Modeling for Process Operations. Introduction to Stochastic Programming. Jeffrey Kantor. Jeffrey Kantor. October 30 November 1, 2006

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1 October 30 November 1, 2006

2 Class Projects Class Projects: Sean Electric Power Operation and risk management for a gas powered generation facility. Kim Robust portfolio optimization. Josh Real Options for imperfect markets. David Nate

3 Project Elements Core Idea: What is the key idea, and how is related to the theme of process operations? Review: Where did the idea originate? What are the seminal papers, applications? To whom should the ideas be attributed? Formulation: Present a mathematical (or other concise) synopsis of the idea. Application: Demonstrate through models and numerical computation an illustrative application. Deliverables: 20 minute class presentation Short paper, 3-5 pages

4 Project Ideas Ecological Planning Models Hedging in Supply/Demand driven markets Electricity Planning Models Real Option Analysis for Natural Resource Approximate Dynamic Others? Open to Ideas!

5 Agenda Power Utility Expansion Example (Bischopp, 1999) Scenarios Expected Value with Perfect Information Value of a Stochastic Solution

6 Power System Expansion Example (Bisschop, 1999) An existing electric power utility company foresees a large increase in demand. The utility currently operates Coal and Hydro powered plants buys power as needed from the grid to satisfy customer demand can add capacity with Coal, Hydro, or Nuclear generating capacity. If the demand forecast is uncertain, how does the utility decide how much to invest in additional capacity?

7 Demand Scenarios The utility is expecting to see increased demand for electric power from its customers. There are two categories of demand, a 24 hour Base Load period, and a 6 hour Peak Load period. Four possible scenarios:

8 Demand Scenarios The utility is expecting to see increased demand for electric power from its customers. There are two categories of demand, a 24 hour Base Load period, and a 6 hour Peak Load period: Scenarios Base (24 hr) Peak (6 hr) (s) [GW] [GW] Scenario Scenario Scenario Scenario Average

9 Additional Cost Information Units Coal Hydro Nuclear Grid Existing Cap. e p [GW] Capital Cost C p [$k/gw] Oper. Cost O p [$k/gwh]

10 Model Sets p Plant Types (Coal, Hydro, Nuclear, Grid) k Demand Categories (Base, Peak) Parameters e p Existing capacities [GW] C p Capital Cost [k$/gw] Operating Cost [k$/gwh] O p D k T k Demand for Category k [GW] Duration of Category k [hr] Decision Variables x p Additional Capacity [GW] Allocation of Capacity to Categories [GW] y pk

11 Operating Model For a plant capacities, choose an operating policy y pk to minimize Cost where Cost = C p e p + T k O p y pk p Plants p Plants k Demand Subject to Constraints k Demand p Plant y pk e p p y pk = D k k y pk 0

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16 Operating Costs with No Additional Capacity Scenario Cost ($1,000 s) Scenario 1 $27,690 Scenario 2 $35,490 Scenario 3 $24,090 Scenario 4 $30,090 Average $29,340 Much of the cost is for purchased power. Will adding capacity reduce costs?

17 Design Model Choose additional capacity x p and operating policy y pk to minimize Cost where Cost = C p (e p + x p ) + T k O p y pk p Plants p Plants k Demand Subject to Constraints k Demand p Plant y pk e p + x p p y pk = D k k x p, y pk 0

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22 Operating Costs with Additional Capacity Summarizing the designs required to minimize cost for each scenario Scenario Coal Hydro Nuclear Grid Cost ($1,000 s) Scenario $7,055 Scenario $8,160 Scenario $6,500 Scenario $7,275 Average 1 4 $7,247.50

23 Expected Value with Perfect Information Expected Value with Perfect Information Each scenario results in a different plan for additional capacity. If we had perfect information, we d have an Expected Value with Perfect Information of $7, But we have to make the decision before we know the demand. So which plan do we choose?

24 Plan 1, Scenario 2

25 Evaluating Plans 1 4 for Scenarios 1 4 We evaluate each plan for each scenario. Scen. 1 Scen. 2 Scen. 3 Scen. 4 Ave. Plan 1 7,055 9,500 6,785 7,415 7,689 Plan 2 7,620 8,160 7,350 7,710 7,710 Plan 3 7,805 10,250 6,500 7,910 8,116 Plan 4 7,350 9,615 6,825 7,275 7,766 EVM 7,146 9,370 7,169 7,349 7,758 The last line is computed using the average of the demand scenarios (8.6875,+2.125), computing a design, then evaluating for all scenarios. It s a useful base case for comparison.

26 Evaluating Plans 1 4 for Scenarios 1 4 We evaluate each plan for each scenario. Scen. 1 Scen. 2 Scen. 3 Scen. 4 Ave. Plan 1 7,055 9,500 6,785 7,415 7,689 Plan 2 7,620 8,160 7,350 7,710 7,710 Plan 3 7,805 10,250 6,500 7,910 8,116 Plan 4 7,350 9,615 6,825 7,275 7,766 Plan 1 results in the minimum expected cost.

27 Evaluating Plans 1 4 for Scenarios 1 4 We evaluate each plan for each scenario. Scen. 1 Scen. 2 Scen. 3 Scen. 4 Ave. Plan 1 7,055 9,500 6,785 7,415 7,689 Plan 2 7,620 8,160 7,350 7,710 7,710 Plan 3 7,805 10,250 6,500 7,910 8,116 Plan 4 7,350 9,615 6,825 7,275 7,766 Plan 2 minimizes the maximum cost.

28 Two Stage Decision Making Choose x p to minimize average cost over all possible outcomes. Stage 1: Choose x p averaging over all possible outcomes. Stage 2: After the uncertainty has been resolved, choose y pk to minimize operating cost.

29 Model Sets p Plant Types (Coal, Hydro, Nuclear, Grid) k Demand Categories (Base, Peak) s Scenarios Parameters e p Existing capacities [GW] C p Capital Cost [k$/gw] Operating Cost [k$/gwh] O p D ks T k Demand for Category k [GW] Duration of Category k [hr] Decision Variables x p Additional Capacity [GW] y pks Allocation of Capacity to Categories [GW] Variable cost for scenario s [k$] v s

30 Stochastic Model Choose capacity x p and operating policy y pks to minimize Cost = p Plants C p (e p + x p ) + 1 N s s Scenarios v s Subject to Constraints y pks e p + x p p, s p Plants k Demand k Demand p Plant y pks = D k s k, s T k O p y pks = v s s x p, v s, y pks 0

31 Putting a Value on Information Summary of results: Expected Value with Perfect Information: $7, Stochastic Solution: $7, Expected Value of Mean Design: $7,758 Interpretation of these results: The Value of Stochastic Solution is $7, 758 $7, 573 = $185. The Expected Value of Perfect Information is $7, 573 $7, 248 = $325.

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