Symbol description and noun interpretation unit cost: power generation power; active power flow: power transmission power and direction; DC model: DC optimal trend (DC-OPF); blocked cost, due to line capacity limit, make the original offer low cost The unit has to reduce power generation, and turn it out of other venues, thereby increasing the cost of power generation; relative safety margin: that is, in emergency, the absolute value can exceed the percentage of the limit;
The problem retrospects the reform of the world's power industry, has established a variety of power markets, and its core dedication is: introducing competition, improving efficiency, reducing electricity price, improving services, and sustaining development. Participants in the electricity market are direct purposes, while reducing electricity prices, and improving services will benefit the whole society. With the in-depth of my country's power market reform, there have been many new problems. The distribution of network trends in the power market depends mainly on the distribution of power transactions "and power transactions are maximized to profit", "can inevitably appear some systems. Low lines of transmission or low power vessels around electricity often carry a heavier load "increases the chance of occurrence of electric stiffness. The so-called electrical resistance is assumed that there are several power grids and several main lines, and the active trend (transmission power and direction) on each line depends on the grid structure and the effort of each generator set. The absolute value of the active trend on each line of the grid has a safety limit, and the limit has a certain relative safety margin (ie the upper limit of the percentage of the elevation in the emergency situation). If each unit cost distribution scheme enables an absolute value of the active trend on a certain line to exceed the limit, it is called a transmission barrier. At the same time, electricity has been used as a commodity marketization - "The relationship between the members of the system in the electricity market is pure economic interests", and the traditional method of administrative ordered re-scheduled to eliminate transmission blocks in the electricity market. Again, it is necessary to use the lever of the electricity price to coordinate the market members. Therefore, the substance of the problem is to establish a model that relies on reasonable transactions and scheduling integration. When organizing transactions, scheduling and delivery, the grid company must follow the principle of "safety first" of the grid. At the same time, it is necessary to develop a power market transaction rule to operate according to the economic goal of the purchase cost. The grid company develops a scheduling plan that meets the safe operation of the grid in accordance with the load forecast and transaction rules - the power distribution (power generation power) allocation scheme of each generator set. This article will use the data and rules given in the problem to establish a reasonable mechanism hoping to add a piece of bricks to our own electricity market reform and development.
Model assumptions Basic assumptions: 1. Follow the principle of safe electricity, exceeding the safety limit * (1 security margin), when the transfer is invalid; 2. Do not consider the priority of the user's electricity; 3. No Differences in electricity prices during the day and evening; 4. Do not consider the loss during transmission; 5. There is no need to consider the electricity period; 6. Suppose 6 lines are connected to each other, but there is no need to consider the topology of the grid. Suppose all power schedules are instantaneous.
The establishment and solving model of the model 1: The grid has 8 generator sets, 6 main lines, Table 1 and Table 2's program 0 give the current chart's relationship with the active trend, in the modern power transport network Power Grid Corporation can't see the location of the unit is responsible for power transportation scheduling and distribution, and the power from the unit is no longer transported to a particular route, but the program developed by the grid operator to transmit the entire power grid, which is easy. Understanding changes in a unit will affect all power flows of all grids, Table 1 and the remaining 32 programs of Table 2, and the fine-tuning caused by 6 power grids. As shown in the figure: Table 1 Various units of various units (unit: MW, record MW) program / unit 1 2 3 4 5 6 7 80 120 73 180 80 125 125 81.1 901 133.02 73 180 80 125 125 81.1 902 129.63 73 180 80 125 125 81.1 903 125 125 81.1 904 145.32 73 180 80 120 78.596 180 80 125 125 81.1 906 120 75.45 180 80 125 125 81.1 907 120 90.487 180 80 125 125 81.1 908 120 83.848 180 80 125 125 81.1 909 120 73 231.1 9010 120 73 198.48 80 125 125 81.1 9012 125 125 81.1 9012 120 73 190.55 80 125 125 81.1 9013 120 73 180 75.852 125 125 81.1 9014 120 73 180 65.958 125 125 81.1 9015 120 73 180 87.258 125 125 81.1 97.824 125 125 81.1 9017 120 73 180 80 140.71 125 81.1 9018 120 73 180 80 141.58 125 81.1 9019 120 73 180 80 132.37 125 81.1 9020 120 73 180 80 156.93 125 81.1 9021 120 73 180 81 9022 120 73 180 80 125 131.21 81.1 9023 120 73 180 80 125 120 73 180 80 125 120 73 180 80 125 125 60.582 9026 120 73 180 80 125 125 70.962 9027 120 73180 80 125 125 73 180 80 125 125 75.529 9029 120 73 180 80 125 125 81.1 104.8430 125 73 180 80 125 125 81.1 111.2231 120 73 180 80 125 125 81.1 98.09232 120 73 180 80 125 125 81.1 120.44
Table 2 Trend value of each line (corresponding to Table 1, unit: MW) Scheme / Line 1 2 3 4 5 60 164.78 140.87 - 144.25 119.09 135.44, 157.691 165.63 135.37 160.762 165.51 140.25-144.92 118.7 135.33 159.983 167.93 138.71 117.72 135.41 166.814 166.79 139.45 -146.91 -145.92 -143.84 118.13 135.41 163.645 164.94 141.5 118.43 136.72 164.8 141.13 157.226 118.82 136.02 157.57 165.59 143.03 -144.07 -143.16 -143.49 117.24 139.66 156.598 165.21 142.28 117.96 137.98 156.969 167.43 129.58 132.04 153.610 165.71 140.82 -152.26 140.82 -147.08 -149.33 122.85 134.21 140.82 125.75 133.28 166.45 156.2311 155.0912 156.7713 165.23 140.85 -145.82 121.16 134.75 164.23 140.73 119.12 135.57 157.214 163.04 140.34 -144.18 -144.03 -144.32 119.31 135.97 141.1 118.84 135.06 165.54 156.3115 158.2616 159.2817 134.67 166.88 141.4 118.67 164.07 143.03 -144.34 -140.97 118.75 133.75 158.8318 164.27 142.85 134.57 141.44-143.3 119 134.88 158.0120 163.89 143.61 -140.25 118.64 133.28 159.1221 166.35 139.29 -144.2 119.1 13 6.33 157.5922 157.6723 165.54 140.14 135.81 166.75 138.95 119.09 -144.19 -144.17 -144.14 119.15 136.55 138.07 119.19 137.11 167.69 157.5924 157.6525 154.2626 162.21 141.21 135.5 163.54 116.03 -144.13 -144.16 117.56 135.44 141 155.9327 154.8828 162.7 135.4 116.74 141.14 -144.21 -144.18 118.24 164.06 140.94 135.4 156.6829 164.66 142.27-147.2 120.21 135.7 14.68 135.16 157.6331 164.67 141.56-45.88 119.68 135.29 157.6132 164.69 143.84 -150.34 121.34 135.12 157.64
From the figure above, we can easily see that every certain value for the drain of the unit, the powerful trend of the line is a set of random variables, which may comply with a certain distribution. That is, after the value of the unit is determined, the active power flow of the line cannot be determined [1]. Therefore, we can use the method of regression analysis to determine the relationship between the active trend on each generator set. We use the regress () function provided in MATLAB's statistical toolbox to solve the minimum multiplier estimate of the regression coefficient, establish the linear model of the regression, the error estimate, and its confidence interval, give a coefficient (complex correlation coefficient) statistic value [2 ]. First, establish a M file named new111 (New 111 is the M file established by the line 1's tide function and various units of the unit, the new 112 and other classes): We give the first column of the matrix X plus one list 1, which is return A constant item appears in the equation, and we can see this approach from the results of residual analysis makes the results closer to the real situation. Although there is a bad point in the residual diagram of New 113 and New 114, it is easy to find a bad point residual value NEW113 from the value obtained from EINT, which is very close to 0, so bad point The impact on the entire problem can be ignored. Only one M file is given here, and the rest is given.
New111: y = [164.78 165.81 165.51 167.93 166.79 164.94 164.8 165.59 165.21 167.43 165.71 166.45 165.23 164.23 163.04 165.54 166.88 164.07 164.27 164.57 163.89 166.35 165.54 166.75 167.69 162.21 163.54 162.7 164.06 164.66 164.7 164.67 164.69]; X = [1 120 73 180 80 125 125 81.1 901 133.02 73 180 80 129.63 73 180 80 125 125 81.1 901 158.77 73 180 80 125 125 81.1 90 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 901 120 90.45 81.1 901 120 83.848 180 80 125 125 81.1 901 120 73 198.48 80 125 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 75.857 120 73 180 65.958 125 125 81.1 901 120 73 180 87.258 125 125 81.1 901 125 73 180 97.824 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 132.37 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 120 73 180 80 125 131.21 81.1 901 120 73 180 80 125 149.29 81.1 901 120 73 180 80 125 125 60.582 901 120 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 75.529 90 125 125 81.1 104.841 120 73 180 80 125 125 81.1 111.221 120 73 180 80 125 125 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, STATS] = Regress (Y, X, 0.05) RCOPLOT (E, EINT) regression residual map: new111Y1 = 110.4775 0.0826X1 0.0478X2 0.0528X3 0.1199X4 -0.0257X5 0.1216X6 0.1220 x7-0.0015X8B = 110.4775 0.0826 0.0478 0.0528 0.1199 -0.0257 0.1216 0.1220 -0.0015
New112y2 = 131.3521-0.0547x1 0.1275x2-0.0001x3 0.0332x4 0.0867x5-0.1127x6-0.0186x7 0.0985x8b = 131.3521 -0.0547 0.1275 -0.0001 0.0332 0.0867 -0.1127 -0.0186 0.0985New113y3 = 108.9928 0.0694x1-0.0620x2 0.1565x3 0.0099X4-0.1247X5-0.0024X6 0.0028X7 0.2012X8B = 108.9928 0.0694 -0.0620 0.1565 0.0099 -0.1247 -0.0024 0.0028 0.2012
New114y4 = 77.6116-0.0346x1-0.1028x2 0.2050x3-0.0209x4-0.0120x5 0.0057x6 0.1452x6 0.0763X8B = 77.6116 -0.0.0209 -0.0120 0.0057 0.1452 0.0763
New115Y5 = 133.1334 0.0003x1 0.2428x2-0.0647x3-0.0412x4 -0.0655x5 0.0700x6-0.0039X7-0.0092x8b = 133.1334 0.0003 0.2428 -0.0647 -0700 -0.0039 -0.0092
New116Y6 = 120.8481 0.2376x1-0.0607 x2 -0.0781x3 0.0929X4 0.046664X7 0.0004X8B = 120.8481 0.2376 -0.0607 -0.0781 0.0929 0.0466 -0.0003 0.1664 0.0004
Y1-6 is the relationship between active trends and individual generator sets.
Great Square Group:
Y1 = 110.4775 0.0826x1 0.0478x2 0.0528x3 0.1199x4-0.0257x5 0.1216x6 0.1220 x7-0.0015x8
Y2 = 131.3521-0.0547X1 0.1275x2-0.0001x3 0.0332x4 0.0867 x5-0.1127x6-0.0186 x7 0.0985x8
Y3 = 108.9928 0.0694x1-0.0620x2 0.1565x3 0.0099x4-0.1247 x5-0.0024 x6 0.0028x7 0.2012x8
y4 = 77.6116-0.0346x1-0.1028x2 0.2050x3-0.0209x4-0.0120x5 0.0057 x6 0.1452x6 0.0763x8 y5 = 133.1334 0.0003x1 0.2428x2 -0.0647x3-0.0412x4-0.0655x5 .0700x6-0.0039x7 -0.0092x8
Y6 = 120.8481 0.2376x1-0.0607x2-0.0781x3 0.0929x4 0.0466x5-0.0003x6 0.1664x7 0.0004x8
Y1-6 is a tide current of each line; the processes of each motor set of X1-8. This equation is not --a
Question 2: In the case of completely open, in the case of constraints and restrictions of power transmission, add power transmission occurs in a certain condition due to the constraints and restrictions of power transmission. Introducing a competitive mechanism in the electricity market, the transmission network plays an important role, but in the case of insufficient power transmission network, there will often be blocked phenomena, so that the management of the transmission network is to ensure the reliable and safe operation of the transmission network. Do not limit and constrain power transmission. After these constraints, on the one hand, in terms of certain extent, the distant generators have entered the local power market, and on the other hand, the local power market generators have reducing the competitors of power industry. The common role of both aspects will eventually lead to the weak market competition mechanism, not only from the power industry and the user universal, but the manager of the transmission network is not more beneficial. As a result, there is a contradiction between the free transmission of electricity and the safety of the transmission network. Under a specific transmission network, if it is too emphasizable from the free transmission of electric power, it is possible to cause the system to crash; if it is too emphasizable, the power industry will lose the market, lose the user, lose the profit. To this end, these contradictions have a significant significance for ensuring the good operation of production and life. We first assume that under the ideal market environment, we first do not consider the economic scheduling of network constraints, that is, the following optimization problems (P1) minf = (1) ST (2) i = 1, N: (3), n For the number of system nodes, f is the total purchaser cost, which is the cost function of the node i generator is the load of node i in the system, which represents the actual function of the node i in electrical mechanism and maximum power out force. And the Lagrangian multiplier of the corresponding constraints (1) and (2), respectively. Optimization issues P1 Because there is no test line capacity constraint, it is referred to as unconstrained scheduling. The result is recorded in which the marketer is unconstrained, representing the optimal output vector of each node generator, representing the market liquidation price, the model is all nodes under the model equal. The above results may result in certain lines overload, that is, the economic scheduling of the economic scheduling that is repeated and the branch capacity constraint is required to ensure the safety and economic operation of the system. The optimization problem is (P2)
Minf = (4) ST (5) (5) (6) (7) The problem P2 is actually a DC optimal trend (DC-OPF) general form, where B, H reflects the constant matrix of network characteristics, respectively, respectively, respectively The Lagrangian multiplier vector, the optimal solution of the problem, where c represents the constraint. Optimization Problem P2 Since the line safety limit is considered, this title is simply a constrained schedule. Since the constraint (6) is added, the power generation cost in the target function generates an increment, ie Δ, this is the so-called blocking cost, which makes the original offer with low cost unit due to line capacity limitations. It has to be reduced, and the unit of power generation is multiple, and the increase in power generation costs is the inclusion cost. If a line block occurs, it is often necessary to cut the contract based on certain principles. The principles of the basis can be cut (weighted) and minimal, transaction cut (weighted) and minimum, etc., they all directly or indirectly reflect the cost of blocking scheduling. The weight here is usually reflecting the factor W of the importance of the contract. In a specific market environment, it may be an indicator of the pre-set reaction transaction priority of the system operator, or it is possible to apply for a transaction. In order to avoid the rate of being reached and the original payment. When the weight and minimum target of the transaction cut reduction, the blocking scheduling model is as follows (P3) min (8) ST. (9) (10), representing the original and adjusted contract power, X represents Independent state variables, the above model medium and inequality represent the trend balance equation and the line easy constraints. If the DC model is used, (9), the DC flossing equilibrium equation. Thus, the above model is a secondary planning model, which can be solved by a LagRange method, or transformed into linear planning. The blocking scheduling model directly in the targeted and minimum of the transaction cut, only the target function in (P3) is changed to the following formula, namely: MIN (11) When using a DC model, this will become a standard linearity. Planning issues, so it is convenient to solve. The following principles must be paid to the following principles during transmission schedules [3]: a minimum bidding priority grid power generation principle. Transmission scheduling is carried out in the case of central transactions. The Center Exchange arranged in parallel power generation in the order of a small to large. Under certain power transactions, which power generation enterprise is lowered to power generation, until the given trading volume is met; the bidding power generation enterprise cannot generate electricity. This reflects competitiveness and fairness. B Power Transaction Transaction Summary Principle. In a particular transmission network, transactions are as high as possible to ensure the reliable operation of the transmission network, so that the transmission network operates in the case where the total volume is the largest. C The principle of safe and reliable operation of C-transmission network. To ensure that the transmission network is running in the case where the total volume is the largest, the power transmission network will be unimpeded. However, it is actually difficult to do this. Because the power transmission line always has a safe delivery capacity. If the line delivery capacity exceeds the secure delivery capacity, it will threaten the power operation of the transmission network. To this end, the limit capacity is always developed.
Question 3, 4, 5:
Table 3 Segment capacity (unit: MW) unit / section 1 2 3 4 5 6 7 8 9 101 70 0 50 0 0 30 0 0 0 402 30 0 20 8 15 6 2 0 0 83 110 0 40 0 30 0 20 40 0 404 55 5 10 10 10 10 15 0 0 15 75 5 15 0 15 0 10 10 106 95 0 10 20 0 15 10 20 0 107 50 15 5 15 10 10 5 10 3 28 70 0 20 20 0 20 0 20 10 15 5 Table 4 Section of various units (unit: yuan / MW hours, remembering the metallographic / MWH) unit / section 1 2 3 4 5 6 7 8 9 101 -505 0 124 168 210 252 312 330 363 4892 -560 0 182 203 245 300 320 360 410 4953-610 0 152 189 233 258 308 356 415 5004-500 150 170 435 8005 -590 0 116 146 188 215 250 310 396 5106-607 0 159 173 205 252 305 380 405 5207 -500 120 180 251 260 306 315 335 348 5488 -800 153 183 233 253 283 303 318 400 800
Table 5 Climbing rate (unit: Mw / minute) unit 1 2 3 4 5 6 7 8 Rate 2.2 1 3.2 1.3 1.8 2 1.4 1.8
The forecast load for the next period given in question 3 is 982.4MW. Since the units of each unit is divided into 10 segments and the paragraphs of various units of 10 segments, the data in Table 3 and Table 4 is One-on-one relationship. Therefore, we can re-arrange this 80 sets of data according to this relationship. Its arrangement rules: example: unit number paragraph capacity 1 -505 70
Then, according to the sequential order, it is re-sorted in the order of small to large. After the arrangement, you can get a table (Table 7) in Table 7 refers to the actual segment capacity in this period, that is, because the unit is in front. It can be climbed to the predicted paragraph with the case where it climbs the forecast, and the correction segment capacity is equal to the normal paragraph capacity, and the lower end of the table may be climbed. At the 49th line of time, the unit can only climb to 9.5, while the user gets from the user is 978.5, and the value of 978.5 is defined as a temporary final value, which can be used as a reference point to be used to maribe to reach the required value. position. The meaning of δ is obvious, it is a value that will be taken from the next paragraph, and if the next paragraph can't be taken, you can continue to find it until you find it. We call this method as "sort statistics". Table 7 (Unit: MW) segment paragraph capacity correction segment capacity sequential content Summary The final value forecast load δ8 -800 70 70 3 -610 110 110 6 -607 95 95 5 -590 75 75 2 -560 30 30 -505 70 70 4 -500 55 55 7 -500 50 50 1 0 0 0 2 0 0 0 0 0 0 0 5 0 5 5 6 0 0 0 5 116 15 15 7 120 15 15 1 124 50 50 5 146 0 0 4 150 5 5 3 152 40 40 8 153 0 0 0 0 4 170 10 10 6 173 20 20 7 180 5 5 2 182 20 20 8 183 20 20 5 188 15 15 3 189 0 0 4 200 10 10 2 203 8 6 205 0 0 1 210 0 0 5 215 15 15 8 233 0 0 3 233 30 30 2 245 15 15 5 250 0 0 7 251 15 15 1 252 30 30 6 252 15 15 8 253 20 20 4 255 10 3 258 0 0 7 260 10 10 8 283 0 0 2 300 6 6 4 302 10 9.5 0.5 978.5 8 303 20 7 13 985.5 982.4 3.96 305 10 7 306 10 7.1 2.9 3 308 20 20 5 310 10 10 1 312 0 0 7 315 5 0 5 8 318 10 0 10 2 320 2 2 1034.6 4 325 15 0 15 1 330 0 0 7 335 10 0 10 7 348 3 0 3 3 356 40 28 12 10 62.6 1052.8 18.22 360 0 0 1 363 0 0 4 380 0 0 6 380 20 5 15 5 396 10 10 8 400 15 0 15 6 405 0 0 2 410 0 0 3 415 0 0 4 435 0 0 1 489 40 5 35 2 495 8 7 1 3 500 40 0 40 5 510 10 7 3 6 520 10 0 10 7 548 2 0 2 4 800 1 0 1 8 800 5 0 5
Then, the pressing unit of Table 7 is classified in Table 8: Table 8 (Unit: MW) unit Bag Price Segment Capacity Correction Segment Capacity 1 -505 70 70
1 0 0 0
1 124 50 50
1 168 0 0
1 210 0 0
1 252 30 30
Σ 150 unit full load capacity = base capacity climbing rate * Time first unit full load capacity = 120 2.2 * 15 (minutes) = 155 second unit full load capacity = 73 1 * 15 (minutes) = 88 third unit full load capacity = 180 3.2 * 15 (minutes) = 228 Fourth unit full load capacity = 80 1.3 * 15 (minutes) = 99.5 Fifth unit full load capacity = 125 1.8 * 15 (Minute) = 152 sixth unit full load capacity = 125 2 * 15 (minutes) = 155 seventh unit full load capacity = 81.1 1.4 * 15 (minutes) = 102.1 Eight-unit unit full load capacity = 90 1.8 * 15 (minutes) = 117 Correction Segment capacity sum: Y1 = 150 Symposium The rest of the table is obtained by modifying each unit to modify the correction segment capacity, then you can: Y2 = 79 y3 = 180 Y4 = 99.5 Y5 = 125 Y6 = 140 y7 = 95 y8 = 113.9 Since the above scheme is arranged in the increasing sequence, each selection is performed according to the minimum principle; therefore, it is easy to infer the use of "sort statistics" method. The conclusion is that the minimum bidding is prioritized. The data obtained by the above obtained: x1-6 = 173.9194 141.9034 136.9190 120.514 136.9190 120.514 13983 168.5149 thus evaluating the difference in standard: -8.3084 8.9834 9.0810 34.0966 -4.8083 -6.5149 It is known that the scheme will be blocked. Adjustment according to the following method: 1. The sensitivity of the calculated generator set is negative [4] The unit of sensitivity should be added, and the reduction force set is a balancer. After the unit is added, according to the quotation curve, the price has risen accordingly, and thus the settlement price is increased, and the unified marginal price settlement is used, and thus the increase of the desired increase should be, according to the sensitivity of the generator set, the corresponding quotation curve. Make the price of each unit to allocate the output of various units, which guarantees that the overall purchased cost increases. The specific method is: 1 Calculating the adjustment amount when each generator set is adjusted, if the overloaded branch is ΔPl, then the adjustment amount when the generator set is adjusted separately ΔPi = i = 1, 2, ..., G, i ≠ Vδ Vδ is a balance node 2 given a set of settlement prices, and calculate the output of various units by the quote curve: △, △, ... △,, by correcting △ ρ, until satisfying = 1 That is, the output of each generator set should be added. 3 At this time, the decision of the unit is a balancer, and the number of reduced efforts is determined by the trend calculation. 2. The sensitivity of the calculated generator set is in this case, the plus force set is a balancing machine. In order to ensure the smallest purchase cost, then the reduced unit should select the largest uniterous unit, which adjusts this The smallest power is the smallest, that is, the increased purchase cost is the smallest, both of which are unified. The calculation of the reduction force is as followed by the overload of the branch, the maximum unit of sensitivity is K, and the reduction force of the generator set is (18), the discharge of the unit is subtracted, then the trend is calculated, and the balancer is obtained. Add power.
3. The sensitivity of the calculated generator set is positive, and the sensitivity is an increase in the negative unit, in order to ensure the smallest purchase cost, the reduction force set is still the most positive unit of sensitivity. At this time, the difference between the calculation method and the first) is: 1 Calculating the adjustment amount formula of the generator separately adjusted to i = 1, 2 ... is a unmerged unit, J is the largest unit of sensitivity 2 according to the first 1 case 2 Calculate the increase of power increments of the extension unit. 3 The reduction force of the reduced force set J is the power and the power of the increase in force unit. Note: (1) In the case of adjusting the number of branchs, the adjustment scheme of the above 2 or 3 may occur, and then adjust the lower limit of other branching . In the process of adjusting the above 2 or 3, we will make the adjusted generators are not all in the same quotation point. At this time, if we follow the branch of 1 or 3, we should be on the above 1 Or 3 adjustment schemes are corrected. The correction method is as follows: 1, as in 1 or 3 described above, the amount of adjustment is first adjusted separately from each generator. 2 Adjustment by low to high, the sum of the power increments, the sum of the power increments, and the principle of equalization value, and the adjustment process is overloaded if the adjustment process is overloaded End, then the calculation ends, otherwise the calculation is calculated to eliminate overloads according to the above 1 or 3. (2) When adjusting according to the above method, if there is a set of units that have all overloaded or decreased, the unit has all achieved the lower limit, and the overload is still not eliminated, then the method of pulling the shutter is resolved. The actual significance of the model is theoretically to relieve the pressure of electricity in theory, but in actual production life, people still need to save electricity. But there is still a lot of work to do in actual engineering, but here is only a small part. In the actual situation, the user is classified. According to the general point, users can be divided into three categories: 1. Category, such as heavy-duty industrial enterprises (steel companies), such companies are absolutely unstowered, so when blocking occurs, other types of user power resources can be cut, and the power supply is concentrated; 2. Second Category User This kind of user can accept a short power outage impact (enterprise such as fertilizer plant); Three types of ordinary residents can be electrified and can accept relatively long-term power outages. If the rules of the above type are added to the assignment scheme, the scheme will make the scheme more practical and operability.
Appendix New111.my = [164.78 165.81 165.51 167.93 166.79 164.94 164.8 165.59 165.21 167.43 165.71 166.45 165.23 164.23 163.04 165.54 166.88 164.07 164.27 164.57163.89 166.35 165.54 166.75 167.69 162.21 163.54 162.7 164.06 164.66 164.7 164.67 164.69]; X = [1 120 73 180 80 125 125 81.1 90 80 125 125 81.1 901 129.63 73 180 80 125 125 81.1 90 125 125 81.1 901 145.32 73 180 80 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 901 120 90.45 81.1 901 120 83.81.1 90 80 125 125 81.1 901 120 73 231.39 80 125 125 81.1 901 120 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 75.8525 125 81.1 95.958 125 125 81.1 901 125 125 81.1 901 125 73 180 97.82 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 132.37 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 1 20 73 180 80 125 131.21 81.1 901 120 73 180 80 125 149.29 81.1 901 120 73 180 80 125 125 60.582 901 120 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 73 180 80 125 125 81.1 104.841 120 73 180 80 125 125 81.1 111.22 125 81.1 98.0921 120 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, Stats ] = regress (y, x, 0.05) Rcoplot (E, EINT) execution result: b = 110.4775 0.0826% regression equation is y = b0 b1x ... bpx; 0.0478 0.0528 0.1199 -0.0015bint =
109.5421 111.4129% b0-p confidence intervals 0.0808 0.0844 0.0437 0.0518 0.0514 0.0542 0.1166 0.1231 -0.0277 -0.0237 0.1190 0.1243 0.1189 0.1251 -0.0037 0.0007e = 0.0615 0.0159 -0.0040 0.0088 -0.0201 -0.0458 -0.0356 0.0362 -0.0267 -0.0016 0.0158 0.0083 -0.0455 eint = -0.0092 0.1321% 0.0080 0.0045 residual 0.0123 -0.0485 0.0251 -0.0223 0.0409 -0.0078 -0.0570 0.0660 -0.0013 0.0166 -0.0055 0.0582 -0.0366 0.0211 -0.0360 0.0137 -0.0363 0.0177 percent observations yi its estimated residual vector estimation interval - 0.0591 0.0910 -0.0801 0.0720 -0.0386 0.0563 -0.0875 0.0472 -0.1185 0.0269 -0.1103 0.0392 -0.0061 0.0785 -0.0941 0.0407 -0.0485 0.0453 -0.0589 0.0905 -0.0600 0.0765 -0.1192 0.0282 -0.0657 0.0817 -0.0534 0.0623 -0.1176 0.0207 -0.0236 0.0739 -0.0524 0.0771 -0.0950 0.0503 -0.0334 0.1152 -0.0629 0.0473 -0.1245 0.010 5 -0.0047 0.1368 -0.0698 0.0672 -0.0363 0.0695 -0.0602 0.0492 -0.0111 0.1275 -0.1000 0.0267 -0.0546 0.0967 -0.1078 0.0358 -0.0538 0.0811 -0.1108 0.0383 -0.0338 0.0691stats = 1.0e 003 * 0.0010 5.3768 0% correlation coefficient R = 1, F is the magnitude of statistical significance% 5376.8 root of P = 0 in FIG. analysis: New112.m: y = [140.87 140.13140.25 138.71 139.45 141.5 141.13 143.03 142.28 140.82 140.82 140.82 140.85 140.73 140.34 141.1 141.4 143.03 142.29 141.44 143.61 139.29 140.14 138.95 138.07 141.21 141 141.14 140.94 142.27 142.94 141.56 143.84]; x =
[1 120 73 180 80 125 125 81.1 90 125 125 81.1 901 129.63 73 180 80 125 125 81.1 901 158.77 73 180 80 125 125 81.1 901 145.32 73 180 80 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 180 80 125 125 81.1 901 120 83.848 180 80 125 125 81.1 901 125 125 81.1 901 120 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 65.958 125 125 81.1 901 120 73 180 87.25 125 125 81.1 901 125 73 180 97.824 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 120 73 180 80 125 131.21 81.1 901 120 73 180 80 125 141.71 81.1 901 120 73 180 80 125 149.29 81.1 901 120 73 180 80 125 125 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 75.529 901 120 73 180 80 125 125 81.1 104.841 120 73 180 80 125 120 73 180 80 125 120 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, Stats] = Regress (Y, X, 0.05 RCOPLOT (E, EINT) Execution: b = 131.3521 -0.0547 0.1275 -0.0001 0.0332 0.0867 -0.1127 -0.0186 0.0985% regression equation is Y = B0 B1X ... BPX; bint = 130.5461 132.1580% B0-P confidence interval; -0.0563 -0.0532 0.1240 0.1310 -0.0013 0.0010 0.0304 0.0360 0.0850 0.0884 -0.1150 -0.0110 0.0966 0.1004E =
0.0418 0.0142 0.0032 0.0073 -0.0513 -0.0417 -0.0106 -0.0278 -0.0055 -0.0034 0.0687 -0.0007 0.0234 -0.0264 0.0395 -0.0217 0.0307 -0.0204 0.0249 -0.0269 0.0145 0.0259 0.0116 0.0089 0.0079 0.0048 -0.0211 -0.0007 -0.0172 -0.0203 0.0211 -0.0655 0.0126% observations yi its estimated residual vector; eint = -0.0206 0.1043% confidence interval of the remaining amount; -0.0504 0.0789 -0.1130 0.0105 -0.0378 0.0442 -0.0512 0.0657 -0.1041 0.0207 -0.0761 0.0550 -0.0647 0.0398 -0.0701 -0.0411 0.0091 0.0178 0.1196 0.0591 -0.0623 0.0555 -0.0416 0.0883 -0.0218 0.1007 -0.0706 0.0273 -0.0302 0.0915 -0.0625 0.0218 -0.0812 0.0284 -0.0373 0.0871 -0.0916 0.0378 -0.0326 0.0617 -0.0350 0.0868 -0.0539 0.0771 -0.0542 0.0638 -0.0661 0.0240 -0.0479 0.0464 -0.0802 0.0458 -0.0472 0.0650 -0.0576 0.0735 -0.0830 0.0423 -0.036 6 0.0787 -0.1247 -0.0062 -0.0318 0.0571stats = 1.0e 003 * 0.0010 6.9702 0% complex correlation coefficient R = 1, F statistic value is% 6970.2 Significant root rate P = 0; analysis
New113.m: y = [144.25 145.14 144.92 146.91 145.92 143.84 144.07 143.16 143.49 152.26 147.08 149.33 145.82 144.18 144.03 144.32 144.34 140.97 142.15 143.3 140.25 144.2 144.19 144.17 144.14 144.13 144.16 144.21 144.18 147.2 148.45 145.88 150.34]; X = [1 120 73 180 80 125 125 81.1 90 80 125 125 81.1 901 129.63 73 180 80 125 125 81.1 90 125 125 81.1 901 145.32 73 180 80 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 901 120 90.45 81.1 901 120 83.81.1 90 80 125 125 81.1 901 120 73 231.39 80 125 125 81.1 901 120 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 75.8525 125 81.1 95.958 125 125 81.1 901 125 125 81.1 901 125 73 180 97.82 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 132.37 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 1 20 73 180 80 125 131.21 81.1 901 120 73 180 80 125 149.29 81.1 901 120 73 180 80 125 125 60.582 901 120 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 73 180 80 125 125 81.1 104.841 120 73 180 80 125 125 81.1 111.22 125 81.1 98.0921 120 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, Stats ] = Regress (Y, X, 0.05) RCOPLOT (E, EINT)
Results of: b = 108.9928 0.0694 -0.0620 0.1565 0.0099 -0.1247 -0.0024 0.0028 0.2012% regression equation y = b0 b1X ... bpX; bint = 108.1650 109.8207% b0-p confidence interval; -0.0656 -0.0584 0.1553 0.1577 0.0678 0.0710 0.0070 0.0127 -0.1264 -0.1229 -0.0047 0.0000 0.0000 0.0055 0.1993 0.2031e = 0.0406 0.0272 0.0424 0.0105 -0.0462 -0.0225 0.0125 0.0346 -0.0469 0.0081 -0.0215 0.0125 -0.0404 0.0115 -0.0408 0.0390 -0.0453 -0.0341 0.0077 0.0094 0.0213 0.0233 -0.0047 0.0000 -0.0121 -0.0222 0.0049 -0.0287 0.0426 -0.0211 0.0459 -0.0138 0.0063 percent observations yi its estimated residual vector; eint = -0.0238 0.1051% residual estimation interval; -0.0384 0.0929 -0.0224 0.1073 -0.0314 0.0524 -0.1029 0.0105 -0.0885 0.0435 - 0.0548 0.0798 -0.0024 0.071 - 0.0333 0.0495 -0.0872 0.0442-0.0478 0.0727 -0.1057 0.0248 -0.0536 0.0766 -0.0889 0.0074 -0.0227 0.1007 -0.0850 -0.0056 -0.0 897 0.0215 -0.0571 0.0724 -0.0580 0.0769 -0.0267 0.0693 -0.0394 0.0861 -0.0722 0.0627 -0.0606 0.0606 -0.0590 0.0348 -0.0697 0.0253 -0.0857 0.0435 -0.0084 0.1002 -0.0810 0.0533 -0.0600 0.0699 -0.0873 0.0299 -0.0223 0.1074 -0.0397 0.0522stats = 1.0 E 004 * 0.0001 2.1788 0% complex correlation coefficient r = 1, F statistic value is% 21788 significant root ratio P = 0; analysis: new114.3 118.7 117.72 118.13 118.43 118.82 117.24 117.96 129.58 122.85 125.75 121.1419.84 118.67 118.75 118.8 119.09 119.15 119.19 116.03 117.56 116.74 118.24 120.21 120.68 119.68 121.34: x =
[1 120 73 180 80 125 125 81.1 90 125 125 81.1 901 129.63 73 180 80 125 125 81.1 901 158.77 73 180 80 125 125 81.1 901 145.32 73 180 80 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 180 80 125 125 81.1 901 120 83.848 180 80 125 125 81.1 901 125 125 81.1 901 120 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 65.958 125 125 81.1 901 120 73 180 87.25 125 125 81.1 901 125 73 180 97.824 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 120 73 180 80 125 131.21 81.1 901 120 73 180 80 125 141.71 81.1 901 120 73 180 80 125 149.29 81.1 901 120 73 180 80 125 125 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 75.529 901 120 73 180 80 125 125 81.1 104.841 120 73 180 80 125 120 73 180 80 125 120 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, Stats] = Regress (Y, X, 0.05 RCOPLOT (E, EINT) Execution: b = 77.6116 -0.0346 -0.1028 0.2050 -0.0209 -0.0120 0.0057 0.1452 0.0763% regression equation is Y = B0 B1X ... BPX; bint = 76.8074 78.4158% B0-P confidence interval; -0.0362 -0.0331 -0.1063 -0.0993 0.2039 0.0181 -0.0137 -0.0181 -0.0137 -0.0103 0.0034 0.0080 0.1425 0.1479 0.0745 0.0782E =
0.0441 0.0350 -0.0124 0.0168 -0.0390 -0.0407 0.0259 -0.0086 0.0290 -0.0028 0.0150 0.0117 -0.0490 -0.0124 -0.0291 -0.0543 -0.0037 -0.0221 -0.0249 0.0131 0.0034 0.0427 0.0088 0.0090 0.0058 0.0031 0.0313 0.0143 0.0533 -0.0363 -0.0137 0.0164 -0.0296% observation yi its estimated residual vector; eint = -0.0179 0.1061% of the estimated remaining amount -0.0280 0.0980 -0.0776 0.0528 -0.0235 range 0.0571 -0.0949 0.0169 -0.1031 0.0216 -0.0387 0.0905 -0.0472 0.0299 -0.0284 0.0865 -0.0431 0.0376 -0.0491 0.0792 - 0.0469 0.0702 -0.1111 0.0130 -0.0756 0.0508 -0.0773 0.0190 -0.1119 0.0032 -0.0466 0.0392 -0.0425 0.0687 -0.0596 0.0663 -0.0203 0.1057 -0.0686 0.0243 -0.0857 0.0359 -0.0567 0.0742 -0.0498 0.0677 -0.0400 0.0516 -0.0807 0.0081 -0.0767 0.0494 0.0021 0.1045 - 0.0624 0.0686 -0.0304 0. 0930 -0.0436 0.0722 -0.0487 0.0815 -0.0724 0.0133Stats = 1.0e 004 * 0.0001 2.4424 0% complex phase correlation coefficient r = 1, F statistic value value% 24424 significant root ratio P = 0; analysis: new115.m: y = [135.44 135.37 135.33 135.41 135.41 136.72 136.02 139.66 137.98 132.04 134.21 133.28 134.75 135.57 135.97 135.06 134.67 133.75 134.27 134.88 133.28 136.33 135.81 136.55 137.11 135.5 135.44 135.4 135.4 135.28 135.16 135.29 135.12]; X =
[1 120 73 180 80 125 125 81.1 90 125 125 81.1 901 129.63 73 180 80 125 125 81.1 901 158.77 73 180 80 125 125 81.1 901 145.32 73 180 80 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 180 80 125 125 81.1 901 120 83.848 180 80 125 125 81.1 901 125 125 81.1 901 120 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 65.958 125 125 81.1 901 120 73 180 87.25 125 125 81.1 901 125 73 180 97.824 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 120 73 180 80 125 131.21 81.1 901 120 73 180 80 125 141.71 81.1 901 120 73 180 80 125 149.29 81.1 901 120 73 180 80 125 125 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 75.529 901 120 73 180 80 125 125 81.1 104.841 120 73 180 80 125 120 73 180 80 125 120 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, Stats] = Regress (Y, X, 0.05 ) rcoplot (e, eint) execution results: b = 133.1334 0.0003 0.2428 -0.0647 -0.0412 -0.0655 0.0700 -0.0039 -0.0092% regression equation y = b0 b1X ... bpX; bint = 132.2863 133.9804 -0.0013 0.0020 0.2391 0.2465 -0.0660 -0.0635 -0.0441 -0.0383-0.0673 -0.06370.0676 0.0725 -0.0067 -0.0011 -0.0112 -0.0072% B0-P confidence interval; E =
0.0540 -0.0203 0.0113 -0.0592 0.0275 -0.0403 0.0157 -0.0249 0.0390 -0.0206 0.0198 0.0061 0.0467 0.0183 0.0467 0.0133 0.0054 -0.0270 -0.0308 -0.0237 -0.0162 -0.0280 -0.0109 -0.0062 0.0230 -0.0493 0.0340 -0.0077 0.0300 -0.0315 0.0145 -0.0218 0.0131% observations yi its estimated residual vector; eint = -0.0103 0.1182% confidence interval of the remaining amount; -0.0880 0.0474 -0.1232 0.0048 -0.0316 0.0541 -0.0454 0.0768 -0.0923 0.0424 -0.0280 0.1060 -0.0115 0.0665 -0.0997 0.0191 -0.0621 0.0210 -0.0476 0.0872 -0.0557 0.0679 -0.0194 0.1127 -0.0533 0.0798 -0.0470 0.0578 -0.0914 0.0374 -0.0262 0.0629 -0.0086 0.1020 -0.0958 0.0341 -0.0920 0.0447 -0.0657 0.0333 -0.0919 0.0359 -0.0798 0.0580 -0.0681 0.0558 -0.0242 0.0703 -0.0133 0.0814 -0.0520 0.0809 - 0.1045 0.0058 -0.0766 0.0612 -0.0352 0.09 53 -0.0913 0.0284 -0.0901 0.0465 -0.0337 0.0598Stats = 1.0e 003 * 0.0010 6.4339 0% complex correlation coefficient r = 1, F statistic value is% 6433.9 Significant root rate P = 0; analysis
New116.m: y = [157.69 160.76 159.98 166.81 163.64 157.22 157.5 156.59 156.96 153.6 156.23 155.09 156.77 157.2 156.31 158.26 159.28 158.83 158.37 158.01 159.12 157.59 157.67 157.59 157.65 154.26 155.93 154.88 156.68 157.65 157.63 157.61 157.64] '; X = [1 120 73 180 80 125 125 81.1 90 125 125 81.1 901 129.63 73 180 80 125 125 81.1 90 125 125 81.1 901 145.32 73 180 80 125 125 81.1 901 120 78.596 180 80 125 125 81.1 901 120 75.45 180 80 125 125 81.1 901 120 90.45 81.1 901 120 83.848 180 80 125 125 81.1 901 125 125 81.1 901 120 73 198.48 80 125 125 81.1 901 120 73 212.64 80 125 125 81.1 901 120 73 190.55 80 125 125 81.1 901 120 73 180 75.857 125 73 180 65.958 125 125 81.1 901 125 125 81.1 901 120 73 180 97.82 125 125 81.1 901 120 73 180 80 150.71 125 81.1 901 120 73 180 80 141.58 125 81.1 901 120 73 180 80 132.37 125 81.1 901 120 73 180 80 156.93 125 81.1 901 120 73 180 80 125 138.88 81.1 901 120 73 180 81 901 120 73 180 80 125 141.71 81.1 901 120 73 180 80 125 120 73 180 80 125 125 60.582 901 120 73 180 80 125 125 70.962 901 120 73 180 80 125 125 64.854 901 120 73 180 80 125 125 73 180 80 125 125 81.1 104.841 120 73 180 80 125 125 81.1 111.22 125 81.1 98.0921 120 73 180 80 125 125 81.1 120.44]; [B, BINT, E, EINT, Stats ] = regress (y, x, 0.05) Rcoplot (E, EINT) execution results: b = 120.8481 0.0376 -0.0607 -0.0781 0.0929 0.0466 -0.0003 0.1664 0.0004% regression equation is y = b0 b1x ... bpxbint = 119.8808 121.8154% B0 -p confidence interval;
0.2357 0.2394 -0.0649 -0.0565 -0.0795 -0.0766 0.0896 0.0962 0.0446 0.0487 -0.0031 0.0025 0.1631 0.1696 -0.0019 0.0027e = 0.0628 0.0396 0.0650 -0.0278 -0.0025 -0.0676 0.0215 0.0241 -0.0088 -0.0160 0.0452 0.0105 -0.0338 -0.0424 -0.0128 -0.0415 - 0.0030 0.0038 0.0391 0.0037 -0.0332 -0.0304 0.0446 -0.0324 0.0298 -0.0204 0.0461 -0.0107 -0.0446 0.0170 -0.0055 -0.0204 0.0009 percent observations yi its estimated residual vector; eint = -0.0104 0.1359% residual estimation interval; -0.0364 0.1156 -0.0086 0.1385 -0.0756 0.0199 -0.0727 0.0677 -0.1399 0.0047 -0.0569 0.0998 -0.0213 0.0695 -0.0794 0.0617 -0.0640 0.0320 -0.0299 0.1203 -0.0601 0.0810 -0.1113 0.0438 -0.1164 0.0317 -0.0724 0.0468 -0.1140 0.0311 -0.0547 0.0486 -0.0634 0.0710 -0.1050 0.0442 -0.0380 0.1162 -0.0533 0.0608 -0. 1061 0.0397 -0.0319 0.1210 -0.1018 0.0371 -0.0238 0.0835 -0.0069 0.0991 -0.0867 0.0654 -0.1093 0.0202 -0.0987 0.0578 -0.0586 0.0926 -0.0755 0.0645 -0.0986 0.0579 -0.0528 0.0547stats = 1.0e 004 * 0.0001 1.6029 0% correlation coefficient R = 1, F statistic value is% 16029 significant root rate P = 0; analysis: reference
