GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 55 of 65

Marks: +1, -0

Processing times (including step times) and due dates for six jobs waiting to be processed at a work centre are given in the table. The average tardiness (in days) using shortest processing time rule is (correct to two decimal places).

Job	Processing time (days)	Due date(days)
A	3	8
B	7	16
C	4	4
D	9	18
E	5	17
F	13	19

Your Answer:

Validate

Solution:

By SPT Rule

Job	P.T.	D.D.	Job Flow Time	Tardiness
A	3	8	$0+3=3$	$0$
C	4	4	$3+4=7$	2
E	5	17	$7+5=12$	0
B	7	16	$12+7=19$	3
D	9	18	$19+9=28$	10
F	13	19	$28+13=41$	22
				33

Total tardiness

=38

Average tardiness per job

=\;{\;\text " Total tardiness "\;}/{\;\text " No. of Jobs "\;}=\;{38}/{6}=6.33 \;\text " days "\;

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