# Unit 8 problems | Engineering homework help

Unit 8 Problems

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Unit 8 Problems

Problem 6

A. Sequence the jobs using (1) FCFS, (2) SPT, (3) EDD, and (4) CR. Assume the list is by order of arrival.

1. First      Come, First Served (FCFS)

Sequence

Job   Time

Flow   Time

Due   Date

Lateness

A

14

14

20

0

B

10

24

16

8

C

7

31

15

16

D

6

37

17

20

TOTALS

37

106

44

The sequencing of the jobs based on the First Come, First Served (FCFS) basis is A – B – C – D, and the chart shows that shows that jobs B, C, and D will be late by a total of 44 days.

1. Shortest      Possible Time (SPT)

Sequence

Job   Time

Flow   Time

Due   Date

Lateness

D

6

6

17

0

C

7

13

15

0

B

10

23

16

7

A

14

37

20

17

TOTALS

37

79

24

The sequencing on the jobs based on the Shortest Possible Time is D – C – B – A, and the chart shows that jobs B and A will be late by a total of 24 days.

1. Earliest      Due Date (EDD)

Sequence

Job   Time

Flow   Time

Due   Date

Days   Tardy

C

7

7

15

0

B

10

17

16

1

D

6

23

17

6

A

14

37

20

17

TOTALS

37

84

24

The sequencing of the jobs based on the Earliest Due Date (EDD) is given as C – B – D – A, and the average completion time is 42 days.

1. Critical      Ratio

Sequence

Job   Time

Flow   Time

Due   Date

Tardiness

A

14

37

20

17

C

7

44

15

29

D

6

50

17

33

B

10

60

16

44

TOTALS

37

191

123

All the critical ratios are greater than 1, and this indicates that all the jobs will be completed in time if scheduled as A – C – D – B.

B. For each of the methods in part a, determine (1) the average job flow time, (2) the average tardiness, and (3) the average number of jobs at the work center.

1. Average      time flow

a) FCFS = average completion time/number of jobs = 106/4 = 26.5 days.

b) SPT = average completion time/number of jobs = 79/4 = 19.75 days

c) EDD = average completion time/number of jobs = 84/4 = 21 days

d) CR = average completion time/number of jobs = 191/4 = 47.75 days

1. Average      tardiness

a) FCFS = 44/4 = 11 days

b) SPT = 24/4 = 6 days

c) EDD =24/4 = 6 days

d) CR = 123/4 = 30.75 days

1. Average      number of jobs

a) FCFS = 106/37 = 2.865

b) SPT = 79/37 = 2.135

c) EDD = 84/37 = 2.270

d) CR = 191/37 = 5.162

C. Is one method superior to the others? Explain.

None of the methods is superior than the others because they each have their inherent weaknesses in addition to their strengths. For example, using the FCFS and SPT priority rules results in the increased delays in jobs that require longer times to complete while the EDD approach tends to ignore the processing times during the scheduling. However, the CR method has the most weaknesses in scheduling of the jobs.

Problem 11

Given the operation times provided:

a. Develop a job sequence that minimizes idle time at the two work centers.

The determination of the job sequence that would reduce the idle time at the two work centers is completed using the Johnson’s as shown in the table below:

Developing   the Job Sequence

Iteration

Job   Sequence

A

D

The   shortest job time is 12 minutes for D at Center 2, and thus D is scheduled   last.

B

B

D

D   is then removed from the table of eliminated times and the next least job   time is observed as 16 minutes for B at Center 1. Thus, B is scheduled first.

C

B

A

D

B   is then eliminated from the table and the next least job time is 20 minutes   for A at Center 1 and is there scheduled next.

D

B

A

F

D

A   is then eliminated from the table and the next least time is 24 minutes for F   at Center 2 and this is scheduled next.

E

B

A

E

F

D

F   is then eliminated from the table and the next least time is E at Center 2   and this is scheduled next.

F

B

A

C

E

F

D

E   is then eliminated from the table and the last least time is 43 minutes at   Center 1 and it is scheduled the earliest.

Thus the job sequence that reduces the idle time at Centers 1 and 2 is given as B – A – C- E – F – D

b. Construct a chart of the activities at the two centers, and determine each one’s idle time, assuming no other activities are involved.

The chart of activities for the two centers is shown below:

0

16

0

36

79

114

156

216

B

A

C

E

F

D

B

A

C

E

F

D

16

46

73

79

130

158

218

216

228

From the chart, the idle time at Center is given as 0 minutes while the total idle time at Center 2 is given as ((16 – 0) + (79 – 73) + (216 – 182)) = 16 + 6 + 34 = 56 minutes.

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