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# PWR SYS OPERATNCTRL EEL 6266

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This 69 page Class Notes was uploaded by Tracy Okuneva Sr. on Thursday September 17, 2015. The Class Notes belongs to EEL 6266 at Florida State University taught by Staff in Fall. Since its upload, it has received 80 views. For similar materials see /class/205575/eel-6266-florida-state-university in Electrical Engineering at Florida State University.

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Date Created: 09/17/15

EEL 6266 Power System Operation and Control Chapter 6 Generation with Limited Energy Supply 0 Hard limits and slack variables take account of the limits on the takeorpay generating unit PTmin S PT S PT this may be added to the Lagrangian by using two new functions and two new variables called slack variables 111 19 PT 55 and 912 PT PT 522 maX maX min 0 where S 1 and 52 are the slack variables 0 they may take on any real value including zero the new Lagrangian jmax N jmax N jmax L Z jlloadj ZPij PTj Zltnquj qtotal jl i1 jl i1 jl jmax 2 jmax 2 ZaljPTj PTmaX SljZa2jPTmin PTj S2j 1 j i1 0 Hard limits and slack variables 051 and 052 are Lagrange multipliers 39 the rst partial derivatives for the kth interval are dF all 0 nk 1k 813239 sz k 8L dam o yn 46 a a aka k Tk 1k 2k BL 8a 2 0 205118911 1k BL 8a 2 0 ZaZkSZk 2k 0 when the constrained variable PM is Within bounds 051 062 and SI and 52 are nonzero when PM is limited either SI or 2 is zero and the associated Lagrange multiplier is nonzero O 0 Hard limits and slack variables 39 consider some interval k Where PM P max 0 51k 0 and alki 0 then 161 Tk i051kynk 0 oif amp gt Wk 161 dPT k the value of 051k takes on the value just sufficient to make the equality true 0 Example 39 reconsider the 6interva1 24hour fuel scheduling a maximum generation on PT is reduced to 300 MW I in the original optimal schedule PT 3533 in the third time period I when the limit is reduced to 300 MW the gas red unit bums more fuel in other time periods to meet the 40 M ft3 gas consumption constraint 0 Example 0 resulting optimal schedule with PtmaX 300 MW 3500 3000 594 586 008 5000 3000 630 586 044 595 1405 524 524 0 1214 1786 539 539 O 1 2 3 4 2454 2546 569 569 0 5 6 a shadow price 7 08603 0 total cost 122985 o Fuel scheduling a the major elements in the fuel supply and delivery system c rawfuel suppliers I coal oil and gas companies I longterm contracts with minmax limits on fuel per timeperiod I prices may change subject to renegotiation provisions in contract 0 transportation I railroads river barges gaspipeline companies I represent problems in scheduling of fuel deliveries a fuel storage farms fuel inventory I coal piles oil storage tanks underground gas storage facilities I keeping proper inventory levels to forestall fuel shortages 0 load levels exceed forecast suppliers or shippers unable to deliver a generation stations o Fuel scheduling fuel scheduling problem 0 the total time period is broken into discrete time increments o constraint functions I some constrain functions Will span two or more time steps solution technique 0 linear programming I procedure that minimizes a linear objective function I variables are also subject to linear constraints I any nonlinear functions objective or constrain equations must be approximated by linearization about an operating point or by piecewise linear functions 0 Linear programming 39 optimization function Z 01x1 02x2 chN 0 linear constraints a11x1 aux2 a1NxN S 1 aZIX1 6122x2 aZNxN S 2 aM1x1 aM2x2 aMNxN S 9M and the variables may having speci ed upper and lower limits xinin S x1 S xlmax min maX x2 S x2 S x2 min max xN S xN S xN 0 One of many Linear programming LP techniques upperbound linear programming algorithm 0 variable limits are handled implicitly 0 requires a slack variable with each constraint the slack variables a used primarily in inequality constraints 0 equals the difference or slack between a constraint and its limit a transforms an inequality constraint into an equality constraint 4x1 6x2 S 28 a4x16x2xs 28 OSxS ltltgtltgt I ifx1 3 andx2 2 thenxS4 I for a greater than or equal to constraint the bounds on the slack variable are changed 39 equality constraints may contain a slack variable with bounds at 0 0 Linear programming linear equations structure canonical form 0 the objective function and constraints are arranged in a matrix all 6112 alN 1 x31 b1 x1 a2 6 22 aZN 1 x52 b2 x2 aNl a12 aNN 1 xSN bN xN c1 02 CN 1 Z 0 0 there is at least one variable in each constraint Whose coefficient is zero in all the other constraints I this set of variables is called the basis variables I the remaining variables are called nonbasis variables 0 Linear programming the solution procedure centers on conducting pivot operations 0 a pivot exchanges a nonbasis variable for a basis variable pivoting 0 consider the operation for pivoting variable x1 for x52 I that is equivalent to pivoting column 1 with row 2 0 steps I multiply the elements in row 2 by l 6121 61 aZJa21 j lN b b26121 39 for each row 1 1 7t 2 subtract from row 1 the product of row 2 and an 611 at al1aj j lN 1 bl al1b i1Nl 2 I I c c cla2j j lN 0 Linear programming 39 the result of the pivoting operation 0 a12 alN 1 61152 x51 b1 1 I I 1 I b 6 22 39 aZN x azsz x52 2 I o 0 5 12 aNN x Clst 1 xSN bN N 0 c cg 0amp2 1 Z z 0 x1 is now one of the basis variables 0 Linear programming the dual upperbounding algorithm proceeds in simple steps 0 variables that are in the basis set are exchanged for variables in the nonbasis set using the pivoting operation a the nonbasis variables are held equal to either their upper or lower bounds o the basis variables are allowed to take any value without respect to their bounds 9 the solution terminates when all the basis variables are within their respective limits o The LP algorithm starting steps 0 for each variable that has a nonzero coefficient c in the objective function the cost row must be set according to the following rule I if c gt 0 setxj xjmi else ifcj lt 0 set x xjmaX I if c 0 x can be set to any value for convenience it is set to xjmi 0 add a slack variable x5 to each constraint in the problem I using the values of x assigned in the above step set the sleek variables to make each constraint equal to its limit variable exchange 0 row selection find the basis variable with the greatest violation I this is the row to pivot on I identi ed as row R o The LP algorithm variable exchange using R with the worst violation 0 column selection 1 use if constrain row R is below its minimum I pick column S so that cS aRaS is minimum for all S that satisfy 0 S is not in the current basis 0 aRS is not equal to zero 0 if xS is at its minimum then 61KS lt O and cS 2 0 0 if xS is at its maximum then 61KS gt O and cS S O 0 column selection 2 use if constrain row R is above its maximum I pick column S so that cS aRaS is minimum for all S that satisfy 0 S is not in the current basis 0 aRS is not equal to zero 0 if xS is at its minimum then 61KS gt O and cS S 0 0 if xS is at its maximum then 61KS lt O and cS 2 O o The LP algorithm variable exchange using row R and column S o pivot at the selected row R and column S o the pivot column s variable x5 goes into the basis a if no column ts the column selection criteria the problem has an infeasible solution a no values for x1xN will satisfy all constraints simultaneously setting the variables after pivoting a all nonbasis variables except for x5 hold there current values 0 the most violated variable xR is set to the limit that was violated a all nonbasis variables are determined set each basis variable to Whatever value is required to make the constraints balance I all basis variables may change many may now violate the limits search the basis variables The algorlthm for the one with the worst violation designate it as R 39 any violation Yes moist violated variable is above its maximum most violated variable is below its minimum pick column S using pick column S using the column selection the column selection procedure for most procedure for most violated variable above violated variable below its maximum its minimum 39 infeasible v infeasible solution solution pivot on selected row and column 0 LP example 39 minimize Z 2x1 x2 39 subject to x1 x2 20 constraint 1 14x1 x2 S 2 constraint 2 2Sx1S 12 1imit1 2 S x2 S 16 limit 2 14x l x S2 1 2 cost contours k X x216 0 LP example 39 problem cast into canonic form constraint 1 x1 constraint 2 1 4x1 cost 2x1 minimum 2 present value 2 maximum 12 x2 x3 20 x2 x4 2 x2 Z 0 OSX3SO 03x4lt 2 0 0 2 16 28 6 l6 0 a variable 3 is the worstviolated basis variable R l exceeds its maximum of 0 0 using column selection procedure pivot at column 2 S 2 S 1 a11 gt0 x1x1min 01 gt0 thencla112l 2 s2 a12gt0 x2x2min CZgtO thencZa12lll 0 LP example 0 pivoting results in constraint 1 x1 constraint 2 24x1 cost x1 minimum 2 present value 2 maximum 12 2 2 18 16 x3 0 0 0 0 132 00 22 a variable 4 is the worstviolated basis variable R 2 exceeds its minimum of 0 0 using column selection procedure pivot at column 1 S 1 s 1 6121lt 0 x1x1min 01gt 0 then cl a21124 04166 s 3 6122 lt 0 x3 x3maX 02 lt 0 then x3 is not eligible 0 LP example 0 pivoting results in constraint 1 x2 583x3 4l7 x4 constraint 2 x1 Jr4l7x3 4l7 x4 cost l4l7x3 4l7 x4 minimum 2 2 0 0 present value 75 125 0 0 maximum 12 16 0 00 275 125 75 275 0 3x3 S 0 0 S x4 ltgto o no violations among the basis variables 0 algorithm stops at the optimum o Fuel scheduling by LP a two coalburning generating units 0 both must remain online for a 3 week period c the combined output is to supply the following load I week 1 1200 MW I week 2 1500 MW I week 3 800 MW 0 one coal supplier is under contract to supply 40000 tons per week to be split between the two plants I there are existing inventories at the start of the 3week period nd the following o the operation schedule for each plant for each week 0 the coal delivery amounts to be made each week 0 Fuel scheduling by LP 39 data a coal I heat value 23 Mbtuton I cost 30ton or 13Mbtu I inventories plant 1 70000 tons plant 2 70000 tons I each plant has a maximum coal storage of 200000 ton o generators 39 plant 1 H1P1 3800 8267 P1 150 S 1D1 S 600 MW q1P1 1652 03594 P1 tonsh F1P1 49565 1078 P1 h quot plant 2 H2P2 5833 8167 P2 400 S 1D2 S 1000 MW q2P2 2536 03551 P2 tonsh F2P2 7608 1065 P2 SSh 0 Fuel scheduling by LP solution process 0 assume that the units are operated at a constant rate during each week I coal deliveries are made at the beginning of each week I the problem is set up for lWeek time intervals o then I plant 1 q1P1 27754 60 4 P1 tonsWk F1P1 832692 1811 P1 Wk I plant 2 q2P2 42605 597 P2 tonsWk F2P21278144 1789 P2 Wk objective function I minimize Z F1P11 F2P21 F1P12 F2P22 F11P13 F21P23 0 Fuel scheduling by LP 39 constraints a power delivery P11P21 1200 1912 P22 1500 1913 P23 800 0 coal deliveries 011 021 44000 012 022 44000 013 023 44000 0 volume of coal V11 011 q11 V12 V21 021 921 V22 V12 012 912 V13 V22 022 922 V23 V13 013 913 V14 V23 023 923 V24 EEL 6266 Power System Operation and Control Chapter 3 Numerical Methods for Economic Dispatch o The solution to the optimal dispatch can be approached by graphical methods plot the incremental cost characteristics for each generator the operating points must have minimum cost and satisfy load 0 that is find an incremental cost rate 7 that meets the demand PR 0 graphically P1 MW APZ MW P3 MW 2 J gtPRP1P2P3 0 An iterative process assume an incremental cost rate 7 and nd the sum of the power outputs for this rate 0 the first estimate will be 1ncorrect if the total power output is too low increase the 7 value or if too high decrease the 7 value c with two solutions a closer value of total power can be extrapolated or interpolated the steps are repeated until the desired output is reached error e 1 solution e lt tolerance 0 M2 M3 M1 7 l I 2 N e PR i1 Lambda projection o This procedure can be adopted for a computer implementation the implementation of the power gt fgflfglitfop output calculation is rather independent of the solution method calculate 0 each generator output could be N solved by a different method as an iterative procedure a stopping criterion must be established 0 two general stopping rules are appropriate for this application I total output power is within a speci ed tolerance of the load demand I iteration loop count exceeds a maximum value 0 Example consider the use of cubic functions to represent the input output characteristics of generating plants H MBtuh A BP CP2 D193 P in MW for three generating units nd the optimum schedule for a 2500 MW load demand using the lambdaiteration method 0 generator characteristics A B C D P P max min Unit 1 74955 695 968gtlt10394 127gtlt10397 320 800 Unit 2 12850 7051 7375gtlt10394 6453gtlt10398 300 1200 Unit 3 15310 6531 104gtlt10393 998 gtlt10398 275 1100 o assume that the fuel cost to be 1MBtu 0 set the value of 7 on the second iteration at 10 above or below the starting value depending on the sign of the error Lg Flor c ia University 0 Example 9 initial iteration Xstart 80 o incremental cost functions A 2 136131 2 695 2968gtlt104P1 3127gtlt107Pl2 A dFZdPZ 7051 27375gtlt104P2 36453gtlt108P2 A dF3dP3 6531 2104gtlt10 3P3 3998gtlt10 8P32 0 find the roots of the three incremental cost functions at 7 80 P1 55756 4943 P2 82159 5967 P3 75934 6462 39 use only the positive values Within the range of the generator upper and lower output limits 0 calculate the error e 2500 4943 5967 6462 7629 MWh o with a positive error set second 7 at 10 above ksm M2 88 0 Example 8 second iteration M2 88 0 find the roots of the three incremental cost functions at 7 88 P1 5904 8225 P2 8662 10430133 7906 9586 0 calculate the error e 2500 8225 1043 9586 3240 MWh I error out of tolerance 0 project 7 2 1 AB 2ebmg l 2amp0 324088 85615 6 e 76293240 0 continue with third iteration 0 Example results of all iterations Iteration K Total Generation P1 P2 P3 1 80 17372 4943 5967 6462 2 88 28241 8225 10430 9586 3 85615 25102 7281 9143 8678 4 85537 24999 7250 9101 8648 0 Issues under some initial starting points the lambdaiteration approach exhibits an oscillatory behavior resulting in a non converging solution 0 try the example again with a starting point of km 100 39 Flor c la University 0 Suppose that the cost function is more complex P a4 example FP a0 a1 P a2 P25 a3 e 5 t the lambda search technique requires the solution of the generator output power for a given incremental cost 0 possible with a quadratic function or piecewise linear function 0 hard for complicated functions we need a more basic method 0 The gradient search method uses the principle that the minimum is found by taking steps in a downward direction from any starting point ado one finds the direction of steepest descent by computing the aFdx1 negative gradient of F at ado VF 3cm 2 3 BFdxn to move in the direction of maximum descent from ado to xm xu x0 anx0 o ais a scalar that when properly selected guarantees that the process converges o the best value of amust be determined by experiment a for the economic dispatch N N problem the gradient L I Z pi 11le 2 p technique is applied directly i1 to the Lagrange function aLapl ddagt j11a A the gradient function is VL aLaPN ddeFN gva 1 0 this formulation does not aL 8 1 P Wd ZPI enforce the constraint function 0 Example solve the economic dispatch for a total load of 800 MW using these generator cost functions F1031 16832376Pl 0004686Pf F2002 930 2355P2 000582Pf F3003 2342370P3 001446P32 0 use 06 100 and starting from PE 2 300 MW 13 200 MW and P3 300 MW 0 7c is initially set to the average of the incremental costs of the generators at their starting generation values 3 d 123760009372300 0Z FiPi0 2355001164200 22827 I 1 2370002892300 Example 3E gendata 1683 2376 0004686 930 2355 000582 234 2370 001446 1 Matlab program to perform the power 300 200 300 alpha a 100 plead a 800 grad1ent search method find lambdaO n length gendata lambdaO 0 for i l n lambdaO lambdaO gendatai2 2 gendatai3 poweri end lambdaO lambdaO 3 clear X0 X0 power X0nl lambdaO calculate the gradient for kk l 10 dispkk clear gradient gradient Pgen 0 cost 0 for i l n gradienti gendatai2 2 gendatai3 X0i X0nl Pgen Pgen X0i cost cost gendatail gendatai2 X0i gendatai3 X0i X0i end gradientnl Pload Pgen disp X0 Pgen cost1000 X1 X0 gradient alpha X0 X1 0 Example the progress of the gradient search is shown in the table below Iteration 7 Total Generation P1 P2 P3 Cost 1 2828 8000 3000 2000 3000 23751 2 2828 8000 3017 2024 2959 23726 3 2828 8001 3034 2048 2919 23704 4 2835 8002 3051 2071 2881 23685 5 2857 8007 3068 2095 2844 23676 6 2923 8018 3087 2121 2810 23687 7 3106 8050 3113 2153 2784 23757 8 3608 8137 3157 2203 2777 23983 9 4979 8374 3251 2303 2821 24632 10 8719 9019 3480 2538 3000 26449 note that there is no convergence to a solution Florida University o A simple variation realize that one of the generators is always a dependent variable and remove it from the problem 0 for example picking P3 then P3 800 P1 P2 0 then the total cost function becomes C Fi131EPgF3800Pi 4 2 0 this function stands by itself as a function of two variables with no loadgeneration balance constraint d 1E dF3 I the cost can be minimized d PIC d Pl d Pl by a gradlent method such as VC 2 i C E dP2 dP2 dP2 I note that the gradient goes to zero When the incremental cost at generator 3 is equal to that at generators l and 2 o A simple variation the gradient steps are performed in like manner as before xm xm VC 0 and W x 2 P2 0 Example rework the previous example with the reduced gradient E E VC dp1 dp1 2376 20004686P1 2370 2001446800 P1 192 E g 2355 2000582P2 2370 2001446800 P1 P2 sz sz or is set to 2000 Example 3F gendata 1683 2376 0004686 930 2355 000582 0 Ma ab pI39OgI am to perform the 234 2370 001446 1 power soo zoo zoo 1 s1mp11f1ed grad1ent search method alpha 2000 Pload 800 form lambdaO n length gendata clear X0 X0 powerln l calculate the gradient for kk l 10 dispkk clear gradient gradient Pn Pload for i l n l Pn Pn X0i end cost gendatanl gendatan2 Pn gendatan3 Pn Pn for i l n l gradienti gendatai2 2 gendatai3 X0i gendatan2 2 gendatan3 Pn cost cost gendatail gendatai2 X0i gendatai3 X0i X0i end disp X0 Pn 800 cost1000 X1 X0 gradient alpha X0 Xl 0 Example the progress of the simpli ed gradient search is shown in the table below Iteration Total Generation P1 P2 P3 Cost 1 8000 3000 2000 3000 23751 2 8000 4161 3300 540 23269 3 8000 3681 2874 1445 23204 4 8000 3815 3071 1114 23194 5 8000 3733 3030 1237 23193 6 8000 3736 3070 1193 23192 7 8000 3714 3076 1210 23192 8 8000 3706 3090 1204 23192 9 8000 3698 3097 1207 23192 10 8000 3689 3104 1206 23192 note that there is a solution convergence by the 6th iteration Florida University o The solution process can be taken one step further observe that the aim is to always drive the gradient to zero VLx 0 since this is just a vector function Newton s method nds the correction that exactly drives the gradient to zero 0 Review of Newton s method suppose it is desired to drive the function gx to zero 0 the first two terms of the Taylor s series suggest the following gxAxgxg xle0 g1xljjxn o the objective function gx is defined as gx E gn xl xn c then the Jacobian is agla xl agla xn g39ltxgt i l a the adjustment at each iteration step is Ax g39x gx if the function g is the gradient vector VLx then Ax 1 AL Bx M2 N o For economic dispatch problems L ZEB 4PM 2 39 d2L d2L i1 i1 and a d2L d2L a xVLx dxtdxl we diL diL dldx1 dldxz 0 note that in general one Newton step solves for a correction that is closer to the minimum than would the gradient method 0 Example solve the previous economic dispatch problem example using the Newton s method 0 the gradient function is the same as in the first example I let the initial value of szl A be equal to zero 12 0 0 1 o the Hessian matrix O dZFZ O 1 takes the following form H l 01P22 11332 0 the initial generation values 1 1 1 0 are also the same as in the first example Example 3G gendata 1683 2376 004686 0 930 2355 000582 0 Ma ab pI39OgI am to perform the 234 2370 001446 1 7 Newton 5 method power 300 200 300 Pload 800 form H n length gendata H zerosnlnl for i l n Hii gendatai3 2 Hinl l Hnli 1 end X0 zerosnll XOlnl transpose power calculate the gradient and Hessian matrices for kk l lO dispkk gradient zerosnll gradientnll Pload for i l n gradientil gendatai2 2 gendatai3 XOil X0nll gradientnll gradientnll X0il end dX H gradient cost O for i l n cost cost gendatail gendatai2 X0i gendatai3 XOi XOi end disp XO cost1000 X0 x0 dX end 0 Example the progress of the gradient search is shown in the table below Iteration 7 Total Generation P1 P2 P3 Cost 1 000 8000 3000 2000 3000 23751 2 2719 8000 3663 3130 1207 23192 3 2719 8000 3663 3130 1207 23192 4 2719 8000 3663 3130 1207 23192 note the quick convergence to a solution compare with the solution of the previous example 39 Flor c la University EEL 6266 Power System Operation and Control Chapter 5 Unit Commitment 0 Dynamic programming chief advantage over enumeration schemes is the reduction in the dimensionality of the problem a in a strict priority order scheme there are only N combinations to try for an N unit system a strict priority list would result in a theoretically correct dispatch and commitment only if c the noload costs are zero 0 unit inputoutput characteristics are linear 0 there are no other limits constraints or restrictions 0 startup costs are a fixed amount 0 Dynamic programming the following assumptions are made in this implementation of the DP approach a a state consists of an array of units I With speci ed units operating and the rest decommitted offline I a feasible state is one in Which the committed units can supply the required load and meets the minimum capacity for each period a startup costs are independent of the off line or downtime I ie it is a xed amount Wrt time o no unit shuttingdown costs a a strict priority order will be used Within each interval a a specified minimum amount of capacity must be operating within each interval o The forward DP approach runs forward in time from the initial hour to the nal hour 0 the problem could run from the final hour back to the initial hour 0 the forward approach can handle a unit s startup costs that are a function of the time it has been off line temperature dependent I the forward approach can readily account for the system s history 0 initial conditions are easier to specified when going forward the minimum cost function for hour K with combination I F KI LinP KIS K 1L K1F K 1L cost cost cost cost 0 costK D least total cost to arrive at state K I o PcostK D production cost for state K I o ScostK l L K I transition cost from state K l L to K I o The forward DP approach state K I is the 1th commitment combination in hour K s a strategy is the transition or path from one state at a given hour to a state at the next hour 0 X is defined as the number of states to search each period c N is defined as the number of strategies to be saved at each step I these variable allow control of the computational effort I for complete enumeration the maximum value of X or N is 2N l I for a simple prioritylist ordering the upper bound on X is n the number of units 0 reducing N means that information is discarded about the highest cost schedules at each interval and saving only the lowest N paths or strategies I there is no assurance that the theoretical optimum Will be found o The forward DP approach restricted o 0 search paths 0 Q N 3 K I X 5 lt39i igt Xr GO Interval Interval K 1 K X3 0 5 0 Example consider a system with 4 units to serve an 8 hour load pattern Incremental NoIoad Fullload Min Time Unit Pmax Pmin Heat Rate Cost Ave Cost h MW MW Btu kWh h mWh Up Down 1 80 25 10440 21300 2354 4 2 2 250 60 9000 58562 2034 5 3 3 300 75 8730 68474 1974 5 4 4 60 20 11900 25200 2800 1 1 Initial Condition Startup Costs Hour Load MW Unit off on Hot Cold Cold start 1 450 h h 2 530 1 5 150 350 4 3 600 2 8 170 400 5 4 540 3 8 500 1100 5 5 400 4 6 0 002 0 6 280 7 290 8 500 0 Example to simplify the generator cost function a straight line incremental curve is used a the units in this example have linear F P functions dF Eta load o the units must operate Within their limits Mama F noJoad min max 0 Case 1 Strict prioritylist ordering the only states examined each hour State Un39t Status Capac39ty 5 o o 1 o 300 MW cons1st of the 11sted four 12 O 1 1 0 550 MW state 5 unit 3 state 12 3 2 14 1 1 1 O 630 MW state 14 3 2 1 state 15 all four 15 1151 1 690 MW all possible commitments start from state 12 initial condition minimum unit up and down times are ignored in hour 1 0 possible states that meet load demand 450 MW 12 14 amp 15 Pcost1 15 E25 F105 113300 E20 Economic Dispatch Eq 173536 20882518001051746300 238020 986136 115 P l l5 S 012 115 DP State Transition Eq cost cost F cost 986136 350002 1021138 o Casel in hour 1 K Pcost Scost Foost 15 9861 350 10211 14 9493 350 9843 12 9208 0 9208 a minimum at state 12 9208 in hour 2 0 possible states that meet load demand 530 MW 12 14 amp 15 P cost F cost 215 E25172185E300E20 1735 20882518001851746300238020 11301 215P 2 15 Scost1 L 215 DP State Transition Eq 3509208 cost 20859 11301mm 09843 010211 0 Case 1 DP diagram state units tatus hourO hour1 hour2 hour3 hour4 hour5 hour6 hour hour8 ca gacitx 450 MW 530 MW 600 MW 540 MW 400 MW 280 MW 290 MW 500 MW 5 V 39 t i I 1 1 1 1 690 W 0 350 350 350 350 350 750 750 14 0 1 1 1 O v 1 39 630 MW 35 35 35 0 350 350 585131000000 total cost 73439 0 priority order list up times and downtimes neglected 0 Case 2 complete enumeration 256 X 109 possibilities o fortunately most are not feasible because they do not supply sufficient capacity in this case the true optimal commitment is found 0 the only difference in the two trajectories occurs in hour 3 I it is less expensive to turn on the less ef cient peaking unit 4 for three hours than to start up the more ef cient unit 1 for that same time period 0 only minor improvement to the total cost quot case 1 73439 39 case 2 73274 0 Case 2 DP diagram t t uni t taetus hourO hour1 hour2 hour3 hour4 03930 450 MW 530 MW 600 MW 540 MW 15 V g 1111 690MW 35 35 35 5 0010 300MW o Lagrange Relaxation dual variables and dual optimization 0 consider the classical constrained optimization problem I primal problem minimize x1 xn subject to ax1 xn 0 I the Lagrangian function Lx1 xn fx1 xn ax1 xn 0 define a dual function qi min Lx1x2i I then the I dilal problem is to nd 61W I335 W1 0 the solution involves two separate optimization problems I in the case of convex functions this procedure is guaranteed to solve the problem 0 Example 39 minimize x1x2 025x12 x22 subject to 03x1x2 5 x1 x2 0 the Lagrangian function Lx1 x2 1 025x12 x 15 x1 x2 39 the dual function ql min Lx1x2l gt x1 21 amp x2 2 ql 12 51 the dual problem q1 r5190 q1 gt 41 gt 5 0 xi 2 o Iterative form of Lagrange relaxation method the optimization may contain nonlinear or nonconvex functions iterative process based on incremental improvements of xi is required to solve the problem 0 select a arbitrary starting xi 0 solve the dual problem such that ql becomes larger 0 update tusing a gradient adjustment 1 1 iqM a 0 find closeness to the solution by comparing d1 the gap between the primal function and the dual function I primal function f min L 39 relative duality gap J j q in practice the gap never reaches zero o Lagrange relaxation for unit commitment 39 loading constraint N Pgad ZP U 0 Vt1T i1 39 unit limits UfPimi 3 13 s Ummi Vi1Nandt1T unit minimum uptime and down time constraints the objective function iilt13it Sstart upit Uz39t FPtUt i1 i1 0 Formation of the Lagrange function e in a similar way to the economic dispatch problem T N mam FP U Exiled Pi tl i1 unit commitment requires that the minimization of the Lagrange function subject to all the constraints a the cost function and the unit constraints are each separated over the set of units I What is done With one unit does not affect the cost of running another unit as far as the cost function unit limits and the uptime and downtime constraints are concerned 0 the loading constraint is a coupling constraint across all the units the Lagrange relaxation procedure solves the unit commitment by temporarily ignoring the coupling constraint o The dual procedure attempts to reach the constrained optimum by maximizing the Lagrangian with respect to the Lagrange multiplier gill max qi where qi LPU A done in two basic steps 1 I 0 Step 1 find a value for each xl which moves ql towards a larger value 0 Step 2 assuming that xl found in Step 1 is fixed find the minimum of L by adjusting the values of Pt and U minimizing L T N T T N LZZlEltHgtSmlUfZIBZQd2113 U tl 39 tl i1 M II Mz Ms M M II iii113 SmlUi Mat Uil iazad Il separation of the units from one another the inside term can now be solved independently for each generating unit T ZEB SulU we U tl the minimum of the Lagrangian is found by solving for the minimum for each generating unit over all time periods N T min qltzgt Emil EB SulU 2 13 UJ i1 tl subject to the uptime and downtime constraints and Ufgmi s P s UfPimi Vt 1T U 10 a this is easily solved as a two state dynamic programming U1 0 A A problem of one variable 1 I 2 I 3 I 4

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