CN106228276A - Single pumping plant list water reservoir system water resource optimal allocation method of canal is directly mended under the conditions of insufficient irrigation - Google Patents

Abstract

The single moisturizing pumping plant of canal and single reservoir cooperation dispatching method is directly mended under the conditions of the present invention relates to insufficient irrigation, it is object function to the maximum with intake area crop annual production, crop each growing stage reservoir, pumping plant output is decision variable, it is available for water inventory with reservoir year, moisturizing pumping plant stage available water, without pumping plant, the reservoir water balance of canal is directly mended by reservoir, reservoir capacity and moisturizing pumping plant year permission water lift total amount etc. are constraints, set up the single pumping plant list water reservoir system water resources optimal operation model directly mending canal, dynamic programming successive approximation method is used to solve, crop maximum annual production under the conditions of insufficient irrigation can be obtained, and each stage reservoir optimal water supply of correspondence, abandon the water yield and pumping plant rate of water make-up process.Single pumping plant list water reservoir system water resources optimal operation of canal is directly mended in this invention under the conditions of can realizing insufficient irrigation, to improving, water in a canal resource efficiency mended by pumping plant, Irrigating Guarantee Rate of Irrigated Area has important practical significance.

Description

Single pumping plant-mono-water reservoir system the water resource directly mending canal under the conditions of insufficient irrigation is excellent Change collocation method

Technical field

Under the conditions of the present invention relates to insufficient irrigation, single moisturizing pumping plant and the method for single reservoir cooperation scheduling, belong to Technical field is distributed rationally in Water Resources Irrigation.

Background technology

Currently uneven due to water resource spatial and temporal distributions, restrict the socio-economic development in many areas, for cimited irrigation Irrigate area, in the case of being available for gross amount of water resources deficiency year, to be target to the maximum by water comprehensive benefit, reinforced region water resource United Dispatching and management, use the hydraulic engineering of irrigation system and regulate as unified entirety, using built engineering (such as water Storehouse and pumping plant combined dispatching run) make it play bigger effect, it is the main path solving irrigated area water shortage problem.Directly mend canal Single pumping plant-mono-water reservoir system traffic control be a content of water resources management, the most reasonably use system water resource to adjust Degree, makes certain goal systems benefit optimal, is problem relatively conventional during water project management uses.

Summary of the invention

The present invention is directed to directly mend under insufficient irrigation single pumping plant-mono-water reservoir system of canal, it is considered under different water frequencies Reservoir basin hydropenia situation, by moisturizing pumping plant and reservoir Joint Replenishment channel hydropenia, set up annual-storage reservoir-pumping plant combined optimization Scheduling mathematic model.Single pumping plant-mono-reservoir cooperation the dispatching patcher of canal year regulation is directly mended, known crop for specific The time of infertility number of stages, reservoir initial storage, minimum capacity of a reservoir, storage capacity that flood control is corresponding, be available for water inventory, each stage year Coming the water yield, evaporation and leakage, moisturizing pumping plant year allows water lift total amount, each stage available water, and intake area crops are filled Divide under each stage water requirement under irrigation conditions and maximum these constraints of annual production, use dynamic programming successive approximation method to ask Solve, can obtain crop maximum annual production under the conditions of insufficient irrigation, and each stage reservoir optimal water supply of correspondence, abandon the water yield With pumping plant rate of water make-up.

The present invention program is as follows:

Single pumping plant-mono-water reservoir system water resource optimal allocation method of canal is directly mended under the conditions of a kind of insufficient irrigation, by Moisturizing pumping plant and reservoir are combined to canal water supply, channel supplement intake area hydropenia, comprise the following steps:

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

1. it is target to the maximum with intake area crop annual production, sets up following object function:

$F=\max Y=\max Y_m\·\underset{i=1}{\overset{N}{\Σ}}\[\frac{(X_i+Y_i)\·K_i}{{\mathrm{YS}}_i}\]---\left(1\right)$

In formula: F is intake area crop maximum annual production (kg)；Y is the actual annual production of intake area crop (kg)；In N is year The crop growth number of stages divided；I is stage numbering, i=1,2 ..., N；Y_mFor the biggest year of the crop under abundant irrigation conditions Yield (kg), K_iFor the i-th stage of crop hydropenia water sensitive coefficient to yield effect；X_i、Y_iIt is respectively insufficient irrigation condition Reservoir yield (ten thousand m in lower i-th stage³), pumping plant rate of water make-up (ten thousand m³), it is called for short reservoir yield X_i, pumping plant rate of water make-up Y_i, YS_i For the water demand of crop (ten thousand m in the i-th stage under abundant irrigation conditions³)。

2. constraints is set

Including being available for water inventory constraints reservoir year, moisturizing pumping plant year allows water lift total amount constraints, moisturizing pumping plant Stage available water constraints, the reservoir water yield equilibrium constraint without pumping plant, directly being mended canal by reservoir and reservoir capacity are about Bundle condition.

Two, model solution

First carry out data preparation, specifically include: crop is divided into N number of stage in time of infertility；According to reservoir initial water Position, searches water level-capacity curve, determines reservoir initial storage V₀；According to the practical operation situation of reservoir, determine that year is available for Water inventory SK, minimum capacity of a reservoir V_min, storage capacity V that flood control is corresponding_P；According to reservoir locality meteorological model data, determine reservoir Each stage water yield LS_i, evaporation with leakage EF_i；According to pump performance characteristic and pumping plant real work situation, determine small pump Stand and year allow water lift total amount BZ, and combine water sources water level year-end drawdown level, determine each stage available water JD_i；According to by water District's variety of crops, planting scale, multiple crop index, determine intake area each stage water requirement YS_i, under abundant irrigation conditions, crop is Big annual production Y_m；Wherein, i=1,2 ..., N.

Next carries out dynamic programming Approach by inchmeal and solves.

Further, described constraints includes:

(1) it is available for water inventory reservoir year to retrain:

X₁+X₂+…+X_N≤SK (2)

In formula: SK be reservoir be available for water inventory (ten thousand m year³)；

(2) constraint of water lift total amount is allowed moisturizing pumping plant year:

Y₁+Y₂+…+Y_N≤BZ (3)

In formula: BZ is moisturizing pumping plant year to allow water lift total amount (ten thousand m³)。

(3) moisturizing pumping plant stage available water constraint:

Y_i≤JD_i (4)

In formula: JD_iIt is the i-th stage pumping plant available water (ten thousand m³)。

(4) without pumping plant, the reservoir water yield Constraints of Equilibrium of directly being mended canal by reservoir:

V_i=V_i-1+LS_i-PS_i-EF_i-X_i (5)

In formula: V_i、V_i-1It is respectively reservoir i-th and the reservoir storage of i-1 stage Mo (ten thousand m³)；LS_i、PS_i、EF_iIt is respectively reservoir The i-th stage come the water yield, abandon the water yield, evaporation with leakage (ten thousand m³)。

(5) reservoir capacity constraint:

V_min≤V_i≤V_P (6)

In formula: V_min、V_PIt is respectively reservoir minimum capacity of a reservoir storage capacity corresponding with flood control (ten thousand m³)；V_iFor reservoir the i-th rank Reservoir storage (ten thousand m at section end³)。

Further, dynamic programming Approach by inchmeal solves and specifically comprises the following steps that

(1) determine that the daily output of reservoir and pumping plant moisturizing scale meet formula (2)～(3) requirement, and occur without reservoir Stage storage capacity is less than minimum capacity of a reservoir V_min；Reservoir stage output X with intake area insufficient irrigation condition_1iAs primary iteration value, Substituted into formula (1), then formula (1)～(6) are converted into each stage pumping plant rate of water make-up Y_iFor decision variable, front i stage pumping plant Moisturizing total amount λ_iOne-dimensional dynamic programming model for state variable；

(2) according to one-dimensional dynamic programming evaluation principle, corresponding recurrence equation is obtained:

1) stage i=1

$g_1\left({\mathrm{\λ}}_1\right)=\max Y_m\·\frac{(X_{11}+Y_{11})\·K_1}{{\mathrm{YS}}_1}---\left(7\right)$

This stage reservoir yield X₁₁Given by initial value, state variable λ₁In corresponding feasible zone discrete: λ₁=0, W₁,W₂,…,BZ；To each discrete λ₁, decision variable i.e. pumping plant rate of water make-up Y₁₁In corresponding feasible zone discrete, centrifugal pump is minimum It is 0, is JD to the maximum₁；By given X₁₁With each discrete Y₁₁Correspondence, by crop field test, determines water sensitive coefficient K₁； The most also should meet Y₁₁≥λ₁；Y by satisfied requirement₁₁Substitution formula (7), respectively obtains each centrifugal pump λ₁Corresponding pumping plant is optimum Rate of water make-up Y₁₁And the g of correspondence₁(λ₁)。

Then, according to formula (5), the 1st stage end pondage V₁=V₀+LS₁-EF₁-X₁₁, the most not yet consider that reservoir is abandoned Water, uses formula (6) inspection, if exceeding the storage capacity V corresponding to flood control_P, then the water yield is abandoned beyond part as reservoir PS₁₁, now V₁ ^*=V_P；Otherwise, if without departing from, then PS₁₁=0, now V₁ ^*=V₁。

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

$g_i\left({\mathrm{\λ}}_i\right)=max\[Y_m\·\frac{(X_{1i}+Y_{1i})\·K_i}{{\mathrm{YS}}_i}+g_{i-1}\left({\mathrm{\λ}}_{i-1}\right)\]---\left(8\right)$

This stage reservoir yield X_1iGiven by initial value, state variable λ_iCarry out discrete the most respectively: λ_i=0, W₁, W₂,…,BZ；To each discrete λ_i, decision variable i.e. pumping plant rate of water make-up Y_1iDiscrete same step 1)；Equally, by given X_1iWith Each discrete Y_1iCorrespondence, by crop field test, determines water sensitive coefficient K respectively_i, and should meet:

State transition equation: λ_i-1=λ_i-Y_1i (9)

By each discrete Y_1iValue substitutes in formula (8) respectivelyBy state transition equation formula (9), look into The i-1 stage is looked for meetThe g required_i-1(λ_i-1) value, it is derived from meeting λ_iThe pumping plant optimum rate of water make-up Y required_1iAnd The g of its correspondence_i(λ_i)；Equally, according to formula (5), the i-th stage end pondage V_i=V_i-1+LS_i-EF_i-X_1i, the most not yet examine Considering reservoir and abandon water, using formula (6) to test, if exceeding the storage capacity V corresponding to flood control_P, then beyond part conduct Water yield PS abandoned by reservoir_1i, now V_i ^*=V_P；Otherwise, if without departing from, then PS_1i=0, now V_i ^*=V_i；Thus derive, it is thus achieved that corresponding Reservoir abandon water yield PS_1i, wherein i=2,3 ..., N-1.

3) stage i=N:

$g_N\left({\mathrm{\λ}}_N\right)=max\[Y_m\·\frac{(X_{1N}+Y_{1N})\·K_N}{{\mathrm{YS}}_N}+g_{N-1}\left({\mathrm{\λ}}_{N-1}\right)\]---\left(10\right)$

This stage reservoir yield X_1NGiven by initial value, state variable λ_N=BZ；Decision variable i.e. pumping plant rate of water make-up Y_1NDiscrete in corresponding feasible zone equally；By given X_1NWith each discrete Y_1NCorrespondence, by crop field test, determines respectively Water sensitive coefficient K_N, and λ should be met_N-1=λ_N-Y_1N；Use step 2) described method, final acquisition meets this λ_NThe pump required Stand optimum rate of water make-up Y_1NAnd the reservoir of correspondence abandons water yield PS_1N, wherein, i=1,2 ..., N.

(3) the pumping plant rate of water make-up Y that step (2) is obtained_1iAs initial set-point, substitute into formula (1), then formula (1)～(6) turn Turn to each stage reservoir yield X_iFor decision variable, front i stage reservoir is for water inventory λ '_iOne-dimensional flow for state variable State plan model, with reference to step (2), uses one-dimensional dynamic programming to solve, it is thus achieved that to meet this λ_N' require reservoir optimum supply water Amount X_2i, and correspondence abandon water yield PS_2i, wherein i=1,2 ..., N.

(4) reservoir yield X that step (3) is obtained_2iAs 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 X optimizing and obtaining_eiWith pumping plant rate of water make-up Y_eiAs model optimal solution, obtain target simultaneously Optimized value, and reservoir optimum abandons water yield PS_ei, wherein i=1,2 ..., N, e are that dynamic programming Approach by inchmeal number of times is compiled Number.

The water resource optimization of single pumping plant-mono-water reservoir system that this invention directly mends canal under the conditions of can realizing insufficient irrigation is adjusted Degree, to improving, water in a canal resource efficiency mended by pumping plant, Irrigating Guarantee Rate of Irrigated Area has important practical significance, to this type of Reservoir-irrigated District water Priority scheduling of resource has some reference value.

Accompanying drawing explanation

Fig. 1 is that the single pumping plant-mono-reservoir water resource of year regulation generally changes system schematic.

Detailed description of the invention

Single pumping plant-mono-reservoir reservoir water resource generally changes system schematic as shown in Figure 1.

Being object function to the maximum with intake area crop annual production, crop each growing stage reservoir, pumping plant output are decision-making Variable, to be available for water inventory, reservoir operation criterion, reservoir water balance, minimum capacity of a reservoir, flood control pair in reservoir, pumping plant year Answer storage capacity, pumping plant stage available water etc. for constraints, set up the single pumping plant-mono-water reservoir system water resource optimization directly mending canal Scheduling model, specific as follows:

One, model construction

1. object function

Use the Blank model in different bearing stage Crop-water Production Functions model, with intake area crop annual production It is target to the maximum, sets up following object function:

$F=\max Y=\max Y_m\·\underset{i=1}{\overset{N}{\Σ}}{K}_i\left(\frac{{\mathrm{ET}}_i}{{\mathrm{ET}}_{m,i}}\right)---(1-1)$

In formula: F is intake area crop maximum annual production (kg)；Y is the actual annual production of intake area crop (kg)；In N is year The crop growth number of stages divided；I be the stage numbering (i=1,2 ..., N)；Y_m、ET_i、ET_m,iIt is respectively crop fully to irrigate Under the conditions of maximum annual production (kg), the actual transpiration quantity in the i-th stage and maximum transpiration quantity (ten thousand m³)；It is the i-th rank Section crop is relative to transpiration quantity；K_iFor the i-th stage of crop hydropenia water sensitive coefficient to yield effect.

In above formula, Y_m、ET_m,iCrop specific to particular locality is it is known that ET_iFor the stage reservoir under the conditions of insufficient irrigation With pumping plant output sum, i.e. ET_i=X_i+Y_i, ET_m,iEach stage water demand of crop YS being under abundant irrigation conditions_i, and Given each stage ET_iIn the case of, K for specific crop_iAlso can determine that, then object function can be converted into:

$F=\max Y=\max Y_m\·\underset{i=1}{\overset{N}{\Σ}}\[\frac{(X_i+Y_i)\·K_i}{{\mathrm{YS}}_i}\]---(1-2)$

In formula, X_i、Y_iRespectively output (ten thousand m in the i-th stage of reservoir under the conditions of insufficient irrigation³), the i-th stage of pumping plant Rate of water make-up (ten thousand m³), it is called for short reservoir yield X_i, pumping plant rate of water make-up Y_i, YS_iFor the crop in the i-th stage under abundant irrigation conditions Water requirement (ten thousand m³), the same formula of remaining variables implication (1-1).

2. constraints

(1) it is available for water inventory reservoir year to retrain:

X₁+X₂+…+X_N≤SK (2)

In formula: SK be reservoir be available for water inventory (ten thousand m year³)；

(2) constraint of water lift total amount is allowed moisturizing pumping plant year:

Y₁+Y₂+…+Y_N≤BZ (3)

In formula: BZ is moisturizing pumping plant year to allow water lift total amount (ten thousand m³)。

(3) moisturizing pumping plant stage available water constraint:

Y_i≤JD_i (4)

In formula: JD_iIt is the i-th stage pumping plant available water (ten thousand m³)。

(4) without pumping plant, the reservoir water yield Constraints of Equilibrium of directly being mended canal by reservoir:

V_i=V_i-1+LS_i-PS_i-EF_i-X_i (5)

In formula: V_i、V_i-1It is respectively reservoir i-th and the reservoir storage of i-1 stage Mo (ten thousand m³)；LS_i、PS_i、EF_iIt is respectively reservoir The i-th stage come the water yield, abandon the water yield, evaporation with leakage (ten thousand m³)。

(5) reservoir capacity constraint:

V_min≤V_i≤V_P (6)

In formula: V_min、V_PIt is respectively reservoir minimum capacity of a reservoir storage capacity corresponding with flood control (ten thousand m³)；V_iFor reservoir the i-th rank Reservoir storage (ten thousand m at section end³)。

Two, model feature

(1) in view of insufficient irrigation condition, therefore in the constraints of model, the addition moisturizing pumping plant stage can supply water Amount constraint, i.e. formula (4).

(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.If certain stage end pondage of idle is less than the dead storehouse of reservoir Hold, then this stage is considered as small pump and stands erectly to connect and carry out moisturizing to channel；Prevent if certain stage end pondage exceedes reservoir Corresponding to flood limiting water level during storage capacity, then need to take out the dispatching requirement abandoning water to ensure reservoir capacity.Busy is according to reservoir filling Amount situation, then consider to be combined by reservoir, pumping plant to supply water to intake area, to meeting plant growth demand as far as possible.

Three, model solution

Model above Chinese style (1-2)～(6) are the nonlinear mathematical models that can divide in a stage, Y in object function_m、YS_iRight If the specific crop of particular locality is it is known that each stage gives X_i、Y_iIn the case of, K for specific crop_iAlso can determine that, model is real Border be with by crop growth divided stages reservoir supply water stage i (i=1,2 ..., N) be stage variable, each stage reservoir supply Water yield X_i, pumping plant rate of water make-up Y_iStage for decision variable can divide nonlinear model, uses dynamic programming successive approximation method to solve.

Assuming that the initial reservoir capacity V of annual-storage reservoir₀Couple about it is known that model Chinese style (2)～(4) are dynamic programming Bundle, formula (5) is the constraint of reservoir operation each stage water balance；Use dynamic programming successive approximation method to solve, utilize formula (6) simultaneously Storage capacity is tested, revises each stage end reservoir capacity, finally can obtain crop maximum annual production under the conditions of insufficient irrigation, And each stage reservoir optimal water supply X of correspondence_i, abandon water yield PS_iWith pumping plant rate of water make-up Y_i(i=1,2 ..., N), for using Single pumping plant-mono-water reservoir system place the Researching on Water Resources Optimal Management directly mending canal provides foundation.

First, carry out data preparation, specifically include:

Single pumping plant-mono-reservoir cooperation that regulates in year for specific direct benefit canal a certain under the conditions of insufficient irrigation is dispatched System, is divided into N number of stage by crop the time of infertility；According to reservoir initial water level, search water level-capacity curve, determine Reservoir initial storage V₀；According to the practical operation situation of reservoir, determine and be available for water inventory SK, minimum capacity of a reservoir V year_min, flood control limit water The storage capacity V that position is corresponding_P；According to reservoir locality meteorological model data, determine reservoir each stage water yield LS_i, evaporation and leakage EF_i；According to pump performance characteristic and pumping plant real work situation, determine and allow water lift total amount BZ moisturizing pumping plant year, and combination takes Water water level year-end drawdown level, determines each stage available water JD_i；Refer to according to intake area variety of crops, planting scale, multiple cropping The data such as number, calculate and determine intake area each stage water requirement YS_i(i=1,2 ..., N), under abundant irrigation conditions, this crop is Big annual production Y_m。

Secondly, carry out dynamic programming Approach by inchmeal and solve, specific as follows:

(1) intake 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 V_min.With specific intake area daily non-fully The reservoir stage output X of irrigation conditions_1i(i=1,2 ..., N) as primary iteration value, substituted into formula (1-2), the most former Model Chinese style (1-2)～(6) are converted into each stage pumping plant rate of water make-up Y_iFor decision variable, front i stage pumping plant moisturizing total amount λ_i For the one-dimensional dynamic programming model of state variable, one-dimensional dynamic programming can be used to solve.

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

1) stage i=1

$g_1\left({\mathrm{\λ}}_1\right)=\max Y_m\·\frac{(X_{11}+Y_{11})\·K_1}{{\mathrm{YS}}_1}---\left(7\right)$

This stage reservoir yield X₁₁Given by initial value, state variable λ₁In corresponding feasible zone discrete: λ₁=0, W₁,W₂,…,BZ.To each discrete λ₁, decision variable (i.e. pumping plant rate of water make-up Y₁₁) discrete, such as 00,000 m in corresponding feasible zone³、 50000 m³, 100,000 m³, 150,000 m³、…Y_11,max(Y_11,maxIt is the 1st stage pumping plant maximum moisturizing ability JD₁).By given X₁₁With Each discrete Y₁₁Correspondence, by crop field test, determines this exsiccosis water sensitive coefficient K to yield effect respectively₁.With Time also should meet: Y₁₁≥λ₁.Y by satisfied requirement₁₁Substitute into formula (7) respectively, respectively obtain each centrifugal pump λ₁Time, pumping plant is optimum Rate of water make-up Y₁₁And the g of correspondence₁(λ₁)。

Then, according to formula (5), the 1st stage end reservoir capacity V₁=V₀+LS₁-EF₁-X₁₁, the most not yet consider that reservoir is abandoned Water, uses formula (6) inspection, if exceeding the storage capacity V corresponding to flood control_P, then the water yield is abandoned beyond part as reservoir PS₁₁, now V₁ ^*=V_P；Otherwise, if without departing from, then PS₁₁=0, now V₁ ^*=V₁。

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

$g_i\left({\mathrm{\λ}}_i\right)=max\[Y_m\·\frac{(X_{1i}+Y_{1i})\·K_i}{{\mathrm{YS}}_i}+g_{i-1}\left({\mathrm{\λ}}_{i-1}\right)\]---\left(8\right)$

This stage reservoir yield X_1iGiven by initial value, state variable λ_iCarry out discrete the most respectively: λ_i=0, W₁, W₂,…,BZ.To each discrete λ_i, decision variable (pumping plant rate of water make-up Y_1i) discrete ibid.Equally, by given X_1iWith respectively from Dissipate Y_1iCorrespondence, by crop field test, determines this exsiccosis water sensitive coefficient K to yield effect respectively_i, and should expire Foot:

State transition equation: λ_i-1=λ_i-Y_1i (9)

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

By each discrete Y_1iValue substitutes in formula (8) respectivelyBy state transition equation formula (9), look into The i-1 stage is looked for meetThe g required_i-1(λ_i-1) value, it is derived from meeting λ_iThe pumping plant optimum rate of water make-up Y required_1iAnd The g of its correspondence_i(λ_i).Equally, according to formula (5), the i-th stage end reservoir capacity V_i=V_i-1+LS_i-EF_i-X_1i, the most not yet consider Water abandoned by reservoir, uses formula (6) to test, if exceeding the storage capacity V corresponding to flood control_P, then beyond part as water Water yield PS is abandoned in storehouse_1i, now V_i ^*=V_P；Otherwise, if without departing from, then PS_1i=0, now V_i ^*=V_i.Thus derive, thus can obtain Water yield PS abandoned by corresponding reservoir_1i, wherein i=2,3 ..., N-1.

3) stage i=N:

$g_N\left({\mathrm{\λ}}_N\right)=max\[Y_m\·\frac{(X_{1N}+Y_{1N})\·K_N}{{\mathrm{YS}}_N}+g_{N-1}\left({\mathrm{\λ}}_{N-1}\right)\]---\left(10\right)$

This stage reservoir yield X_1NGiven by initial value, state variable λ_N=BZ；Decision variable (pumping plant rate of water make-up Y_1N) discrete in corresponding feasible zone equally.By given X_1NWith each discrete Y_1NCorrespondence, by crop field test, determines respectively This exsiccosis water sensitive coefficient K to yield effect_N, and should meet: λ_N-1=λ_N-Y_1N。

Use step 2) described method, final acquisition meets this λ_NThe pumping plant optimum rate of water make-up Y required_1N, and correspondence Water yield PS abandoned by reservoir_1N(i=1 ... N).

(3) the pumping plant rate of water make-up Y that step (2) is obtained_1iAs initial set-point, substitute into formula (1-2), then master mould Chinese style (1-2)～(6) are converted into each stage reservoir yield X_iFor decision variable, front i stage reservoir is for water inventory λ '_iFor state The one-dimensional dynamic programming model of variable, with reference to step (2), uses one-dimensional dynamic programming to solve, it is thus achieved that to meet this λ_N' require Reservoir optimal water supply X_2i, and correspondence abandon water yield PS_2i(i=1,2 ..., N).

(4) reservoir yield X that step (3) is obtained_2iAs initial set-point, substitute into formula (1-2), 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 X optimizing and obtaining_eiWith pumping plant rate of water make-up Y_eiAs master mould optimal solution, the most also can obtain Obtain object function optimal value, and reservoir optimum abandons water yield PS_ei(i=1,2 ..., N) (wherein, e is that dynamic programming is gradually forced Nearly number of times numbering).

More than invention solves conveniently, and precision is reliable, uses single pumping plant-mono-reservoir to supply water under the conditions of being available for insufficient irrigation The popularization and application of Large Or Medium Type Irrigation Areas management unit, reach the purpose that Water Resources Irrigation is distributed rationally, improve irrigated area, society and life State benefit.

Claims (3)

1. directly mend single pumping plant-mono-water reservoir system water resource optimal allocation method of canal under the conditions of insufficient irrigation, by mending Pump works and reservoir are combined to canal water supply, channel supplement intake area hydropenia, it is characterised in that specifically include following steps:

One, model construction, comprises the following steps:

1. it is target to the maximum with intake area crop annual production, sets up following object function:

$F=\max Y=\max Y_m\·\underset{i=1}{\overset{N}{\mathrm{\Σ}}}\[\frac{(X_i+Y_i)\·K_i}{{\mathrm{YS}}_i}\]---\left(1\right)$

In formula: F is intake area crop maximum annual production (kg)；Y is the actual annual production of intake area crop (kg)；N is division in year Crop growth number of stages；I is stage numbering, i=1,2 ..., N；Y_mFor the crop maximum annual production under abundant irrigation conditions (kg), K_iFor the i-th stage of crop hydropenia water sensitive coefficient to yield effect；X_i、Y_iIt is respectively under the conditions of insufficient irrigation the Reservoir yield (ten thousand m in i stage³), pumping plant rate of water make-up (ten thousand m³), it is called for short reservoir yield X_i, pumping plant rate of water make-up Y_i, YS_iFor filling Divide the water demand of crop (ten thousand m in the i-th stage under irrigation conditions³)。

2. constraints is set, including being available for water inventory constraints reservoir year, allows water lift total amount constraint bar moisturizing pumping plant year Part, moisturizing pumping plant stage available water constraints, the reservoir water yield equilibrium constraint without pumping plant, directly being mended canal by reservoir and Reservoir capacity constraints.

Two, model solution

First carry out data preparation, specifically include: crop is divided into N number of stage in time of infertility；According to reservoir initial water level, look into Water detection position-capacity curve, determines reservoir initial storage V₀；According to the practical operation situation of reservoir, determine that year can supply water always Amount SK, minimum capacity of a reservoir V_min, storage capacity V that flood control is corresponding_P；According to reservoir locality meteorological model data, determine each rank of reservoir Section carrys out water yield LS_i, evaporation with leakage EF_i；According to pump performance characteristic and pumping plant real work situation, determine moisturizing pumping plant year Allow water lift total amount BZ, and combine water sources water level year-end drawdown level, determine each stage available water JD_i；According to intake area agriculture Crop varieties, planting scale, multiple crop index, determine intake area each stage water requirement YS_i, crop the biggest year under abundant irrigation conditions Yield Y_m；Wherein, 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) it is available for water inventory reservoir year to retrain:

X₁+X₂+…+X_N≤SK (2)

In formula: SK be reservoir be available for water inventory (ten thousand m year³)；

(2) constraint of water lift total amount is allowed moisturizing pumping plant year:

Y₁+Y₂+…+Y_N≤BZ (3)

In formula: BZ is moisturizing pumping plant year to allow water lift total amount (ten thousand m³)。

(3) moisturizing pumping plant stage available water constraint:

Y_i≤JD_i (4)

In formula: JD_iIt is the i-th stage pumping plant available water (ten thousand m³)。

(4) without pumping plant, the reservoir water yield Constraints of Equilibrium of directly being mended canal by reservoir:

V_i=V_i-1+LS_i-PS_i-EF_i-X_i (5)

In formula: V_i、V_i-1It is respectively reservoir i-th and the reservoir storage of i-1 stage Mo (ten thousand m³)；LS_i、PS_i、EF_iIt is respectively reservoir i-th The individual stage come the water yield, abandon the water yield, evaporation with leakage (ten thousand m³)。

(5) reservoir capacity constraint:

V_min≤V_i≤V_P (6)

In formula: V_min、V_PIt is respectively reservoir minimum capacity of a reservoir storage capacity corresponding with flood control (ten thousand m³)；V_iFor the i-th stage Mo of reservoir Reservoir storage (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) determine that the daily output of reservoir and pumping plant moisturizing scale meet formula (2)～(3) requirement, and occur without the reservoir stage Storage capacity is less than minimum capacity of a reservoir V_min；Reservoir stage output X with intake area insufficient irrigation condition_1iAs primary iteration value, by it Substitution formula (1), then formula (1)～(6) are converted into each stage pumping plant rate of water make-up Y_iFor decision variable, front i stage pumping plant moisturizing Total amount λ_iOne-dimensional dynamic programming model for state variable；

(2) according to one-dimensional dynamic programming evaluation principle, corresponding recurrence equation is obtained:

1) stage i=1

$g_1\left({\mathrm{\λ}}_1\right)=\max Y_m\·\frac{(X_{11}+Y_{11})\·K_1}{{\mathrm{YS}}_1}---\left(7\right)$

This stage reservoir yield X₁₁Given by initial value, state variable λ₁In corresponding feasible zone discrete: λ₁=0, W₁, W₂,…,BZ；To each discrete λ₁, decision variable i.e. pumping plant rate of water make-up Y₁₁In corresponding feasible zone discrete, centrifugal pump is minimum 0, it is JD to the maximum₁；By given X₁₁With each discrete Y₁₁Correspondence, by crop field test, determines water sensitive coefficient K₁；With Time also should meet Y₁₁≥λ₁；Y by satisfied requirement₁₁Substitution formula (7), respectively obtains each centrifugal pump λ₁Corresponding pumping plant optimum is mended Water yield Y₁₁And the g of correspondence₁(λ₁)。

Then, according to formula (5), the 1st stage end pondage V₁=V₀+LS₁-EF₁-X₁₁, the most not yet consider that water abandoned by reservoir, Employing formula (6) is checked, if exceeding the storage capacity V corresponding to flood control_P, then water yield PS is abandoned beyond part as reservoir₁₁, this Time V₁ ^*=V_P；Otherwise, if without departing from, then PS₁₁=0, now V₁ ^*=V₁。

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

$g_i\left({\mathrm{\λ}}_i\right)=max\[Y_m\·\frac{(X_{1i}+Y_{1i})\·K_i}{{\mathrm{YS}}_i}+g_{i-1}\left({\mathrm{\λ}}_{i-1}\right)\]---\left(8\right)$

This stage reservoir yield X_1iGiven by initial value, state variable λ_iCarry out discrete the most respectively: λ_i=0, W₁, W₂,…,BZ；To each discrete λ_i, decision variable i.e. pumping plant rate of water make-up Y_1iDiscrete same step 1)；Equally, by given X_1iWith Each discrete Y_1iCorrespondence, by crop field test, determines water sensitive coefficient K respectively_i, and should meet:

State transition equation: λ_i-1=λ_i-Y_1i (9)

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

By each discrete Y_1iValue substitutes in formula (8) respectivelyBy state transition equation formula (9), search i- 1 stage metThe g required_i-1(λ_i-1) value, it is derived from meeting λ_iThe pumping plant optimum rate of water make-up Y required_1iAnd it is right The g answered_i(λ_i).Equally, according to formula (5), the i-th stage end pondage V_i=V_i-1+LS_i-EF_i-X_1i, the most not yet consider water Water is abandoned in storehouse, uses formula (6) to test, if exceeding the storage capacity V corresponding to flood control_P, then beyond part as reservoir Abandon water yield PS_1i, now V_i ^*=V_P；Otherwise, if without departing from, then PS_1i=0, now V_i ^*=V_i；Thus derive, it is thus achieved that corresponding water Water yield PS is abandoned in storehouse_1i, wherein i=2,3 ..., N-1.

3) stage i=N:

$g_N\left({\mathrm{\λ}}_N\right)=max\[Y_m\·\frac{(X_{1N}+Y_{1N})\·K_N}{{\mathrm{YS}}_N}+g_{N-1}\left({\mathrm{\λ}}_{N-1}\right)\]---\left(10\right)$

This stage reservoir yield X_1NGiven by initial value, state variable λ_N=BZ；Decision variable i.e. pumping plant rate of water make-up Y_1NWith Sample is discrete in corresponding feasible zone；By given X_1NWith each discrete Y_1NCorrespondence, by crop field test, determines moisture respectively Sensitivity coefficient K_N, and λ should be met_N-1=λ_N-Y_1N；Use step 2) described method, final acquisition meets this λ_NThe pumping plant required is Excellent rate of water make-up Y_1NAnd the reservoir of correspondence abandons water yield PS_1N, wherein, i=1,2 ..., N.

(3) the pumping plant rate of water make-up Y that step (2) is obtained_1iAs initial set-point, substitute into formula (1), then formula (1)～(6) are converted into With each stage reservoir yield X_iFor decision variable, front i stage reservoir is for water inventory λ '_iOne-dimensional dynamic rule for state variable Draw model, with reference to step (2), use one-dimensional dynamic programming to solve, it is thus achieved that to meet this λ '_NThe reservoir optimal water supply required X_2i, and correspondence abandon water yield PS_2i, wherein i=1,2 ..., N.

(4) reservoir yield X that step (3) is obtained_2iAs 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 model optimization terminates；With Last reservoir yield X optimizing acquisition_eiWith pumping plant rate of water make-up Y_eiAs model optimal solution, obtain object function simultaneously Optimal value, and reservoir optimum abandons water yield PS_ei, wherein i=1,2 ..., N, e are dynamic programming Approach by inchmeal number of times numbering.