CN106295893A - Fully directly mend single pumping plant list water reservoir system water resource optimal allocation method in storehouse under irrigation conditions - Google Patents

Abstract

The present invention relates to the single moisturizing pumping plant directly mending storehouse under abundant irrigation conditions and single reservoir cooperation dispatching method, the minimum target of quadratic sum with the output of day part in single seat annual-storage reservoir year Yu the difference of intake area water requirement, the day part reservoir yield divided in year is decision variable, it is available for water inventory with reservoir moisturizing pumping plant year, reservoir operation criterion, reservoir water balance, reservoir minimum capacity of a reservoir, flood control correspondence storage capacity etc. are constraints, set up the single pumping plant list water reservoir system water resources optimal operation model directly mending storehouse under abundant irrigation conditions, one-dimensional dynamic programming method is used to solve, intake area minimum water deficit in certain delivery period can be obtained, and the reservoir optimal water supply of correspondence, abandon the water yield and pumping plant rate of water make-up process.The present invention can realize directly mending under abundant irrigation conditions single pumping plant list water reservoir system water resources optimal operation in storehouse, and to improving, water in a canal resource efficiency mended by pumping plant, Irrigating Guarantee Rate of Irrigated Area has important practical significance.

Description

Fully directly mend single pumping plant-mono-water reservoir system water resource optimization in storehouse 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 in benefit storehouse 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 in water project management.

Summary of the invention

The present invention is directed to the single pumping plant directly mending storehouse under abundant irrigation conditions and single water reservoir system, it is considered to different water frequencies Under reservoir basin hydropenia situation, consider first to be supplemented reservoir hydropenia by lift pumping station, set up annual-storage reservoir-pumping plant combined optimization Scheduling mathematic model.

Directly mend the single pumping plant-mono-reservoir cooperation dispatching patcher of storehouse year regulation for specific, supply water at known reservoir Divide time hop count, initial storage, minimum capacity of a reservoir, utilizable capacity, storage capacity that flood control is corresponding, be available for water inventory, each year Period process water, evaporation and leakage process, allow water lift total amount, and day part intake area water requirement moisturizing pumping plant year Under process condition, use one-dimensional dynamic programming method to solve, intake area minimum water deficit in certain delivery period can be obtained, and right The reservoir optimal water supply answered, 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 in storehouse is directly mended, by mending under a kind of abundant irrigation conditions Pump works directly supplies water to reservoir, reservoir, pumping plant combine and supply water to intake area, comprise the following steps:

One, model construction, comprises the following steps 1～step 2:

1. minimum with the quadratic sum of the difference of intake area water requirement with the output of day part in single seat annual-storage reservoir year Target, sets up following object function:

$F=\min Z=\min \underset{i=1}{\overset{N}{\mathrm{\Σ}}}{(X_i-{\mathrm{YS}}_i)}^2---\left(1\right)$

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 The quadratic sum of the difference of the confession water requirement of day part；N is the year interior time hop count divided；Segment number when i is (i=1,2 ... N)；X_i Output (ten thousand m for the i-th period of reservoir³)；YS_iWater requirement (ten thousand m for the i-th period of intake area³)；Object function employing square With expression is to accelerate to reduce the deviation between reservoir yield and intake area water requirement.

2. constraints is set

Including being available for water inventory constraints reservoir-pumping plant year, reservoir operation criterion constraints and reservoir capacity constraint Condition.

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 V_min, storage capacity V that flood control is corresponding_P, and utilizable capacity V_min+Δ₁；According to reservoir locality meteorological model data, Calculating determines that reservoir day part carrys out water yield LS_i, evaporation with leakage EF_i；According to pumping plant working system, determine that moisturizing pumping plant year permits Permitted water lift total amount BZ；According to data such as intake area variety of crops, planting scale, multiple crop indexes, calculate and determine that day part is by water Water demand of crop YS in district_i；Wherein, i=1,2 ..., N；

Secondly one-dimensional dynamic programming method is used to solve.

Further, described constraints includes:

(1) water inventory constraint it is available for reservoir-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+…+X_N≤SK+BZ (2)

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) reservoir operation criterion constraint: according to reservoir-pumping station system water yield equation:

V_i=V_i-1+LS_i+Y_i-PS_i-EF_i-X_i, (i=1,2, N) (3)

1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir V_minTime, then the i-th period should be fed water by lift pumping station Storehouse moisturizing, moisturizing to utilizable capacity (Δ more than reservoir minimum capacity of a reservoir₁), it may be assumed that

V_i＜ V_minTime: Y_i=V_min-V_i+Δ₁；Water PS abandoned by this period reservoir_i=0 (4)

2. when meeting with flood, the i-th period end pondage is more than pondage V corresponding to flood control_PTime, Then the i-th period reply reservoir carries out abandoning water, abandons water to flood control by reservoir regulation limiting water level, it may be assumed that

V_i＞ V_PTime: PS_i=V_i-V_P；This period pumping plant moisturizing Y_i=0 (5)

3. when the i-th period end pondage is between minimum capacity of a reservoir V_minWith flood control corresponding to pondage V_P Between, then the i-th period reservoir need not abandon water, and pumping plant is also without moisturizing, it may be assumed that

V_min≤V_i≤V_PTime: Y_i=PS_i=0； (6)

In formula: V_i、V_i-1It is respectively reservoir i-th, the reservoir storage of i-1 period Mo (ten thousand m³)；Y_iIt it is the pumping plant benefit storehouse of the i-th period The water yield (ten thousand m³)；LS_i、PS_i、EF_iThe water yield (ten thousand m are carried out for the i-th period of reservoir³), abandon the water yield (ten thousand m³), reservoir evaporation and leakage (ten thousand m³)；V_nim、V_PFor storage capacity (ten thousand m corresponding to the minimum capacity of a reservoir of reservoir and flood control³)。

(3) reservoir capacity constraint: the pondage of day part should be corresponding with flood control between reservoir minimum capacity of a reservoir Between storage capacity, it may be assumed that

V_min≤V_i≤V_P, (i=1,2, N) (7).

Further, with reference to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is:

(1) stage i=1:

g₁(λ₁)=min (X₁-YS₁)² (8)

State variable λ₁, it can be discrete in corresponding feasible zone: λ₁=0, W₁,W₂,…,SK+BZ.To each discrete λ₁, Decision variable (reservoir yield X₁) can be discrete, such as 00,000 m in corresponding feasible zone³, 200,000 m³, 400,000 m³, 600,000 m³、…X_1,max Deng (X_1,maxIt is the 1st stage reservoir maximum water supply capacity), should meet: X₁≥λ₁.X by satisfied requirement₁Substitute into formula (8) respectively, point Do not obtain each discrete λ₁During value, optimum X₁And the g of correspondence₁(λ₁)。

Then, according to formula (3), the 1st stage end reservoir capacity V₁=V₀+LS₁-EF₁-X₁, the most not yet consider pumping plant moisturizing Or reservoir abandons water, formula (4)～(6) should be used to test and revise:

1. V is worked as₁＜ V_min, then consider reservoir is carried out moisturizing, Y₁=V_min-V₁+Δ₁, now revise storage capacity V₁ ^*=V_min+ Δ₁。

2. V is worked as₁＞ V_P, then the dispatching requirement abandoning water to ensure reservoir capacity, PS are needed₁=V₁-V_P, now revise storage capacity V₁ ^*=V_P。

3. V is worked as_min≤V₁≤V_P, then Y₁=PS₁=0, now V₁ ^*=V₁。

By step 1.～3., revise and determine the 1st stage end reservoir capacity V₁ ^*, water abandoned by the reservoir that simultaneously can obtain correspondence Amount PS₁, or pumping plant rate of water make-up Y₁。

(2) stage i=2,3 ... N-1:

g_i(λ_i)=min [(X_i-YS_i)²+g_i-1(λ_i-1)] (9)

State variable λ_i, carry out discrete the most respectively: λ_i=0, W₁,W₂,…,SK+BZ.To each discrete λ_i, decision-making becomes Amount (reservoir yield X_i) discrete ibid, and should meet:

State transition equation: λ_i-1=λ_i-X_i (10)

In formula: i=2,3 ..., N-1

By each discrete X_iValue substitutes into the (X in formula (9) respectively_i-YS_i)², by state transition equation formula (10), search i-1 Stage meetsThe g required_i-1(λ_i-1) value, it is derived from meeting this λ_iThe optimum X required_iProcess (i=1 ... i) and The g of its correspondence_i(λ_i)。

Equally, also need to utilize formula (3) to determine the i-th stage end storage capacity V_i, then use formula (4)～(6) to test, revise Determine the i-th period end storage capacity V_i ^*, process water PS abandoned by the reservoir that simultaneously can obtain correspondence_i, and pumping plant rate of water make-up process Y_i(i= 1,2 ... the most 1.～3. i), course synchronization.

(3) stage N:

$g_N\left({\mathrm{\λ}}_N\right)=\min \[{(X_N-{\mathrm{YS}}_N)}^2+g_{N-1}\left({\mathrm{\λ}}_{N-1}\right)\]---\left(11\right)$

State variable λ_N=SK+BZ；Decision variable (reservoir yield X_N) discrete in corresponding feasible zone equally, should meet: λ_N-1=λ_N-X_N。

Step (2) described method, final acquisition is used to meet this λ_NThe reservoir optimum water supply process X required_i(i=1 ... N), process water PS abandoned by corresponding reservoir_i, pumping plant rate of water make-up process Y_i(i=1,2 ... N), and master mould object function is Figure of merit F=g_N(λ_N)。

The present 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 storehouse list pumping plant-mono-reservoir water resource.

Detailed description of the invention

The minimum mesh of quadratic sum with the output of day part in single seat annual-storage reservoir year Yu the difference of intake area water requirement Mark, the day part reservoir yield divided in year is decision variable, and to be available for water inventory system year, (i.e. reservoir year is available for water inventory Draw water lift total amount sum with pumping plant year), reservoir operation criterion, reservoir water balance, reservoir minimum capacity of a reservoir, flood control corresponding Storage capacity etc. are constraints, set up the single pumping plant-mono-water reservoir system water resources optimal operation directly mending storehouse under abundant irrigation conditions Model.

One, model construction

1. object function

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 The quadratic sum of the difference of the confession water requirement of day part；N is the year interior time hop count divided；Segment number when i is (i=1,2 ... N)；X_i Output (ten thousand m for the i-th period of reservoir³)；YS_iWater requirement (ten thousand m for the i-th period of intake area³)；Object function employing square With expression is to accelerate to reduce the deviation between reservoir yield and intake area water requirement.

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+…+X_N≤SK+BZ (2)

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) reservoir operation criterion constraint: according to reservoir-pumping station system water yield equation:

V_i=V_i-1+LS_i+Y_i-PS_i-EF_i-X_i, (i=1,2, N) (3)

1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir V_minTime, then the i-th period should be fed water by lift pumping station Storehouse moisturizing, moisturizing to utilizable capacity (Δ more than reservoir minimum capacity of a reservoir₁), it may be assumed that

V_i＜ V_minTime: Y_i=V_min-V_i+Δ₁；Water PS abandoned by this period reservoir_i=0 (4)

2. when meeting with flood, the i-th period end pondage is more than pondage V corresponding to flood control_PTime, Then the i-th period reply reservoir carries out abandoning water, abandons water to flood control by reservoir regulation limiting water level, it may be assumed that

V_i＞ V_PTime: PS_i=V_i-V_P；This period pumping plant moisturizing Y_i=0 (5)

3. when the i-th period end pondage is between minimum capacity of a reservoir V_minWith flood control corresponding to pondage V_P Between, then the i-th period reservoir need not abandon water, and pumping plant is also without moisturizing, it may be assumed that

V_min≤V_i≤V_pTime: Y_i=PS_i=0； (6)

In formula: V_i、V_i-1It is respectively reservoir i-th, the reservoir storage of i-1 period Mo (ten thousand m³)；Y_iIt it is the pumping plant benefit storehouse of the i-th period The water yield (ten thousand m³)；LS_i、PS_i、EF_iThe water yield (ten thousand m are carried out for the i-th period of reservoir³), abandon the water yield (ten thousand m³), reservoir evaporation and leakage (ten thousand m³)；V_nim、V_PFor storage capacity (ten thousand m corresponding to the minimum capacity of a reservoir of reservoir and flood control³)。

(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

V_min≤V_i≤V_P, (i=1,2, N) (7)

Two, 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) pumping plant year.

(2) constraints considers the constraint of reservoir operation criterion, it is possible to achieve lift pumping station " idle mends storehouse, busy supplies water " Reservoir-pumping station system water resources optimal operation mode.If idle pondage is less than reservoir minimum capacity of a reservoir, then consider reservoir Carry out moisturizing, benefit storehouse water yield Y of lift pumping station_iFor the difference of reservoir minimum capacity of a reservoir Yu i period end pondage, add minimum capacity of a reservoir Above Δ₁, i.e. Y_i=V_min-V_i+Δ₁；If reservoir capacity exceedes reservoir storage corresponding to flood control by reservoir regulation limiting water level, then need to abandon Water ensures the dispatching requirement of reservoir capacity, and water yield PS abandoned by reservoir_iFor i period end pondage and flood control by reservoir regulation limiting water level The difference of corresponding reservoir storage, i.e. PS_i=V_i-V_P.Busy according to pondage situation, then considers to be combined to being subject to by reservoir, pumping plant Supply water in pool, to meeting the water demand of water user as far as possible.

Three, model solution

It is the nonlinear mathematical model that can divide in a stage for model above Chinese style (1)～(7), each stage in object function Intake area water requirement YS_iFor it is known that therefore can with reservoir delivery period divide period i (i=1,2 ..., N) be stage variable, Day part reservoir available water X_iFor decision variable, front i stage reservoir is state variable λ for water inventory_i, use one-dimensional dynamically Planing method solves.

Assuming that the initial storage V of annual-storage reservoir₀It is known that model Chinese style (2) is dynamic programming coupling constraint, formula (3) is Reservoir operation day part water balance criterion；By to each stage reservoir yield X_i(i=1,2 ... N) carry out discrete, adopt Solve with one-dimensional dynamic programming, utilize formula (4)～(7) that storage capacity is tested simultaneously, and then each stage initial storage is entered Row is revised, and finally determines each stage reservoir optimal water supply process X_i, abandon process water PS_i, and pumping plant rate of water make-up process Y_i(i =1,2 ... N).

Model (1)～(7) are though being single decision variable Optimized model, but use one-dimensional dynamic programming method to solve, and not only may be used Obtain optimal decision variate-value each stage reservoir yield X in object function_i, and by using reservoir operation criterion To each stage reservoir capacity examination and correction, the reservoir optimum that also can obtain correspondence abandons water yield PS_iWith pumping plant rate of water make-up process Y_i, real Solve on border and obtain three decision variable values, enrich such complex nonlinear model solution method.

Meanwhile, by solving above, the minimum water deficit in certain delivery period can be obtained, and the reservoir optimum of correspondence supplies The water yield, abandon the water yield and pumping plant rate of water make-up process, to the single pumping plant-mono-water reservoir system institute using directly benefit storehouse under abundant irrigation conditions Foundation is provided at Researching on Water Resources Optimal Management.

First, carry out data preparation, specifically include:

Single pumping plant-mono-reservoir cooperation the scheduling that regulates in year for specific direct benefit storehouse a certain under abundant irrigation conditions is System, was divided into N number of period by 1 year；According to reservoir initial water level, search water level-capacity 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 V year_min, storage capacity V that flood control is corresponding_P、 And utilizable capacity V_min+Δ₁；According to reservoir locality meteorological model data, calculate and determine that reservoir day part carrys out water yield LS_i, evaporation With leakage EF_i；According to pumping plant working system, determine that moisturizing pumping plant year allows water lift total amount BZ；According to intake area farming article The data such as kind, planting scale, multiple crop index, calculate water demand of crop YS determining day part intake area_i(i=1,2 ... N).

With reference to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is:

(1) stage i=1:

g₁(λ₁)=min (X₁-YS₁)² (8)

State variable λ₁, it can be discrete in corresponding feasible zone: λ₁=0, W₁,W₂,…,SK+BZ.To each discrete λ₁, Decision variable (reservoir yield X₁) can be discrete, such as 00,000 m in corresponding feasible zone³, 200,000 m³, 400,000 m³, 600,000 m³、…X_1,max Deng (X_1,maxIt is the 1st stage reservoir maximum water supply capacity), should meet: X₁≥λ₁.X by satisfied requirement₁Substitute into formula (8) respectively, point Do not obtain each discrete λ₁During value, optimum X₁And the g of correspondence₁(λ₁)。

Then, according to formula (3), the 1st stage end reservoir capacity V₁=V₀+LS₁-EF₁-X₁, the most not yet consider pumping plant moisturizing Or reservoir abandons water, formula (4)～(6) should be used to test and revise:

1. V is worked as₁＜ V_min, then consider reservoir is carried out moisturizing, Y₁=V_min-V₁+Δ₁, now revise storage capacity V₁ ^*=V_min+ Δ₁。

2. V is worked as₁＞ V_P, then the dispatching requirement abandoning water to ensure reservoir capacity, PS are needed₁=V₁-V_P, now revise storage capacity V₁ ^*=V_P。

3. V is worked as_min≤V₁≤V_P, then Y₁=PS₁=0, now V₁ ^*=V₁。

By step 1.～3., revise and determine the 1st stage end reservoir capacity V₁ ^*, water abandoned by the reservoir that simultaneously can obtain correspondence Amount PS₁, or pumping plant rate of water make-up Y₁。

(2) stage i=2,3 ... N-1:

g_i(λ_i)=min [(X_i-YS_i)²+g_i-1(λ_i-1)] (9)

State variable λ_i, carry out discrete the most respectively: λ_i=0, W₁,W₂,…,SK+BZ.To each discrete λ_i, decision-making becomes Amount (reservoir yield X_i) discrete ibid, and should meet:

State transition equation: λ_i-1=λ_i-X_i (10)

In formula: i=2,3 ..., N-1

By each discrete X_iValue substitutes into the (X in formula (9) respectively_i-YS_i)², by state transition equation formula (10), search i-1 Stage meetsThe g required_i-1(λ_i-1) value, it is derived from meeting this λ_iThe optimum X required_iProcess (i=1 ... i) and The g of its correspondence_i(λ_i)。

Equally, also need to utilize formula (3) to determine the i-th stage end storage capacity V_i, then use formula (4)～(6) to test, revise Determine the i-th period end storage capacity V_i ^*, process water PS abandoned by the reservoir that simultaneously can obtain correspondence_i, and pumping plant rate of water make-up process Y_i(i= 1,2 ... the most 1.～3. i), course synchronization.

(3) stage N:

g_N(λ_N)=min [(X_N-YS_N)²+g_N-1(λ_N-1)] (11)

State variable λ_N=SK+BZ；Decision variable (reservoir yield X_N) discrete in corresponding feasible zone equally, should meet: λ_N-1=λ_N-X_N。

Step (2) described method, final acquisition is used to meet this λ_NThe reservoir optimum water supply process X required_i(i=1 ... N), process water PS abandoned by corresponding reservoir_i, pumping plant rate of water make-up process Y_i(i=1,2 ... N), and master mould object function is Figure of merit F=g_N(λ_N)。

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.

Claims (3)

1. directly mend single pumping plant-mono-water reservoir system water resource optimal allocation method in storehouse under abundant irrigation conditions, by moisturizing Pumping plant directly supplies water to reservoir, reservoir, pumping plant combine and supply water to intake area, it is characterised in that specifically include following steps:

One, model construction, comprises the following steps:

1. with the minimum target of quadratic sum of the output of day part in single seat annual-storage reservoir year Yu the difference of intake area water requirement, Set up following object function:

$F=\min Z=\min \underset{i=1}{\overset{N}{\Σ}}{(X_i-{\mathrm{YS}}_i)}^2---\left(1\right)$

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 The quadratic sum of the difference of the confession water requirement of section；N is the year interior time hop count divided；Segment number when i is, i=1,2 ... N；X_iFor water Output (ten thousand m of the i-th period of storehouse³)；YS_iWater requirement (ten thousand m for the i-th period of intake area³)。

2. constraints is set, including being available for water inventory constraints, reservoir operation criterion constraints and water reservoir-pumping plant year Kuku holds 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 V_min, storage capacity V that flood control is corresponding_P, and utilizable capacity V_min+Δ₁；According to reservoir locality meteorological model data, calculate Determine that reservoir day part carrys out water yield LS_i, evaporation with leakage EF_i；According to pumping plant working system, determine that moisturizing pumping plant year allows to carry Water inventory BZ；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 YS_i；Wherein, i=1,2 ..., N；

Secondly one-dimensional dynamic programming method is used to solve.

Method the most according to claim 1, it is characterised in that the constraints of setting is specific as follows:

(1) it is available for water inventory constraint reservoir-pumping plant year: in the case of varying level year difference fraction, it is considered to need water requirement, supplies The water yield that Hydraulic Projects can be provided that.

X₁+X₂+…+X_N≤SK+BZ (2)

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) reservoir operation criterion constraint: according to reservoir-pumping station system water yield equation:

V_i=V_i-1+LS_i+Y_i-PS_i-EF_i-X_i, i=1,2, N (3)

1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir V_minTime, then the i-th period should be mended to reservoir by lift pumping station Water, moisturizing is to utilizable capacity, it may be assumed that

V_i＜ V_minTime: Y_i=V_min-V_i+Δ₁；Water PS abandoned by this period reservoir_i=0 (4)

2. when meeting with flood, the i-th period end pondage is more than pondage V corresponding to flood control_PTime, then i-th Period reply reservoir carries out abandoning water, abandons water to flood control by reservoir regulation limiting water level, it may be assumed that

V_i＞ V_PTime: PS_i=V_i-V_P；This period pumping plant moisturizing Y_i=0 (5)

3. when the i-th period end pondage is between minimum capacity of a reservoir V_minWith flood control corresponding to pondage V_PBetween, Then the i-th period reservoir need not abandon water, and pumping plant is also without moisturizing, it may be assumed that

V_min≤V_i≤V_pTime: Y_i=PS_i=0； (6)

In formula: V_i、V_i-1It is respectively reservoir i-th, the reservoir storage of i-1 period Mo (ten thousand m³)；Y_iIt it is the pumping plant benefit storehouse water yield of the i-th period (ten thousand m³)；LS_i、PS_i、EF_iThe water yield (ten thousand m are carried out for the i-th period of reservoir³), abandon the water yield (ten thousand m³), reservoir evaporation and leakage (ten thousand m³)；V_min、V_PFor storage capacity (ten thousand m corresponding to the minimum capacity of a reservoir of reservoir and flood control³)。

(4) reservoir capacity constraint: the pondage of day part should be between reservoir minimum capacity of a reservoir storage capacity corresponding with flood control Between, it may be assumed that

V_min≤V_i≤V_P, i=1,2, N (7).

Method the most according to claim 2, it is characterised in that with reference to one-dimensional dynamic programming evaluation principle, obtain corresponding recursion Equation is:

(1) stage i=1:

g₁(λ₁)=min (X₁-YS₁)² (8)

State variable λ₁, it can be discrete in corresponding feasible zone: λ₁=0, W₁,W₂,…,SK+BZ.To each discrete λ₁, decision-making Variable X₁In corresponding feasible zone discrete, should meet: X₁≥λ₁；X by satisfied requirement₁Substitute into formula (8) respectively, respectively obtain every Individual discrete λ₁During value, optimum X₁And the g of correspondence₁(λ₁)。

Then, according to formula (3), the 1st stage end reservoir capacity V₁=V₀+LS₁-EF₁-X₁, the most not yet consider pumping plant moisturizing or water Water is abandoned in storehouse, and formula (4)～(6) should be used to test and revise:

1. V is worked as₁＜ V_min, then consider reservoir is carried out moisturizing, Y₁=V_min-V₁+Δ₁, now revise storage capacity V₁ ^*=V_min+Δ₁。

2. V is worked as₁＞ V_P, then the dispatching requirement abandoning water to ensure reservoir capacity, PS are needed₁=V₁-V_P, now revise storage capacity V₁ ^*= V_P。

3. V is worked as_min≤V₁≤V_P, then Y₁=PS₁=0, now V₁ ^*=V₁。

By step 1.～3., revise and determine the 1st stage end reservoir capacity V₁ ^*, the water yield abandoned by the reservoir that simultaneously can obtain correspondence PS₁, or pumping plant rate of water make-up Y₁。

(2) stage i=2,3 ... N-1:

g_i(λ_i)=min [(X_i-YS_i)²+g_i-1(λ_i-1)] (9)

State variable λ_i, carry out discrete the most respectively: λ_i=0, W₁,W₂,…,SK+BZ.To each discrete λ_i, decision variable X_i Discrete ibid, and should meet:

State transition equation: λ_i-1=λ_i-X_iI=2,3 ..., N-1 (10)

By each discrete X_iValue substitutes into the (X in formula (9) respectively_i-YS_i)², by state transition equation formula (10), search the i-1 stage full FootThe g required_i-1(λ_i-1) value, it is derived from meeting this λ_iThe optimum X required_iProcess (i=1 ... i) and corresponding G_i(λ_i)。

Equally, also need to utilize formula (3) to determine the i-th stage end storage capacity V_i, then use formula (4)～(6) to test, correction determines I-th period end storage capacity V_i ^*, process water PS abandoned by the reservoir that simultaneously can obtain correspondence_i, and pumping plant rate of water make-up process Y_i, wherein i= 1,2 ... the most 1.～3. i, course synchronization.

(3) stage N:

g_i(λ_i)=min [(X_i-YS_i)²+g_i-1(λ_i-1)] (11)

State variable λ_N=SK+BZ；Decision variable (reservoir yield X_N) discrete in corresponding feasible zone equally, should meet: λ_N-1 =λ_N-X_N。

Step (2) described method, final acquisition is used to meet this λ_NThe reservoir optimum water supply process X required_i, corresponding reservoir is abandoned Process water PS_i, pumping plant rate of water make-up process Y_i, and master mould object function optimal value F=g_N(λ_N), wherein i=1,2 ... N。