
Water Resource Management Systems Overview
Explore the intricate network of multipurpose water resource systems, delve into the concept of reservoirs in series, and discover the Highland Lakes in Colorado River basin. Learn how these systems are crucial for various purposes such as water supply, hydropower, and environmental conservation, and understand the complex dynamics involved in managing water resources effectively.
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Presentation Transcript
Multipurpose Water Resource Systems Water Resources Planning and Management Daene C. McKinney
Reservoirs in Series Q2,t Q1,t S1,t S2,t R1,t R2,t K2 K1 T ( ) B1, B2 benefits from various purposes Municipal water supply Agricultural water supply Hydropower Environmental Recreation Flood protection + = t Maximize B B , 1 , 2 t t 1 = = + + S S S S Q Q R L , 1 R + , 1 1 , 1 , 1 , 1 R , 1 t t t t t + L + , 2 St 1 , 2 , 2 , 2 , 2 t t t t t t K , 1 1 , 2 St K 2
Reservoirs in Series Cascade of Reservoirs Sometimes, cascades or reservoirs are constructed on rivers Reservoir 1 R1,t R1_Hydro,t R1_Spill,t Some of the reservoirs may be pass- through Reservoir 2 R2,t R2_Hydro,t R2_Spill,t Flow through turbines may be limited Reservoir 3 R3,t R3_Hydro,t i= Ri_Hydrot+Ri_Spillt Rt R3_Spill,t Reservoir 4 R4,t R4_Hydro,t i Pi,Max i= kH t iRt kW Pt R4_Spill,t Pi,Max Reservoir 5 Ri_Hydrot R5,t R5_Hydro,t i k'H t R5_Spill,t
Highland Lakes Buchannan 918,000 acre-feet Inks LBJ . Marble Falls 1,170,000 acre-feet Travis Lake Austin
Highland Lakes Colorado R. Q1,t Lake Buchannan S1,t K1 = 918 kaf R1,t S2,t Inks Lake Q2,t R2,t =R1,t Llano R. S3,t Lake LBJ R3,t =R2,t + Q2,t S4,t Lake Marble Falls Pedernales R. R4,t =R3,t 300,000 Q3,t Flow, Demand (AF/month) Flow Total Demand Austin Irrigation Downstream 250,000 S5,t K5 = 1,170 kaf Lake Travis 200,000 R5,t 150,000 Austin M&I 100,000 Channel Losses 50,000 Incremental Flow Rice Irrigation 0 0 10 20 30 40 50 60 Bay & Estuary Month
Highland Lakes t i S , = + + K Continuity t i S t i S Q R t i R t i L Capacity + , 1 i , 1 , t i , , , i t t i 4 = Marble Falls, 5 = Travis, 6=Austin) St,l Storage in lake i in period t (AF) Qt,l Inflow to lake i in period t (AF) Ll Loss from lake i in period t (AF) Rt,l Release from lake i in period t (AF) Time period (month) Lake (1 = Buchannan, 2 = Inks, 3 = LBJ, Ki Capacity of lake i Head vs Storage + ( ) ( ) H t i S H t i S + , , 1 i i = H , t i 2 Hi,t elevation of lake i + + CL + + R I X X X Release , 5 , , , t t A t I t t B t Energy R5,t XA,t XI,t CLt XB,t Release from Lake Travis in period t (AF) Diversion to Austin (AF/month) Diversion to irrigation (AF/month) Channel losses in period t (AF/month) Bay & Estuary flow requirement (AF/month) Ei,t= kiH i,tRi,t E,t i Energy (kWh) efficiency (%) X f T , , A t A t T A X f , , I t I t I TA TI fA,t fI,t target for Austin water demand (AF/year) target for irrigation water demand (AF/year) monthly Austin water demand (%) monthly irrigation water demand (%)
Objective 2 2 TA,t-XA,t TA,t TI,t-XI,t TI,t TR,i,t-hi,t TR,i,t T +wI +wR Minimize wA t=1 i=1&5 Municipal Water Supply Benefits: Try to meet targets Irrigation Water Supply Benefits: Try to meet targets Recreation (Buchanan & Travis) Benefits: Try to meet targets wA wT wR TA,t TI,t TR,i,t monthly target for lake levels, i = Buchanan, Travis weight for Austin demand weight for Austin demand weight for Lake levels monthly target for Austin demand monthly target for irrigation demand ZA ZI ZR penalty for missing target in month t penalty for missing target in month t penalty for missing target in month t minimum minimum minimum XA,t XI,t TA,t release hi,t TI,t TR,i,t target target target release elevation Municipal Irrigation Recreation
Results K1 = Buchannan = 918 kaf K5 = Travis = 1,170 kaf Austin Irrigation No Storage Deficits No Storage Deficits 1200 40 Release Deficit (1000 AF) 1000 30 Storage (1000 AF) 800 20 600 400 10 S_Buch S_Trav 200 0 0 0 12 24 36 48 60 0 12 24 36 48 60 Month Month No Release Deficits No Release Deficits 1200 700 600 1000 Storage Deficit (1000 AF) Buchanan Storage (1000 AF) 500 800 400 600 300 400 S_Buch S_Trav 1170 200 200 100 0 0 0 20 40 60 0 12 24 36 48 60 Month Month 1,000 acre feet = 1,233,482 m3
Weights 100-100-1 10-10-1 1-1-1 1-1-2 1-1-10 Release Storage 4423.79 4341.56 3531.82 2747.64 465.86 Results 1.61 16.41 162.45 307.62 752.12 Water Supply vs Recreation Tradeoff 5000 Storage Deficit (1000 AF) 4000 y = -5.271x + 4409.6 3000 2000 1000 0 0 200 400 600 800 Release Deficit (1000 AF)
What s Going On Here? Multipurpose system Conflicting objectives Tradeoffs between uses: Recreation vs. irrigation No unique solution Let each use j have an objective Zj(x) We want to Z ( x )=[Z1( x ),Z2( x ),...,Zp( Maximize x )] x =(x1,x2,...,xn) X subject to
Multiobjective Problem Single objective problem: Identify optimal solution, e.g., feasible solution that gives best objective value. That is, we obtain a full ordering of the alternative solutions. Maximize Z1( x ) subjectto x =(x1,x2,...,xn) X Multiobjective problem We obtain only a partial ordering of the alternative solutions. Solution which optimizes one objective may not optimize the others Noninferiority replaces optimality Z ( x )={Z1( x =(x1,x2,...,xn) X x ),Z2( Zp( Maximize x ), x )} subjectto
Example Flood control project for historic city with scenic waterfront Alternative Net Benefit $120k Method Effects 1 Increase channel capacity Construct flood bypass Construct detention pond Construct levee Change riverfront, remove historic bldg s Create greenbelt 2 $700k 3 $650k Destroy recreation area Isolate riverfront 4 $800k
Alternative Net Benefit 1 $120k Example 2 $700k 3 $650k 4 $800k Does gain in scenic beauty outweigh $100k loss in NB? (Alt 4 2) Objective 1 Objective 2 Maximize Net Benefit Alternative Maximize Scenic Beauty Alternative Alternative 2 is better than Alternatives 1 and 3 with respect to both objectives. Never choose 1 or 3. They are inferior solutions. 4 2 2 3 3 1 Alternatives 2 and 4 are not dominated by other alternatives. They are noninferior solutions. 1 4
Noninferior Solutions (Pareto Optimal) Noninferior Solutions 40? A feasible solution is noninferior if there exists no other feasible solution that will yield an improvement in one objective w/o causing a decrease in at least one other objective (A & B are noninferior, C is inferior) 35? 30? Irriga on? Supply? 25? 20? feasible region 15? 10? All interior solutions are inferior move to the boundary by increasing one objective w/o decreasing another C is inferior 5? 0? 0? 5? 10? 15? 20? 25? Energy? Produc on? Alternative Energy Production Irrigation Supply Northeast rule: A feasible solution is noninferior if there are no feasible solutions lying to the northeast (when maximizing) A 22 20 B 10 35 C 20 32 D 12 21 E 6 25 Vilfredo Federico Damaso Pareto
Example = Z x x x Maximize ( ) [ ( ), ( )] Z Z 1 2 Decision Space subject to x2 = x ( ) 5 2 Z x x 1 1 x 2 x = x + x ( ) 4 Z 2 x 1 2 x + x ) 1 ( 3 E D 1 + 2 4 ) 2 ( 8 1 2 1 2 F ) 3 ( 6 x C 1 ) 4 ( 4 x Feasible Region 2 3 x1 0 x2 0 B A A B 6 0 x1 C 6 2 D 4 4 Evaluate the extreme points in decision space (x1, x2) and get objective function values in objective space (Z1, Z2) E 1 4 F 0 3 Cohen & Marks, WRR, 11(2):208-220, 1973
Z1 0 Z2 0 Example A B 30 -6 C 26 2 D 12 12 Noninferior set contains solutions that are not dominated by other feasible solutions. Objective Space E -3 15 Z2 F -6 12 E F D Noninferior solutions are not comparable: C: 26 units Z1; 2 units Z2 D: 12 units Z1; 12 units Z2 Noninferior set Feasible Region Which is better? Is it worth giving up 14 units of Z2 to gain 10 units of Z1 to move from D to C? C A Z1 B
Example = Z x x x Maximize ( ) [ ( ), ( )] Z Z 1 2 Decision Space subject to Noninferior set x2 = x ( ) 5 2 Z x x 1 1 x 2 x Z2 = x + x ( ) 4 Z 2 x 1 2 x + x ) 1 ( 3 E D 1 + 2 4 ) 2 ( 8 1 2 1 2 Z1 F ) 3 ( 6 x C 1 ) 4 ( 4 x Feasible Region 2 3 x1 0 x2 0 B A A B 6 0 x1 C 6 2 D 4 4 Evaluate the extreme points in decision space (x1, x2) and get objective function values in objective space (Z1, Z2) E 1 4 F 0 3 Cohen & Marks, WRR, 11(2):208-220, 1973
30? Maximize Z = Z1+wZ2 Z2=(-1/w)Z1+Z /w St. line: Slope = -1/w Intercept = Z /w e.g., w = 5 Slope = -(1/w) = -(1/5) 20? E D F Z2? 10? C 0? A -10? 0? 10? 20? 30? 40? B Z = 10 -10? Z1? Z = 20 Z = 5
30? 20? w = E D w = 0 F Z2? 10? C 0? A -10? 0? 10? 20? 30? 40? B -10? Z1?
Tradeoffs Tradeoff = Amount of one objective sacrificed to gain an increase in another objective, i.e., to move from one noninferior solution to another Z2 E D x ( ) Z i C x ( ) Z j A Z1 Example: Tradeoff between Z1 and Z2 in moving from D to C is 14/10, i.e., 7/5 unit of Z1 is given up to gain 1 unit of Z2 and vice versa B
Multiobjective Methods Information flow in the decision making process Top down: Decision maker (DM) to analyst (A) Preferences are sent to A by DM, then best compromise solution is sent by A to DM Preference methods Bottom up: A to DM Noninferior set and tradeoffs are sent by A to DM Generating methods
Methods Generating methods Present a range of choice and tradeoffs among objectives to DM Weighting method Constraint method Others Preference methods DM must articulate preferences to A. The means of articulation distinguishes the methods Noninterative methods: Articulate preferences in advance Goal programming method, Surrogate Worth Tradeoff method Iterative methods: Some information about noninferior set is available to DM and preferences are updated Step Method
Weighting Method = Z x x x x Mazimize ( ) { ( ), ( ), , ( )} Z Z Z 1 2 p subject to x ( ) g b 1 1 x ( ) g b m m = x ( , ,..., ) 0 x x x 1 2 n = + + + x x x x Mazimize ( ) ( ) ( ) ... ( ) Z w Z w Z w Z 1 1 2 2 p p subject to x ( ) g b Vary the weights over reasonable ranges to generate a wide range of alternative solutions reflecting different priorities. 1 1 x ( ) g x b m m = ( , ,..., ) 0 x x x 1 2 n
Constraint Method = Z x x x x Mazimize ( ) { ( ), ( ), , ( )} Z Z Z 1 2 p subject to x ( ) g b 1 1 x ( ) g b m m = x ( , ,..., ) 0 x x x 1 2 n x Mazimize ( ) Z k subject to x Optimize one objective while all others are constrained to some particular bound Solutions are noninferior solutions if correct values of the bounds (Lk) are used = ( ) ,..., 1 , Z L i p i k i i x ( ) g b 1 1 x ( ) g b m m = x ( , ,..., ) 0 x x x 1 2 n