利用ATP做大型电网仿真 一直出现如下错误 1 T: U7 E0 ]7 i0 l2 T; b3 \9 r% @
++++ Type-59 S.M. number 44 begins operation on segments 0 and 0. , C9 `/ ]( j7 p1 L# w( T* D REDUCT. M, N, J, IJ = 0 3 3 6" X) g F0 j) | U
4 B# u8 R, K. O: v. R. _( u8 P------------------------------------------------------------------------------------------------------------------------------------ $ l- q# N q* j ~% v* `" X; [ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ 9 _, D v3 C3 I% k/ s) I9 YERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ERROR/ ) I7 q/ q6 u& z" T------------------------------------------------------------------------------------------------------------------------------------5 l4 H- \9 k1 p# B6 L" j
You lose, fella. The EMTP logic has detected an error condition, and is now going to terminate program execution. The following 2 b& x/ w& ]0 amessage summarizes the circumstances leading to this situation. Where an otherwise-unidentified data card is referred to, or where ' A9 G( k* `/ f# a, L) ]' Nthe "last" card is mentioned, it is the most recently read card of the input data that is meant. The 80-column image of this card! q) w2 v6 y; s; A* P+ N. h
is generally the last one printed out prior to this termination message. But possibly this last-read card has not yet been% d; ^6 E k- ?( e
displayed, so a copy follows:+ B; W6 \( Q0 C' @7 n
" " ! z+ b* V, k! {' h, M& i7 P6 ] KILL code number Overlay number Nearby statement number / m8 w- `+ {" A- Q 86 13 4233 $ F+ I1 _; l. A3 D/ NKILL = 86. Singularity has been detected within SUBROUTINE REDUCT, which does Gaussian elimination. The diagonal element for7 a8 |9 o& C) F3 A- K, i- X! Q
row 3 is equal to -1.#IND0000E+000, which is less than the tolerance EPSILN of the floating-point miscellaneous data card. : q8 J* U: O n. U! [: M: A. Y9 U% A. WOther local variables that might be valuable to Program Maintenance in interpreting the trouble are as follows : N = 3 M = 06 v4 |+ S3 ?2 O# f+ ^2 O
.+ o' _: E9 m n1 S
这是哪里的问题???