CN106327065A - Water resource optimization configuration method for single pumping station - single reservoir system for direct canal supplement under full irrigation condition - Google Patents

Water resource optimization configuration method for single pumping station - single reservoir system for direct canal supplement under full irrigation condition Download PDF

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CN106327065A
CN106327065A CN201610663218.4A CN201610663218A CN106327065A CN 106327065 A CN106327065 A CN 106327065A CN 201610663218 A CN201610663218 A CN 201610663218A CN 106327065 A CN106327065 A CN 106327065A
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程吉林
龚懿
陈兴
蒋晓红
张礼华
袁承斌
周建康
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Yangzhou University
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Abstract

The invention relates to a combined operation dispatching method for a single pumping station - single reservoir system for the direct canal supplement under a full irrigation condition, and the method comprises the steps: taking the minimum quadratic sum of the sum of water supply of a single regulating reservoir and a single water supplement pumping station in each time period of a year and the difference of water demands of water receiving regions as the target, wherein the water supply of the reservoir in each time period and the water supplement of the pumping station in each time period are taken as decision making variables; taking the yearly allowed water pumping quantity of the reservoir, the yearly water supply quantity of the pumping station, a reservoir dispatching rule, a water balance rule, a dead storage and the storage corresponding to a flood control and limiting water level as the constraint conditions, and building a water resource optimization dispatching model for the single pumping station - single reservoir system for direct canal supplement; employing a dynamic planning successive approximation method for solving, and obtaining the minimum water shortage of the water receiving regions in a certain period and the corresponding optimal water supply and surplus water of the reservoir and the water supplement of the pumping station. The method can achieve the water resource optimization dispatching for the single pumping station - single reservoir system for direct canal supplement, and is of great practical significance to the improvement of the canal supplement water resource efficiency of the pumping station and the probability of irrigation of an irrigation region.

Description

Fully directly mend single pumping plant-mono-water reservoir system water resource optimization of canal under irrigation conditions Collocation method
Technical field
The present invention relates to single small pump station and the method for single reservoir cooperation scheduling under abundant irrigation conditions, belong to Water Resources Irrigation distributes technical field rationally.
Background technology
Currently uneven due to water resource spatial and temporal distributions, restrict the socio-economic development in many areas, for fully irrigating For the irrigated area of condition, under limited water resources total amount and water source project scale, to be target to the maximum by water comprehensive benefit, strengthen The United Dispatching of regional water resources and management, use the hydraulic engineering of irrigation system and regulate as unified entirety, using Built engineering (as reservoir and pumping plant combined dispatching run) makes it play bigger effect, is the main of solution irrigated area water shortage problem Approach.Directly single pumping plant-mono-water reservoir system traffic control of benefit canal is as one content of water resources management, the most reasonably uses The scheduling of system water resource reaches certain target makes system benefit optimal, is problem relatively conventional during water project management uses.
Summary of the invention
The present invention is directed to the single pumping plant directly mending canal under abundant irrigation conditions and single water reservoir system, it is considered to different water frequencies Under intake area hydropenia situation, consider directly to be supplemented by lift pumping station channel hydropenia first, set up the associating of annual-storage reservoir-pumping plant Optimized Operation mathematical model.Single pumping plant-mono-reservoir cooperation the dispatching patcher of canal year regulation is directly mended, known for specific Reservoir supply water divide time hop count, initial storage, minimum capacity of a reservoir, storage capacity that flood control is corresponding, be available in year water inventory, each time Section comes process water, evaporation and leakage process, allows water lift total amount, and day part intake area crop water moisturizing pumping plant year Under amount process condition, use dynamic programming successive approximation method to solve, intake area minimum water deficit in the regular period can be obtained, and Corresponding reservoir optimal water supply, abandon the water yield and pumping plant rate of water make-up process.
The present invention program is as follows:
Single pumping plant-mono-water reservoir system water resource optimal allocation method of canal is directly mended, by carrying under a kind of abundant irrigation conditions Pump works directly supplements channel hydropenia, comprises the following steps:
One, model construction, comprises the following steps 1～step 2:
1. with single seat annual-storage reservoir and the output sum of day part and intake area water requirement in single seat moisturizing pumping plant year The minimum target of quadratic sum of difference, set up following object function:
F = min Z = min Σ i = 1 N ( X i + Y i - YS i ) 2 - - - ( 1 )
In formula: F be the difference of the confession water requirement of day part in object of study year least square and；In Z is object of study year Day part is for the quadratic sum of the difference of water requirement；N is the year interior time hop count divided；Segment number when i is (i=1,2 ... N)；Xi、Yi It is respectively reservoir, the output of the i-th period of pumping plant and rate of water make-up (ten thousand m3)；YSiWater requirement (ten thousand m for the i-th period of intake area3)； It is to accelerate to reduce the deviation between system water supply amount and intake area water requirement that object function uses quadratic sum to express.
2. constraints is set
It is available for water inventory constraints including reservoir, moisturizing pumping plant year, directly mends the reservoir water balance in storehouse without pumping plant about Bundle and reservoir capacity constraints.
Two, model solution
First carry out data preparation, specifically include: be divided into N number of period by 1 year；According to reservoir initial water level, search water Position-capacity curve, determines reservoir initial storage V0；Specify with reference to reservoir operational management, determine and be available for water inventory SK, dead year Storage capacity Vmin, and storage capacity V corresponding to flood controlP；According to reservoir locality meteorological model data, calculate and determine that reservoir is each Period water yield LSi, evaporation with leakage EFi；According to pumping plant working system, determine that moisturizing pumping plant year allows water lift total amount BZ；Root According to data such as intake area variety of crops, planting scale, multiple crop indexes, calculate and determine day part intake area water demand of crop YSi； Wherein, i=1,2 ..., N；
Next carries out dynamic programming Approach by inchmeal and solves.
Further, described constraints includes:
(1) reservoir, be available for water inventory constraint moisturizing pumping plant year: in the case of varying level year difference fraction, it is considered to need Water requirement, the water yield that water supply project can be provided that.
X1+X2+…+XN≤SK (2)
Y1+Y2+…+YN≤BZ (3)
In formula: SK be reservoir be available for water inventory (ten thousand m year3)；BZ is moisturizing pumping plant year to allow water lift total amount (ten thousand m3)。
(2) the reservoir water yield Constraints of Equilibrium in storehouse is directly mended without pumping plant:
Vi=Vi-1+LSi-PSi-EFi-Xi, (i=1,2, N) (4)
In formula: Vi、Vi-1It is respectively reservoir i-th and the reservoir storage of i-1 period Mo (ten thousand m3)；LSi、PSi、EFiFor reservoir i-th Period carry out the water yield (ten thousand m3), abandon the water yield (ten thousand m3), evaporation with leakage (ten thousand m3)。
(3) reservoir capacity constraint: day part end pondage should be right between reservoir minimum capacity of a reservoir and flood control institute Between the storage capacity answered, it may be assumed that
Vmin≤Vi≤VP, (i=1,2, N) (5)
In formula: Vmin、VPFor reservoir minimum capacity of a reservoir storage capacity corresponding with flood control (ten thousand m3)。
Further, dynamic programming Approach by inchmeal solves and specifically comprises the following steps that
(1) study area being carried out investigation, collect the statistics daily output of reservoir and pumping plant moisturizing scale, this data should expire Foot formula (2)～(3) requirement, and reservoir stage storage capacity will not occur less than minimum capacity of a reservoir Vmin.This specific intake area is met with actual The fully reservoir stage output X of irrigation conditions1iAs primary iteration value, substituted into formula (1), then master mould (1)～(5) turn Turn to mend water in a canal amount Y with each stage pumping plantiFor decision variable, front i stage pumping plant moisturizing total amount λiOne-dimensional flow for state variable State plan model, uses one-dimensional dynamic programming to solve；Wherein, i=1,2 ... N.
(2) with reference to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is:
1) stage i=1:
g1(λ1)=min (X11+Y11-YS1)2 (6)
This period reservoir yield X11Given by initial value, state variable λ1, it can be discrete in corresponding feasible zone: λ1 =0, W1,W2,…,BZ.To each discrete λ1, decision variable (pumping plant rate of water make-up Y11) can be discrete, such as 0 in corresponding feasible zone Ten thousand m3, 50,000 m3, 100,000 m3, 150,000 m3、…Y11,maxDeng (Y11,maxIt is the 1st stage pumping plant maximum moisturizing ability), should meet: Y11≥ λ1.Y by satisfied requirement11Substitute into formula (6) respectively, respectively obtain each discrete λ1During value, optimum Y11And the g of correspondence1(λ1)。
Then, according to formula (4), the 1st stage end reservoir capacity V1=V0+LS1-EF1-X11, the most not yet consider that reservoir is abandoned Water, uses formula (5) inspection, if exceeding the storage capacity V corresponding to flood controlp, then the water yield is abandoned beyond part as reservoir PS11, now V1 *=VP；Otherwise, without departing from, then PS11=0, now V1 *=V1。
2) stage i=2,3 ... N-1:
gi(λi)=min [(X1i+Y1i-YSi)2+gi-1(λi-1)] (7)
This period reservoir yield X1iGiven by initial value, state variable λiCarry out discrete the most respectively: λi=0, W1, W2,…,BZ.To each discrete λi, decision variable (pumping plant rate of water make-up Y1i) discrete ibid, and should meet:
State transition equation: λi-1=λi-Y1i (8)
In formula: i=2,3 ..., N-1.
By each discrete Y1iValue substitutes into the (X in formula (7) respectively1i+Y1i-YSi)2, by state transition equation formula (8), search The i-1 stage meetsThe g requiredi-1(λi-1) value, it is derived from meeting this λiThe optimum Y required1iProcess and correspondence thereof Gi(λi).Equally, according to formula (4), the i-th period end reservoir capacity Vi=Vi-1+LSi-EFi-X1i, the most not yet consider that reservoir is abandoned Water, uses formula (5) to test, if exceeding the storage capacity V corresponding to flood controlP, then water is abandoned beyond part as reservoir Amount PS1i, now Vi *=VP；Otherwise, without departing from, then PS1i=0, now Vi *=Vi.Thus pushing over, the reservoir that can obtain correspondence is abandoned Process water PS1i.Wherein, i=1 ... i.
3) stage N:
This period reservoir yield X1NGiven by initial value, state variable λN=BZ；Decision variable (pumping plant rate of water make-up Y1N) discrete in corresponding feasible zone equally, should meet: λN-1=λN-Y1N。
Use step 2) described method, final acquisition meets this λNPumping plant optimum moisturizing process Y required1i, and corresponding Reservoir abandon process water PS1i, wherein, i=1 ... N.
(3) canal process water Y mended by pumping plant step (2) obtained1iAs initial set-point, substitute into formula (1), then master mould ～(5) are converted into each stage reservoir yield X (1)iFor decision variable, front i stage reservoir is for water inventory λi' become for state The one-dimensional dynamic programming model of amount, with reference to step (2), uses one-dimensional dynamic programming to solve, it is thus achieved that to meet this λN' require water Storehouse optimum water supply process X2i(i=1 ... N), and correspondence abandon process water PS2i, wherein, i=1 ... N.
(4) reservoir yield process X that step (3) is obtained2iAs initial set-point, substitute into formula (1), repeat step ～(3) (2), Approach by inchmeal solves repeatedly, until adjacent twice object function optimal value error precision is less than 1%, then model is excellent Change terminates.With last reservoir yield process X optimizing and obtainingmiCanal process water Y is mended with pumping plantmiAs master mould Excellent solution, the most also can obtain object function optimal value, and reservoir optimum abandons process water PSmi, wherein, i=1 ... N, m are State planning Approach by inchmeal iterations numbering.
This invention solves conveniently, and precision is reliable, is available under abundant irrigation conditions using the big-and-middle of single pumping plant-mono-reservoir water supply Type irrigated areas administration unit popularization and application, reach the purpose that Water Resources Irrigation is distributed rationally, improve irrigated area, society and Ecological Effect Benefit.
Accompanying drawing explanation
Fig. 1 generally changes system schematic for directly mending canal list pumping plant-mono-reservoir water resource.
Detailed description of the invention
Single pumping plant-mono-reservoir reservoir water resource generally changes system schematic as shown in Figure 1.
With single seat annual-storage reservoir and in single seat moisturizing pumping plant year the output sum of day part and intake area water requirement it The minimum target of quadratic sum of difference, day part reservoir yield, pumping plant rate of water make-up are decision variable, to be available in reservoir, pumping plant year Water inventory, reservoir operation criterion, water balance criterion, minimum capacity of a reservoir, flood control correspondence storage capacity etc. are constraints, set up Directly mend single pumping plant-mono-water reservoir system water resources optimal operation model of canal, specific as follows:
F = min Z = min Σ i = 1 N ( X i + Y i - YS i ) 2 - - - ( 1 )
In formula: F be the difference of the confession water requirement of day part in object of study year least square and；In Z is object of study year Day part is for the quadratic sum of the difference of water requirement；N is the year interior time hop count divided；Segment number when i is (i=1,2 ... N)；Xi、Yi It is respectively reservoir, the output of the i-th period of pumping plant and rate of water make-up (ten thousand m3)；YSiWater requirement (ten thousand m for the i-th period of intake area3)； It is to accelerate to reduce the deviation between system water supply amount and intake area water requirement that object function uses quadratic sum to express.
3.1.2 constraints
(1) water inventory constraint it is available for year: in the case of varying level year difference fraction, it is considered to need water requirement, waterman The water yield that journey can be provided that.
X1+X2+…+XN≤SK (2)
Y1+Y2+…+YN≤BZ (3)
In formula: SK be reservoir be available for water inventory (ten thousand m year3)；BZ is moisturizing pumping plant year to allow water lift total amount (ten thousand m3)。
(2) the reservoir water yield Constraints of Equilibrium in storehouse is directly mended without pumping plant:
Vi=Vi-1+LSi-PSi-EFi-Xi, (i=1,2, N) (4)
In formula: Vi、Vi-1It is respectively reservoir i-th and the reservoir storage of i-1 period Mo (ten thousand m3)；LSi、PSi、EFiFor reservoir i-th Period carry out the water yield (ten thousand m3), abandon the water yield (ten thousand m3), evaporation with leakage (ten thousand m3)。
(3) reservoir capacity constraint: day part end pondage should be right between reservoir minimum capacity of a reservoir and flood control institute Between the storage capacity answered, it may be assumed that
Vmin≤Vi≤VP, (i=1,2, N) (5)
In formula: Vmin、VPFor reservoir minimum capacity of a reservoir storage capacity corresponding with flood control (ten thousand m3)
3.2 model feature
(1) water total amount control should strictly be carried out, therefore in the constraints of model in view of in water resources development and utilization It is available for water inventory middle addition reservoir year and allows the constraint of water lift total amount, i.e. formula (2)～(3) pumping plant year.
(2) constraints considers reservoir water balance and storage capacity constraint, " idle reservoir moisturizing, station, busy storehouse can be realized Co-supplying " reservoir-pumping station system water resources optimal operation mode.If certain period end pondage of idle is dead less than reservoir Storage capacity, then this period is considered as small pump and stands erectly to connect and carry out moisturizing to channel；If certain period reservoir capacity exceedes flood control by reservoir regulation Corresponding to limiting water level during storage capacity, then need to take out the dispatching requirement abandoning water to ensure reservoir capacity.Busy is according to pondage Situation, then consider to be combined by reservoir, pumping plant to supply water to intake area, to meeting the water demand of water user as far as possible.
3.3 model solution
Being the nonlinear mathematical model that can divide in a stage for model above (1)～(5), in object function, each stage is by water District's water requirement YSiFor it is known that model be with reservoir delivery period divide period i (i=1,2 ..., N) be stage variable, each rank Section reservoir yield Xi, pumping plant rate of water make-up YiFor decision variable, dynamic programming successive approximation method is used to solve.
Assuming that the initial reservoir capacity V of annual-storage reservoir0Couple about it is known that model Chinese style (2)～(3) are dynamic programming Bundle, formula (4) is reservoir operation day part water balance criterion；Solved by dynamic programming successive approximation method, use formula (5) simultaneously Storage capacity is tested, revises each stage end reservoir capacity, finally can obtain intake area minimum water deficit in the regular period, and Corresponding reservoir optimal water supply Xi, abandon water yield PSiWith pumping plant rate of water make-up process Yi(i=1,2 ..., N), to fully irrigating bar Single pumping plant-mono-water reservoir system place the Researching on Water Resources Optimal Management directly mending canal is used to provide foundation under part.
It is divided into N number of period by 1 year；According to reservoir initial water level, search water level-capacity curve, determine at the beginning of reservoir Beginning storage capacity V0；Specify with reference to reservoir operational management, determine and be available for water inventory SK, minimum capacity of a reservoir V yearmin, and flood control pair The storage capacity V answeredP；According to reservoir locality meteorological model data, calculate and determine that reservoir day part carrys out water yield LSi, evaporation and leakage EFi；According to pumping plant working system, determine that moisturizing pumping plant year allows water lift total amount BZ；According to intake area variety of crops, plantation rule The data such as mould, multiple crop index, calculates and determines day part intake area water demand of crop YSi(i=1,2 ... N).
Dynamic programming step-by-step process is as follows:
(1) study area being carried out investigation, collect the statistics daily output of reservoir and pumping plant moisturizing scale, this data should expire Foot formula (2)～(3) requirement, and reservoir stage storage capacity will not occur less than minimum capacity of a reservoir Vmin.This specific intake area is met with actual The fully reservoir stage output X of irrigation conditions1i(i=1,2 ... N) as primary iteration value, substituted into formula (1), the most former Model (1)～(5) are converted into mends water in a canal amount Y with each stage pumping plantiFor decision variable, front i stage pumping plant moisturizing total amount λiFor shape The one-dimensional dynamic programming model of state variable, uses one-dimensional dynamic programming to solve.
(2) with reference to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is:
1) stage i=1:
g1(λ1)=min (X11+Y11-YS1)2 (6)
This period reservoir yield X11Given by initial value, state variable λ1, it can be discrete in corresponding feasible zone: λ1 =0, W1,W2,…,BZ.To each discrete λ1, decision variable (pumping plant rate of water make-up Y11) can be discrete, such as 0 in corresponding feasible zone Ten thousand m3, 50,000 m3, 100,000 m3, 150,000 m3、…Y11,maxDeng (Y11,maxIt is the 1st stage pumping plant maximum moisturizing ability), should meet: Y11≥ λ1.Y by satisfied requirement11Substitute into formula (6) respectively, respectively obtain each discrete λ1During value, optimum Y11And the g of correspondence1(λ1)。
Then, according to formula (4), the 1st stage end reservoir capacity V1=V0+LS1-EF1-X11, the most not yet consider that reservoir is abandoned Water, uses formula (5) inspection, if exceeding the storage capacity V corresponding to flood controlP, then the water yield is abandoned beyond part as reservoir PS11, now V1 *=VP；Otherwise, without departing from, then PS11=0, now V1 *=V1。
2) stage i=2,3 ... N-1:
gi(λi)=min [(X1i+Y1i-YSi)2+gi-1(λi-1)] (7)
This period reservoir yield X1iGiven by initial value, state variable λiCarry out discrete the most respectively: λi=0, W1, W2,…,BZ.To each discrete λi, decision variable (pumping plant rate of water make-up Y1i) discrete ibid, and should meet:
State transition equation: λi-1=λi-Y1i (8)
In formula: i=2,3 ..., N-1.
By each discrete Y1iValue substitutes into the (X in formula (7) respectively1i+Y1i-YSi)2, by state transition equation formula (8), search The i-1 stage meetsThe g requiredi-1(λi-1) value, it is derived from meeting this λiThe optimum Y required1iProcess (i=1 ... And the g of correspondence i)i(λi).Equally, according to formula (4), the i-th period end reservoir capacity Vi=Vi-1+LSi-EFi-X1i, the most not yet Considering that water abandoned by reservoir, using formula (5) to test, if exceeding the storage capacity V corresponding to flood controlP, then make beyond part Water yield PS is abandoned for reservoir1i, now Vi *=VP；Otherwise, without departing from, then PS1i=0, now Vi *=Vi.Thus pushing over, it is right to obtain Process water PS abandoned by the reservoir answered1i(i=1 ... i).
3) stage N:
gN(λN)=min [(X1N+Y1N-YSN)2+gN-1(λN-1)] (9)
This period reservoir yield X1NGiven by initial value, state variable λN=BZ；Decision variable (pumping plant rate of water make-up Y1N) discrete in corresponding feasible zone equally, should meet: λN-1=λN-Y1N。
Use step 2) described method, final acquisition meets this λNPumping plant optimum moisturizing process Y required1i(i=1 ... And process water PS abandoned by the reservoir of correspondence N),1i(i=1 ... N).
(3) canal process water Y mended by pumping plant step (2) obtained1iAs initial set-point, substitute into formula (1), then master mould ～(5) are converted into each stage reservoir yield X (1)iFor decision variable, front i stage reservoir is for water inventory λi' become for state The one-dimensional dynamic programming model of amount, with reference to step (2), uses one-dimensional dynamic programming to solve, it is thus achieved that to meet this λN' require water Storehouse optimum water supply process X2i(i=1 ... N), and correspondence abandon process water PS2i(i=1 ... N).
(4) reservoir yield process X that step (3) is obtained2iAs initial set-point, substitute into formula (1), repeat step ～(3) (2), Approach by inchmeal solves repeatedly, until adjacent twice object function optimal value error precision is less than 1%, then model is excellent Change terminates.With last reservoir yield process X optimizing and obtainingmiCanal process water Y is mended with pumping plantmiAs master mould Excellent solution, the most also can obtain object function optimal value, and reservoir optimum abandons process water PSmi(i=1 ... N).Wherein, m is Dynamic programming Approach by inchmeal iterations is numbered.
More than inventing and solve conveniently, precision is reliable, is available under abundant irrigation conditions using the big of single pumping plant-mono-reservoir water supply Medium-sized irrigated areas administration unit popularization and application, reach the purpose that Water Resources Irrigation is distributed rationally, improve irrigated area, social and ecological Benefit.

Claims (3)

1. directly mend single pumping plant-mono-water reservoir system water resource optimal allocation method of canal under abundant irrigation conditions, by water lift Pumping plant directly supplements channel hydropenia, it is characterised in that specifically include following steps:
One, model construction, comprises the following steps:
1. with single seat annual-storage reservoir and the difference of the output sum of day part and intake area water requirement in single seat moisturizing pumping plant year The minimum target of quadratic sum, set up following object function:
F = min Z = min Σ i = 1 N ( X i + Y i - YS i ) 2 - - - ( 1 )
In formula: F be the difference of the confession water requirement of day part in object of study year least square and；When Z is each in object of study year Section is for the quadratic sum of the difference of water requirement；N is the year interior time hop count divided；Segment number when i is (i=1,2 ... N)；Xi、YiRespectively For reservoir, the output of the i-th period of pumping plant and rate of water make-up (ten thousand m3)；YSiWater requirement (ten thousand m for the i-th period of intake area3)；Target It is to accelerate to reduce the deviation between system water supply amount and intake area water requirement that function uses quadratic sum to express；
2. constraints is set, is available for water inventory constraints including reservoir, moisturizing pumping plant year, directly mends the reservoir in storehouse without pumping plant Water balance constraint and reservoir capacity constraints.
Two, model solution
First carry out data preparation, specifically include: be divided into N number of period by 1 year；According to reservoir initial water level, search water level-storehouse Hold relation curve, determine reservoir initial storage V0；Specify with reference to reservoir operational management, determine and be available for water inventory SK, minimum capacity of a reservoir year Vmin, and storage capacity V corresponding to flood controlP；According to reservoir locality meteorological model data, calculate and determine reservoir day part Carry out water yield LSi, evaporation with leakage EFi；According to pumping plant working system, determine that moisturizing pumping plant year allows water lift total amount BZ；According to being subject to The data such as pool variety of crops, planting scale, multiple crop index, calculate and determine day part intake area water demand of crop YSi；Its In, i=1,2 ..., N.
Next carries out dynamic programming Approach by inchmeal and solves.
Method the most according to claim 1, it is characterised in that the constraints of setting is specific as follows:
(1) reservoir, be available for water inventory constraint moisturizing pumping plant year: in the case of varying level year difference fraction, it is considered to need water to want Ask, the water yield that water supply project can be provided that.
X1+X2+…+XN≤SK (2)
Y1+Y2+…+YN≤BZ (3)
In formula: SK be reservoir be available for water inventory (ten thousand m year3)；BZ is moisturizing pumping plant year to allow water lift total amount (ten thousand m3)。
(2) the reservoir water yield Constraints of Equilibrium in storehouse is directly mended without pumping plant:
Vi=Vi-1+LSi-PSi-EFi-Xi, (i=1,2 ..., N) (4)
In formula: Vi、Vi-1It is respectively reservoir i-th and the reservoir storage of i-1 period Mo (ten thousand m3)；LSi、PSi、EFiFor the i-th period of reservoir Carry out the water yield (ten thousand m3), abandon the water yield (ten thousand m3), evaporation with leakage (ten thousand m3)。
(3) reservoir capacity constraint: day part end pondage should be between corresponding to reservoir minimum capacity of a reservoir and flood control Between storage capacity, it may be assumed that
Vmin≤Vi≤VP, (i=1,2 ..., N) (5)
In formula: Vmin、VPFor reservoir minimum capacity of a reservoir storage capacity corresponding with flood control (ten thousand m3)。
Method the most according to claim 2, it is characterised in that what dynamic programming Approach by inchmeal solved specifically comprises the following steps that
(1) study area being carried out investigation, collect the statistics daily output of reservoir and pumping plant moisturizing scale, this data should meet formula ～(3) requirement, and reservoir stage storage capacity will not occur less than minimum capacity of a reservoir V (2)min.To meet this specific intake area abundant with actual The reservoir stage output X of irrigation conditions1iAs primary iteration value, substituted into formula (1), then master mould (1)～(5) are converted into Water in a canal amount Y is mended with each stage pumping plantiFor decision variable, front i stage pumping plant moisturizing total amount λiOne-dimensional dynamic rule for state variable Draw model, use one-dimensional dynamic programming to solve；Wherein, i=1,2 ... N.
(2) with reference to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is:
1) stage i=1:
g1(λ1)=min (X11+Y11-YS1)2 (6)
This period reservoir yield X11Given by initial value, state variable λ1, it can be discrete in corresponding feasible zone: λ1=0, W1,W2,…,BZ.To each discrete λ1, decision variable Y11In corresponding feasible zone discrete, should meet: Y11≥λ1.To meet and want The Y asked11Substitute into formula (6) respectively, respectively obtain each discrete λ1During value, optimum Y11And the g of correspondence1(λ1)。
Then, according to formula (4), the 1st stage end reservoir capacity V1=V0+LS1-EF1-X11, the most not yet consider that water abandoned by reservoir, adopts Check by formula (5), if exceeding the storage capacity V corresponding to flood controlp, then water yield PS is abandoned beyond part as reservoir11, now V1 *=VP；Otherwise, without departing from, then PS11=0, now V1 *=V1。
2) stage i=2,3 ... N-1:
gi(λi)=min [(X1i+Y1i-YSi)2+gi-1(λi-1)] (7)
This period reservoir yield X1iGiven by initial value, state variable λiCarry out discrete the most respectively: λi=0, W1, W2,…,BZ.To each discrete λi, decision variable Y1iDiscrete ibid, and should meet:
State transition equation: λi-1=λi-Y1i (8)
In formula: i=2,3 ..., N-1.
By each discrete Y1iValue substitutes into the (X in formula (7) respectively1i+Y1i-YSi)2, by state transition equation formula (8), search i-1 rank Section meetsThe g requiredi-1(λi-1) value, it is derived from meeting this λiThe optimum Y required1iProcess and the g of correspondence thereofi (λi).Equally, according to formula (4), the i-th period end reservoir capacity Vi=Vi-1+LSi-EFi-X1i, the most not yet consider that water abandoned by reservoir, Employing formula (5) is tested, if exceeding the storage capacity V corresponding to flood controlp, then the water yield is abandoned beyond part as reservoir PS1i, now Vi *=VP；Otherwise, without departing from, then PS1i=0, now Vi *=Vi.Thus pushing over, water abandoned by the reservoir that can obtain correspondence Amount process PS1i.Wherein, i=1 ... i.
3) stage N:
gN(λN)=min [(X1N+Y1N-YSN)2+gN-1(λN-1)] (9)
This period reservoir yield X1NGiven by initial value, state variable λN=BZ；Decision variable Y1NFeasible in correspondence equally In territory discrete, should meet: λN-1=λN-Y1N。
Use step 2) described method, final acquisition meets this λNPumping plant optimum moisturizing process Y required1i, and the water of correspondence Process water PS is abandoned in storehouse1i, wherein, i=1 ... N.
(3) canal process water Y mended by pumping plant step (2) obtained1iAs initial set-point, substitute into formula (1), then master mould (1) ～(5) are converted into each stage reservoir yield XiFor decision variable, front i stage reservoir is for water inventory λi' for state variable One-dimensional dynamic programming model, with reference to step (2), uses one-dimensional dynamic programming to solve, it is thus achieved that to meet this λN' the reservoir that requires Excellent water supply process X2i(i=1 ... N), and correspondence abandon process water PS2i, wherein, i=1 ... N.
(4) reservoir yield process X that step (3) is obtained2iAs initial set-point, substitute into formula (1), repeat step (2)～ (3), Approach by inchmeal solves repeatedly, until adjacent twice object function optimal value error precision is less than 1%, then and model optimization knot Bundle.With last reservoir yield process X optimizing and obtainingmiCanal process water Y is mended with pumping plantmiAs master mould optimal solution, The most also can obtain object function optimal value, and reservoir optimum abandons process water PSmi, wherein, i=1 ... N, m are dynamically to advise Draw Approach by inchmeal iterations numbering.
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CN107742166A (en) * 2017-10-19 2018-02-27 扬州大学 More storehouse multiple station systems water resource optimal allocation methods in storehouse are directly mended under a kind of fully irrigation conditions
CN107748930A (en) * 2017-10-19 2018-03-02 扬州大学 Single storehouse multiple station systems water resource optimal allocation method of canal is directly mended under a kind of fully irrigation conditions
CN107798471A (en) * 2017-10-19 2018-03-13 扬州大学 More storehouse multiple station systems water resource optimal allocation methods of canal are directly mended under a kind of fully irrigation conditions
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