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302<br />

( )<br />

C.-L. Tseng et al.rDecision Support Systems 24 1999 297–310<br />

correspond<strong>in</strong>g to the most unsatisfied capacity constra<strong>in</strong>t and <strong>in</strong>creas<strong>in</strong>g only the multiplier mt<br />

of this hour. This<br />

approach can be modified to simultaneously updat<strong>in</strong>g the multipliers m t’s of it all hours that the sp<strong>in</strong>n<strong>in</strong>g<br />

capacity requirement is violated. Such a modification can speed up the feasibility phase at the cost of<br />

over<strong>commitment</strong> w5,14 x. Under the three-phase scheme, this over<strong>commitment</strong> effect can nevertheless be<br />

corrected <strong>in</strong> the f<strong>in</strong>al <strong>unit</strong> de<strong>commitment</strong> phase to be addressed <strong>in</strong> a later section.<br />

The feasibility phase presented <strong>in</strong> this paper conta<strong>in</strong>s two parts. Part 1 seeks a reserve-feasible <strong>commitment</strong><br />

by <strong>in</strong>creas<strong>in</strong>g the multipliers mt<br />

of all hours with <strong>in</strong>sufficient sp<strong>in</strong>n<strong>in</strong>g capacities. Start<strong>in</strong>g from this reserve-<br />

feasible <strong>commitment</strong> obta<strong>in</strong>ed, Part 2 solves a l<strong>in</strong>ear program Ž LP.<br />

for each hour to verify if this <strong>commitment</strong> is<br />

also dispatchable at this hour. If not, <strong>in</strong>formation can be extracted from the solution of the LP to determ<strong>in</strong>e how<br />

much additional sp<strong>in</strong>n<strong>in</strong>g capacity at this hour is required so that a dispatchable <strong>commitment</strong> exists. The<br />

sp<strong>in</strong>n<strong>in</strong>g reserve requirement of this hour is thus uplift. Part 1 is then rerun with respect to the new Ž uplift.<br />

sp<strong>in</strong>n<strong>in</strong>g reserve requirements. Suppose Part 1 of the feasibility phase term<strong>in</strong>ates at iteration k with an obta<strong>in</strong>ed<br />

<strong>commitment</strong> u k<br />

4<br />

it . Let qbt<br />

be the variable represent<strong>in</strong>g the total generation to be dispatched to bus b <strong>in</strong> time t.<br />

k<br />

4<br />

m<strong>in</strong> m<strong>in</strong> k max max k<br />

Given a <strong>commitment</strong> uit let qbt sÝig L pi uit and qbt sÝig L pi u it, which form an immediate<br />

b<br />

b<br />

lower bound and upper bound for q bt.<br />

ŽLPŽ t..<br />

Ž .<br />

m<strong>in</strong> Ý B y yq max Ž 19.<br />

q bt, ybt<br />

bs1 bt bt<br />

s.t. qbt m<strong>in</strong> FqbtFy bt, ;b Ž 20a.<br />

q max Fy FÝ p max , ;b Ž 20b.<br />

bt bt ig L b i<br />

Ýbs1 B qbtsDt<br />

Ž 20c.<br />

yFlFÝbs1 B G l bŽ qbtyDbt. FF l , ; l Ž 20d.<br />

q Gr D , ;b Ž 20e.<br />

bt bt bt<br />

Ž Ž ..<br />

max<br />

In LP t , y is a variable upper bound of both q and q . Let y ) be the solution of ŽLPŽ t ...Ifu k<br />

4<br />

bt bt bt bt it is<br />

max<br />

dispatchable <strong>in</strong> time t, y )sq solves ŽLPŽ t.. and the optimal objective value of ŽLPŽ t.. is zero. If u k<br />

4<br />

bt bt it is<br />

not dispatchable <strong>in</strong> time t, ŽLPŽ t..<br />

yields a positive objective value, which <strong>in</strong>dicates that to be dispatchable,<br />

more <strong>unit</strong>s should be committed <strong>in</strong> time t, and the target value of the new bus capacity requirement is now<br />

<strong>in</strong>creased to y bt) for each bus b <strong>in</strong> time t. That is, an additional bus sp<strong>in</strong>n<strong>in</strong>g capacity requirement is imposed<br />

to each bus subproblem d Ž P .:<br />

Ý i it bt<br />

b<br />

p max u Gy ). Ž 21.<br />

igL b<br />

Ž .<br />

Part 1 of the feasibility phase is therefore rerun with respect to the updated capacity requirement 21 . The phase<br />

2 algorithm is summarized below.<br />

3.2.1. Phase 2 algorithm<br />

Step 0: k§0; m 0 and r 0 and other multipliers are from Phase 1.<br />

k k<br />

Ž . Ž k k .<br />

Step 1: Given m and r and other multipliers, solve the dual subproblems 14 to obta<strong>in</strong> u , p .<br />

Ž. Ž.<br />

Step 2: If 3 and 6 are satisfied, go to Step 5 if Step 5 has not been visited, otherwise stop.<br />

kq1 k k<br />

Ž<br />

B max k<br />

.<br />

kq1 k k<br />

Step 3: For ;b,t, m §m qs Pm<strong>in</strong> 0, R yÝ Ý p u ; r §r qs Pm<strong>in</strong>Ž<br />

t t t bs1 i g L i it bt bt 0,sbtRbty<br />

b<br />

max<br />

Ý p u k<br />

..<br />

i g L b i it

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