local minimum and the linear decision rule  introduces “break points” this is a hard problem to solve.  I suggest you do a fair amount of               experimentation. For example, you should consider trying the evolutionary solver . Find the optimal choice of parameters and the optimal average daily profit.