Abstract: . . .  by solving the 4×4 sub-problem (12 multiplicative operations and 12 addition operations) = q p q u p u q q u p p u y y y y q q q q p p p p ? ? ? ? ) ( ) ( ) ( ) ( ) 4 ( )3 ( ) 2 ( )1 ( ) 4 ( )3 ( )2 ( )1 ( 1 1 0 0 0 0 1 1 , (45) Then y r is obtained by (4 multiplicative operations and 4 addition operations) ) ( )4 ( ) ( )3 ( )2 ( ) ( )1 ( q q u p p u r r y r y r y r y r y - - - - ± = ? . (46) Thus, solving 4×4 sub-problem requires 33 elementary operations (16 multiplicative operations and 17 addition operations). (b) From the same plant (30) 1.) Solving p 2 : First p r is obtained trivially from (36). Then p u(p) , p p and p q are obtained by solving the 3×3 subproblem (13multiplicative operations and 14 addition operations) [ ] )3 ( ) 2 ( )1 ( )3 ( )2 ( )1 ( ) ( 1 1 1 q q q p p p q p p u p p p . [ ] r q r p r p u r c r c r c p p p )3 ( ) 2 ( )1 ( ) ( - - - = (47) 2.) Solving y 2 : First y u(p) , y p and y q are obtained by solving the 3×3 sub-problem (10 multiplicative operations and 11 addition operations) = q p p u q p p u y y y q q q p p p ? ? ? ) ( ) ( )3 ( ) 2 ( )1 ( )3 ( )2 ( )1 ( 1 1 1 , (48) Page 15 13 Then y r is obtained by (3 multiplicative operations and 3 addition operations) ) ( )3 ( )2 ( ) ( )1 ( q p p u r r y r y r y r y - - - ± = ? . (49) Thus, solving the 4×4 sub-problem requires 27 elementary operations (13 multiplicative operations and 14 addition operations). (IV) No basic slack/surplus variables. (a) The basic variables are chosen from three different trigeneration plants (31). Solutions are obtained by solving the 6×6 sub-problem. 66 operations are required (9 multiplicative operations, 21 addition operation and 36 operations for solving the general 3× 3 sub-problems). 1.) Solving p 2 : p u(p) , p u(q), p u(r), p p , p q and p r are obtained by solving 6×6 sub-problem [ ] ) 6 ( )5 ( ) 4 ( )3 ( ) 2 ( )1 ( ) 6 ( )5 ( )4 ( )3 ( ) 2 ( )1 ( ) 6 ( )5 ( ) 4 ( )3 ( ) 2 ( )1 ( ) ( ) ( ) ( 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 r r r r r r q q q q q q p p p p p p r q p r u q u p u p p p p p p = [ ] r r u q q u p p u c c c c c c ) ( ) ( ) ( . (50) 2.) Solving y 2 : y u(p) , y p , y u(q) , y q , y u(r) and y r are obtained by solving the 6×6 sub-problem = r q p r u q u p u r r u q q u p p u y y y y y y r r r r r r q q q q q q p p p p p p ? ? ? ? ? ? ) ( ) ( ) ( ) ( ) ( ) ( )6 ( )5 ( ) 4 ( )3 ( )2 ( )1 ( ) 6 ( )5. . .
--3000,1,1500,2455,27796