S M E L B O R P

S M E L B O R P T N E M N G I S AS

GENERAL ASSIGNMENT PROBLEMS (BALANCED) �IT HAS N WORKERS & N JOBS TO BE DONE. �THERE IS ONE-ONE CORRESPONDENCE BETWEEN THE WORKERS AND THE JOBS. �ALOCATION OF JOB IS DONE IN SUCH A WAY THAT THE TOTAL COST IS MINIMISED. �EVERY WORKER CAN DO ALL JOBS AND VISE VERSA.

FORMAT OF AN ASSIGNMENT PROBLEMS. P 1 P 2 P 3 J 1 C 12 C 13 J 2 C 21 C 22 C 23 J 3 C 31 C 32 C 33

HUNGERIAN METHOD OF SOLVING AN ASSIGNMENT PROBLEMS �

EG…. A B C D 1 16 1 6 11 2 25 10 0 10 3 10 25 2 14 4 15 7 14 10 ------ COLUMN REDN ------ ROW REDN A B C D 1 15 0 5 10 2 25 10 0 10 3 8 23 0 12 4 8 0 7 3 A B C D 1 7 0 5 7 2 17 10 0 7 3 0 23 0 9 4 0 0 7 0

�ALL ZEROS ARE LINEAR X OR & EACH ROW AND COLUMN HAS EXACTLY ONE REACTANGLED ZERO. HENCE OPTIMUM SOLUTION IS REACHED. � 1 -> B, 2 -> C, 3 -> A, 4 -> D �MIN COST = 1+0+10+10 = 21 UNITS A 1 7 2 17 3 4 B C D 5 7 10 7 23 9 7

WHERE MORE THAN ONE ZERO COMES IN A ROW A B C D 1 2 10 9 7 1 0 8 7 5 2 13 10 12 2 2 11 0 10 0 3 3 4 6 1 3 2 3 5 0 4 4 15 4 9 4 0 11 0 5 � WE CAN SEE THAT THERE ARE MORE THAN ONE ZEROS IN ROW 2 AND 4. � START WITH ONE ZERO ROW ONLY & IGNOR THE TWO ZERO ROWS. • A 1 2 11 3 2 4 B C 8 7 D • 5 • 10 3 11 5 5 1 -> A, 2 -> B, 3 -> D, 4 -> C IN THE OPTIMUM SOLUTION. MIN COST = 2+2+4+1 = 9 UNITS

EG. OF IMPROVEMENT IN THE SOLUTION A B C D 1 3 4 6 5 1 0 0 1 2 2 5 6 10 9 2 0 0 3 4 3 1 2 3 0 0 0 1 4 4 10 6 4 4 0 5 0 0 �THERE IS NO SINGLE ZEROS IN ANY ROW. SO WE CHECK THE COLUMNS. IN 4 TH COLUMN WE HAVE A SINGLE 0. SO START FROM HERE

A B C D 1 0 0 1 2 2 0 0 3 4 3 0 0 0 1 4 STILL NO SINGLE 0 ROW BUT COLUMN 3 HAS A SINGLE 0 HENCE, NO SINGLE 0 ROW OR COLUMN IS FOUND. 5 A B C D 1 0 0 1 2 2 0 0 3 4 1 5 HENCE, WE HAVE THE OPTIMUM OF MAKING RANDOM ALLOCATION ONLY.

A B C D A 1 1 2 2 3 3 4 4 � RANDOM ALLOCATION AT 1 B AND LAST ALLOCATION AT 2 A B C � OPTIMUM SOLUTION D

PROHIBITED ASSIGNMENT PROBLEM �IN SUCH PROBLEMS ONE OF THE Cij ELEMENT IS MISSING ON SKIPPED. �A PARTICULAR PERSON CANNOT DO A PERTICULAR JOB. A B C D 1 - 5 2 6 2 4 5 3 8 3 6 6 2 5 4 1 6 3 4 MR. A CANNOT DO JOB 1 MODIFICATION: A VERY LARGE NUMBER ‘M’ IS PUT IN PLACE OF THE DASH (-) HENCE, M – ANY NUMBER = M

�Cij NO. OF DAYS TO COMPELE ASSESSMENT A B C D 1 M 5 2 6 2 4 5 3 8 3 6 6 2 5 4 1 6 3 4 A B 1 M 1 2 1 3 4 4 REDUCED COST MATRIX C 2 3 2 B C D 1 M 1 0 1 2 1 0 0 2 3 4 2 0 0 4 0 3 2 0 D 1 2 A 1 -> C 2 -> B 3 -> D 4 -> A IS THE OPTIMUM SOLUTION MIN COST = 2+5+5+1 = 13 UNITS

EG. OF CHANGING THE CIJ VALUE FOR IMPROVEMENT A B C D 1 12 1 11 5 ROW 2 3 11 10 8 3 3 4 6 1 4 2 13 1 7 A B C D 1 11 0 5 4 2 0 8 2 5 3 2 3 0 0 4 0 11 4 5 REDUCED COST MATRIX REDUCED A B C D 1 11 0 10 4 COLUMN 2 0 8 7 5 3 2 3 5 0 4 0 11 9 5 A B C D 5 4 2 5 4 5 1 11 2 3 4 8 2 3 11 REDUCED OPTIMUM SOLUTION IS NOT REACHED

MODIFICATION OF THE MATRIX √ 2 A 1 11 2 3 4 � B 8 2 C D 5 4 2 5 √ 3 4 5 √ 1 3 11

THE MINIMUM ELEMENT OUTSIDE THE LINES IS 2 � SUBSTRACT 2 FROM ALL THE ELEMENTS NOT ON THE LINES. � ADD 2 TO ALL THE ELEMENTS AT THE INTERSECTION OF THE LINES. � KEEP THE REMAINING ELEMENTS AS THEY ARE THIS RESULT. A B C D 1 13 0 5 4 2 0 6 0 3 3 4 0 0 0 4 0 9 2 3 A 1 13 4 C D 5 4 6 2 3 B 3 4 9 2 3 1 -> B 2 -> C 3 -> D 4 -> A MIN COST = 1+10+1+2 = 14 UNITS

MAXIMISATION IN ASSIGNMENT PROBLEM �IT A BE THE MAXIMISATION MATRIX WITH MAXIMUM ELEMENTS SAY M. �FROM A MATRIX B FROM A BY SUBSTRACTING EACH ELEMENT OF A FROM M => (M-x) �SOLVE THE MATRIX B BY MINIMISATION OF ASSIGNMENT PROBLEMS.

UNBALANCED ASSIGNMENT PROBLEMS. � PROBLEMS WITCH HAVE UNEQUAL NUMBER OF PERSON & JOB. � CREATE A DUMMY ROW OR COLUMN TO BALANCED IT. � PROCEED AS USUAL FOR SOLVING THE PROBLEM A B C D 1 3 6 2 7 1 4 4 3 3 8 5 8 4 5 2 4 5 5 7 6 A B C D DUM 1 3 6 2 6 0 2 7 1 4 4 0 3 3 8 5 8 0 3 4 5 2 4 3 0 2 5 5 7 6 2 0 UNBALANCED A 1 REDUCED COST MATRIX 2 B C 5 4 3 D 4 2 2 7 3 6 1 4 2 1 2 5 2 6 4 DUM 1 -> C 2 -> B 3 -> A 4 -> DUM -> JOB 4 IS UNATTENDED 5 -> D MIN COST = 2+1+3+0+2 = 8 UNITS

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