CHAPTER 16: SCHEDULING - DrJimMirabella.com · · 2010-03-21Chapter 16 - Scheduling 16-1 CHAPTER 16: SCHEDULING Solutions: 1. Job A B C A B C 1 5 8 6 row 1 0 3 1 Worker 2 6 7 9

Chapter 16 - Scheduling

16-1

CHAPTER 16: SCHEDULING

Solutions:

1. Job

A B C A B C

1 5 8 6 row 1 0 3 1

Worker 2 6 7 9 reduction 2 0 1 3

3 4 5 3 3 1 2 0

column

reduction

A B C

1 0 2 1 Optimum:

2 0 0 3 Worker 1, Job A

3 1 1 0 2 B

3 C

2. Initial

Job

Initial

revised

A B C A B C A B C

1 5 8 6 1 4 1 3 row 1 3 0 2

Worker 2 6 7 9 2 3 2 0 reduction 2 3 2 0

3 4 5 3 3 5 4 6 3 1 0 2

column

reduction

A B C

Optimum: 1 2 0 2

1–B, 2–C, 3–A 2 2 2 0

3 0 0 2


16-2

Solutions (continued)

3. Route

A B C D E A B C D E

1 4 5 9 8 7 1 0 1 5 4 3

2 6 4 8 3 5 2 3 1 5 0 2

Truck 3 7 3 10 4 6 row 3 4 0 7 1 3

4 5 2 5 5 8 reduction 4 3 0 3 3 6

5 6 5 3 4 9 5 3 2 0 1 6

column

reduction

A B C D E A B C D E

1 0 2 6 4 1 1 0 1 5 4 1

2 3 2 6 0 0 2 3 1 5 0 0

3 3 0 7 0 0 add and 3 4 0 7 1 1

4 2 0 3 2 3 subtract 1 4 3 0 3 3 4

5 2 2 0 0 3 5 3 2 0 1 4

Optimum: 1–A, 2–E, 3–D, 4–B, 5–C

or 1–A, 2–D, 3–E, 4–B, 5–C


16-3


4. Initial + Dummy

Machine

A B C D

1 12 8 11 0

Job

2 13 10 8 0 row [no change due to dummy]

3 14 9 14 0 reduction

4 10 7 12 0 column

reduction

A B C D A B C D

1 1 0 2 0 1 2 1 3 0

Job

2 3 3 0 1 add and 2 3 3 0 0

3 3 1 5 0 subtract 1 3 4 2 6 0

4 0 0 4 1 4 0 0 4 0

Optimum: 1–B, 2–C, 3–D, 4–A

Total cost = $26.00

5. a. Initial revised

Machine

A B C D E A B C D E

1 14 18 20 17 18 1 0 4 6 3 4

2 14 15 19 50 17 2 0 1 5 36 3

Job 3 12 16 15 14 17 row 3 0 4 3 2 5

4 11 13 14 12 14 reduction 4 0 2 3 1 3

5 10 16 15 14 13 5 0 6 5 4 3

column

reduction

A B C D E

1 0 3 3 2 1

Optimum:

1–A, 2–B, 3–C,

4–D, 5–E

2 0 0 2 35 0

3 0 3 0 1 2

4 0 1 0 0 0

5 0 5 2 3 0


16-4


5. b. Initial revised

Machine

A B C D E A B C D E

1 50 18 20 17 18 1 33 1 3 0 1

2 14 15 19 50 17 2 0 1 5 36 3

Job 3 12 16 15 14 17 row 3 0 4 3 2 5

4 11 13 14 12 14 reduction 4 0 2 3 1 3

5 10 16 15 14 13 5 0 6 3 4 3

column

reduction

A B C D E A B C D E

1 34 1 1 0 0 1 38 0 0 0 0

2 0 0 2 35 1 2 0 0 2 36 2

3 0 3 0 1 3 add and 3 0 3 0 2 4

4 0 1 0 0 1 subtract 1 4 0 1 0 1 2

5 0 5 2 3 1 5 0 5 2 4 2

Optimum: 1–E, 2–B, 3–C, 4–D, 5–A


16-5


6. a. FCFS: A–B–C–D

SPT: D–C–B–A

EDD: C–B–D–A

CR: A–C–D–B

FCFS: Job time Flow time Due date Days

Job (days) (days) (days) tardy

A 14 14 20 0

B 10 24 16 8

C 7 31 15 16

D 6 37 17 20

37 106 44

SPT: Job time Flow time Due date Days


D 6 6 17 0

C 7 13 15 0

B 10 23 16 7

A 14 37 20 17

37 79 24

EDD: Job time Flow time Due date Days


C 7 7 15 0

B 10 17 16 1

D 6 23 17 6

A 14 37 20 17

84 24

Critical Ratio

Job

Processing Time

(Days) Due Date Critical Ratio Calculation

A 14 20 (20 – 0) / 14 = 1.43

B 10 16 (16 – 0) /10 = 1.60

C 7 15 (15 – 0) / 7 = 2.14

D 6 17 (17 – 0) / 6 = 2.83

Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After

the completion of Job A, the revised critical ratios are:


16-6


Job

Processing Time


A – – –

B 10 16 (16 – 14) /10 = 0.20

C 7 15 (15 – 14) / 7 = 0.14

D 6 17 (17 – 14) / 6 = 0.50

Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After

the completion of Job C, the revised critical ratios are:

Job

Processing Time


A – – –

B 10 16 (16 – 21) /10 = –0.50

C – – –

D 6 17 (17 – 21) / 6 = –0.67

Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.

The critical ratio sequence is A–C–D–B and the makespan is 37 days.

Critical

Ratio

sequence

Processing

Time (Days) Flow time Due Date Tardiness

A 14 14 20 0

C 7 21 15 6

D 6 27 17 10

B 10 37 16 21

99 37


16-7


jobs 67.237

99jobs of number Average

days 75.244

99time flow Average ;days 25.10

4

41nessTardi Average

b.

times job

time Flowcenter the at jobs of number Average

jobs of Number

tardy Daysnessardit job Average

jobs of Number

time Flowtime flow Average

FCFS SPT EDD CR

26.50 19.75 21.00 24.75

11.0 6.00 6.00 9.25

2.86 2.14 2.27 2.67

c. SPT is superior.


16-8


7. FCFS: a–b–c–d–e

SPT: c–b–a–e–d

EDD: a–b–c–e–d

CR: a–e–b–c–d

FCFS: Operation Flow time Due date Hours

Job time (hr.) (hr.) (hr.) tardy

a 7 7 4 3

b 4 11 10 1

c 2 13 12 1

d 11 24 20 4

e 8 32 15 17

32 87 26

SPT: Operation Flow time Due date Hours


c 2 2 12 0

b 4 6 10 0

a 7 13 4 9

e 8 21 15 6

d 11 32 20 12

32 74 27

EDD: Operation Flow time Due date Hours


a 7 7 4 3

b 4 11 10 1

c 2 13 12 1

e 8 21 15 6

d 11 32 20 12

32 84 23


16-9

Critical Ratio

Job

Processing Time

(Hours) Due Date Critical Ratio Calculation

A (.14 x 45) + .7 = 7 4 (4 – 0) / 7 = .57

B (.25 x 14) + .5 = 4 10 (10 – 0) / 4 = 2.5

C (.10 x 18) + .2 = 2 12 (12 – 0) / 2 = 6

D (.25 x 40) + 1 = 11 20 (20 – 0) / 11 = 1.82

E (.10 x 75) + .5 = 8 15 (15 – 0) / 8 = 1.88


Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4 hours, the

revised critical ratios are:

Job

Processing

Time (Hrs.) Due Date Critical Ratio Calculation

A – – –

B 4 10 (10 – 4) / 4 = 1.5

C 2 12 (12 – 4) / 2 = 4

D 11 20 (20 – 4) / 11 = 1.45

E 8 15 (15 – 4) / 8 = 1.38

Job B has the lowest critical ratio therefore it is scheduled next and it is completed after 11 hours

(7 + 4). After the completion of Job B, the revised critical ratios are:

Job

Processing Time


A – – –

B – – –

C 2 12 (12 – 11) / 2 = 0.5

D 11 20 (20 – 11) / 11 = .81

E 8 15 (15 – 11) / 8 = .5

Job C and Job E are tied for the lowest critical ratio and Job C is arbitrarily selected and is

scheduled next. Job C is completed in 2 hours bringing the total completion time to 11 + 2 = 13.

After the completion of Job C, the revised critical ratios are:


16-10

Job

Processing Time


A – – –

B – –

C – – –

D 11 20 (20 – 13) / 11 = .63

E 8 15 (15 – 13) / 8 = .25


Job E has the lowest critical ratio therefore it is scheduled next. The critical ratio final sequence is

A–B–C–E–D. Total completion of all six jobs (makespan) is 32 hours.

Critical Ratio

sequence

Processing

Time (Days) Flow time Due Date Tardiness

A 7 7 4 3

B 4 11 10 1

C 2 13 12 1

E 8 21 15 6

D 11 32 20 12

32 84 23

jobs 63.232

84jobs of number Average

hours 8.165

84time flow Average ; 6.4

5

23nessardit Average

times job

time Flowcenter the at jobs of number Average

jobs of Number

late oursHnessardit job Average

jobs of Number

time Flowtime flow Average

FCFS SPT EDD CR

17.40 14.80 16.80 16.8

5.20 5.4 4.60 4.6

2.72 2.31 2.63 2.63


16-11


8. a. (1) FCFS: A–B–C–D–E

(2) S/O: B–D–C–A–E OR D–B–C–A–E [see below]

Time Due date Remaining number

Job (days) (days) Slack of operations Ratio Rank

A 8 20 12 2 6.0 4

B 10 18 8 4 2.0 1,2 (tie)

C 5 25 20 5 4.0 3

D 11 17 6 3 2.0 1,2 (tie)

E 9 35 26 4 6.5 5

b. S/O: [Assume B–D–C–A–E]

Time Flow time Due date Days


B 10 10 18 0

D 11 21 17 4

C 5 26 25 1

A 8 34 20 14

E 9 43 35 8

43 134 27

Time Flow time Due date Days


A 8 8 20 0

B 10 18 18 0

C 5 23 25 0

D 11 34 17 17

E 9 43 35 8

43 126 25

FCFS S/O

Average flow time =

flow time: 25.20 26.80

number of jobs

Average number of jobs in the system =

flow time: 2.93 3.12

job times


16-12


9.

Time (hr.)

Order Step 1 Step 2

A 1.20 1.40

B 0.90 1.30

C 2.00 0.80

D 1.70 1.50

E 1.60 1.80

F 2.20 1.75

G 1.30 1.40

Sequence of assignment:

.80 [C] last (i.e., 7th)

.90 [B] first

1.20 [A] 2nd

1.30 [G] 3rd

1.60 [E] 4th

1.50 [D] 6th

1.75 [F] 5th

Thus, the sequence is b-a-g-e-f-d-c.

10. a. Job Machine A Machine B

a 16 5 7

b 3 2 13 Thus, the sequence is e–b–g–h–d–c–a–f.

c 9 6 6

d 8 7 5

e 2 1 14

f 12 4 8

g 18 14 3

h 20 11 4

b.

0 2 5 23 43 51 60 76 88


16-13

e B g h d c a f

e b g h d c a f

0 2 16 29 43 54 61 67 76 81 88 92

c. Original idle time for B: 2 + 9 + 7 = 18 hrs., and original makespan is 92.

The last two tasks in the sequence are a and f. Splitting both of their completion times evenly,

we get the following results.

Machine 1 Machine 2

a1 8 2.5

a2 8 2.5

f1 6 2

f2 6 2


After splitting, we get the following Gantt chart:

0 2 5 23 43 51 60 68 76 82 88

e b g h d c a1 a2 f1 f2

e b g h d c

a1

a2

f1

f2

0 2 16 29 43 54 61 67 68 70.5 76 78.5 84 90

After splitting, idle time is: 2 + 1 + 5.5 + 3.5 + 4 = 16 hours, and the new makespan = 90.

There is a savings of 2 hr.

Time (minutes)

11. a. Job Center 1 Center 2

A 20 2 27

B 16 1 30 Thus, the sequence is B–A–C–E–F–D.

C 43 3 51

D 60 12 6

E 35 28 4

F 42 24 5

b.

0 16 36 79 114 156 216

B A C E F D

B A C E F D

0 16 46 73 79 130 158 182 216 228


16-14

Idle time of 56 hours.


12. a. Job

Station A

Station B

a 27 2 45

b 18 1 33 Thus, the sequence is b–a–c–d–e.

c 70 30 3

d 26 24 4

e 15 10 5

0 18 45 115 141 156

b a c d e

b a c d e

0 18 51 96 115 145 169 179

b. The Idle time for Station B is = 18 + 19 = 37 minutes.

c. Jobs B, A, C, D and E are candidates for splitting in order to reduce throughput time and idle

time.

141 156

0 9 18 31.5 45 80 115 128 148.5

B 18 A 27 C 70 D 26 E 15

B 33 A 45 C 30

D 24 E 10

0 9 42 87 117 128 152 162

Throughput time is 162 minutes, reducing this time by 17 minutes.

The idle time for B of 20 minutes has decreased by 17 minutes.


16-15


13. Determine job times from the schedule table, and then use Johnson’s Rule to sequence the

jobs. The job times are:

Job A B C D E F G

Cutting 2 4 5 4 2 3 1

Polishing 3 3 2 5 3 1 4

Using Johnson’s Rule, we obtain:

Cutting Polishing

Job Start Finish Start Finish

G 0 1 1 5

A 1 3 5 8

E 3 5 8 11

D 5 9 11 16

B 9 13 16 19

C 13 18 19 21

F 18 20 21 22

Note: The order of Jobs A and E can be reversed with no effect on times.

14. a.,b.

SPT Grinding Deburring


C 0 1 1 6

B 1 3 6 10

A 3 6 10 16

D 6 10 16 19

G 10 16 19 21

F 16 24 24 31

E 24 33 33 37

93 140

The Grinding flow time is 93 hours and Deburring flow time is 140 hours. The Total time is 37

hours.


16-16


c. Johnson’s Rule

Grinding Deburring


C 0 1 1 6

B 1 3 6 10

A 3 6 10 16

F 6 14 16 23

E 14 23 23 27

D 23 27 27 30

G 27 33 33 35

107 147

The Grinding flow time is 107 hours and Deburring flow time is 140 hours. The Total time is 35

hours.

d. The tradeoff is between shorter flow time in the Grinding and Deburring departments and shorter

total processing time. Ed would be indifferent if the benefit to be gained by shorter total

processing time was equal to the cost of additional flow time in the Grinding and Deburring

departments.

15. a. FCFS: SPT:

Job Flow Due Days Job Flow Due Days

Job time time date tardy Job time time date tardy

a 4.5 4.5 10 0 d 1.6 1.6 27 0

b 6.0 10.5 17 0 e 2.8 4.4 18 0

c 5.2 15.7 12 3.7 f 3.3 7.7 19 0

d 1.6 17.3 27 0 a 4.5 12.2 10 2.2

e 2.8 20.1 18 2.1 c 5.2 17.4 12 5.4

f 3.3 23.4 19 4.4 b 6.0 23.4 17 6.4

23.4 91.5 10.2 23.4 66.7 14.0

EDD: CR:

Job Flow Due Days Job Flow Due Days

Job time time date tardy Job time time date tardy

a 4.5 4.5 10 0 a 4.5 4.5 10 0

c 5.2 9.7 12 0 c 5.2 9.7 12 0

b 6.0 15.7 17 0 b 6.0 15.7 17 0

e 2.8 18.5 18 .5 e 2.8 18.5 18 0.5

f 3.3 21.8 19 2.8 f 3.3 21.8 19 2.8

d 1.6 23.4 27 0 d 1.6 23.4 27 0

23.4 93.6 3.3 23.4 93.6 3.3


16-17


Critical Ratio

Job

Processing Time


A 4.5 10 (10 – 0) / 4.5 = 2.22

B 6.0 17 (17 – 0) / 6.0 = 2.83

C 5.2 12 (12 – 0) / 5.2 = 2.31

D 1.6 27 (27 – 0) / 1.6 = 16.88

E 2.8 18 (18 – 0) / 2.8 = 6.43

F 3.3 19 (19 – 0) / 3.3 = 5.76

Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4.5 days. The

revised critical ratios are:

Job

Processing


A – – –

B 6.0 17 (17 – 4.5) / 6.0 = 2.08

C 5.2 12 (12 – 4.5) / 5.2 = 1.44

D 1.6 27 (27 – 4.5) / 1.6 = 14.06

E 2.8 18 (18 – 4.5) / 2.8 = 4.82

F 3.3 19 (19 – 4.5) / 3.3 = 4.39

Job C has the lowest critical ratio, therefore it is scheduled next and it is completed after 9.7 days

(4.5 + 5.2). After the completion of Job C, the revised critical ratios are:

Job

Processing


A – – –

B 6.0 17 (17 – 9.7) / 6.0 = 1.22

C – – –

D 1.6 27 (27 – 9.7) / 1.6 = 10.81

E 2.8 18 (18 – 9.7) / 2.8 = 2.96

F 3.3 19 (19 – 9.7) / 3.3 = 2.82


16-18

Job B has the lowest critical ratio, therefore it is scheduled next and it is completed after 15.7

days (9.7 + 6). After the completion of Job B, the revised critical ratios are:

Job

Processing


A – – –

B – – –

C – – –

D 1.6 27 (27 – 15.7) / 1.6 = 7.06

E 2.8 18 (18 – 15.7) / 2.8 = 0.82

F 3.3 19 (19 – 15.7) / 3.3 = 1.0

Job E has the lowest critical ratio, therefore it is scheduled next and it is completed after 18.5

days (15.7 + 2.8). After the completion of Job E, the revised critical ratios are:

Job

Processing


A – – –

B – – –

C – – –

D 1.6 27 (27 – 18.5) / 1.6 = 5.31

E – – –

F 3.3 19 (19 – 18.5) / 3.3 = 0.15

Job F has the lowest critical ratio therefore it is scheduled next and it is completed after 21.8 days

(18.5 + 3.3). The final critical ratio sequence is A–C–B–E–F–D. Total completion of all six jobs

(makespan) is 23.4 days.

Critical

Ratio

sequence

Processing

Time (Days) Flow time Due Date Lateness Tardiness

A 4.5 4.5 10 –5.5 0

C 5.2 9.7 12 –2.3 0

B 6.0 15.7 17 –1.3 0

E 2.8 18.5 18 0.5 0.5

F 3.3 21.8 19 2.8 2.8

D 1.6 23.4 27 –3.6 0

23.4 93.6 –9.4 3.3


16-19


Rule Average =

days late Average flow =

flow time Average =

flow time

tardiness no. of jobs time no. of jobs no. of jobs job time

FCFS 10.2/6 = 1.7 days 91.5/6 = 15.25 days 91.5/23.4 = 3.91 jobs

SPT 14.0/6 = 2.33 days 66.7/6 = 11.117 days 66.7/23.4 = 2.85

EDD 3.3/6 = 0.55 days. 93.6/6 = 15.60 days 93.6/23.4 = 4.00

CR 3.3/6 = .55 days 93.6/6 = 15.60 days 93.6/23.4 = 4.00

b. There are several ways to show this. One is to calculate the ratio of average flow time to average

number of jobs for each rule and then observe that they are equal. Here the ratios are

approximately 3.90. [Slight differences in ratios may arise due to rounding.]

c. S/O job sequence is a-c-b-e-d-f

16.

Job

Remaining

processing

time Due date Slack

Remaining

number of

operations

Slack

Remaining

number of

operations Rank

a 5 8 3 2 1.50 4

b 6 5 –1 4 –0.25 1

c 9 10 1 4 0.25 2

d 7 12 5 3 1.67 5

e 8 10 +2 2 +1.00 3

Using the S/O rule, the sequence is B–C–E–A–D

17. FCFS

Job time Due date Flow time Tardy

Job (hr.) (hr.) (hr.) (hr.)

a 3.5 7 3.5 0

b 2.0 6 5.5 0

c 4.5 18 10.0 0

d 5.0 22 15.0 0

e 2.5 4 17.5 13.5

f 6.0 20 23.5 3.5

23.5 75.0 17.0


16-20


SPT

Job

Job

time

Flow

time

Due

date Tardy

b 2.0 2.0 6 0

e 2.5 4.5 4 0.5

a 3.5 8.0 7 1

c 4.5 12.5 18 0

d 5.0 17.5 22 0

f 6.0 23.5 20 3.5

23.5 68.0 5.0

EDD

Job

Job

time

Flow

time

Due

date Tardy

e 2.5 2.5 4 0

b 2.0 4.5 6 0

a 3.5 8.0 7 1

c 4.5 12.5 18 0

f 6.0 18.5 20 0

d 5.0 23.5 22 1.5

23.5 69.5 2.5

Critical Ratio

Job

Processing Time


A 3.5 7 (7 – 0) / 3.5 = 2.0

B 2.0 6 (6 – 0) / 2.0 = 3.0

C 4.5 18 (18 – 0) / 4.5 = 4.0

D 5.0 22 (22 – 0) / 5.0 = 4.4

E 2.5 4 (4 – 0) / 2.5 = 1.6

F 6.0 20 (20 – 0) / 6 = 3.33


16-21


Job E has the lowest critical ratio, therefore it is scheduled first and completed after 2.5 hours.

The revised critical ratios are:

Job

Processing


A 3.5 7 (7 – 2.5) / 3.5 = 1.29

B 2.0 6 (6 – 2.5) / 2.0 = 1.75

C 4.5 18 (18 – 2.5) / 4.5 = 3.44

D 5.0 22 (22 – 2.5) / 5.0 = 3.90

E – – –

F 6.0 20 (20 – 2.5) / 6 = 2.92

Job A is scheduled next because Job A has the lowest critical ratio. Job A will be completed after

6 hours (2.5 + 3.5). After the completion of Job A, the revised critical ratios are:

Job

Processing


A – – –

B 2.0 6 (6 – 6) / 2.0 = 0

C 4.5 18 (18 – 6) / 4.5 = 2.67

D 5.0 22 (22 – 6) / 5.0 = 3.20

E – – –

F 6.0 20 (20 – 6) / 6 = 2.33

Since Job B has the lowest critical ratio, it is scheduled next and it is completed after 8 hours (6

+ 2). After the completion of Job B, the revised critical ratios are:

Job

Processing


A – – –

B – – –

C 4.5 18 (18 – 8) / 4.5 = 2.22

D 5.0 22 (22 – 8) / 5.0 = 2.80

E – – –

F 6.0 20 (20 – 8) / 6 = 2.00


16-22


Since Job F has the lowest critical ratio, it is scheduled next and it will be completed after 14

hours (8 + 6). After the completion of Job F, the revised critical ratios are:

Job

Processing


A – – –

B – – –

C 4.5 18 (18 – 14) / 4.5 = 0.89

D 5.0 22 (22 – 14) / 5.0 = 1.60

E – – –

F – – –

Since Job C has the lowest critical ratio, it is scheduled next. Job C will be completed after 18.5

hours. The final critical ratio sequence of all jobs is E–A–B–F–C–D. Total completion of all six

jobs (makespan) is 23.5 hours.

Job

Critical

ratio

Job

time

Flow

time

Due

date Tardy

e 1.6 2.5 2.5 4 0

a 2.0 3.5 6.0 7 0

b 3.0 2.0 8.0 6 2

f 3.3 6.0 14.0 20 0

c 4.0 4.5 18.5 18 .5

d 4.4 5.0 23.5 22 1.5

23.5 72.5 4.0

hours 08.126

5.72time flow Average

hours 67.6

4ardinesst Average

FCFS SPT EDD CR

Average flow time 12.5 11.33 11.58 12.08

Average job tardiness 2.83 0.83 0.42 .67


16-23


18. a.

Order Job time

A 16 x 4 = 64

B 6 x 12 = 72

C 10 x 3 = 30

D 8 x 10 = 80

E 4 x 1 = 4

DD

Job

Job

time

Flow

time

Due

date Tardiness

A 64 64 160 0

C 30 94 180 0

D 80 174 190 0

B 72 246 200 46

E 4 250 220 30

250 828 76

b. Average job tardiness = 76/5 = 15.2 minutes

c. Average number of jobs in the system = 828/250 = 3.31

d. SPT

Job

Job

time

Flow

time

Due

date Tardiness

E 4 4 220 0

C 30 34 180 0

A 64 98 160 0

B 72 170 200 0

D 80 250 190 60

60

Average job tardiness = 60/5 = 12 minutes

19. Sequence Setup times Total

A–B–C 2 + 3 + 2 = 7 (best)

A–C–B 2 + 5 + 3 = 10

B–A–C 3 + 8 + 5 = 16

B–C–A 3 + 2 + 4 = 9

C–A–B 2 + 4 + 3 = 9

C–B–A 2 + 3 + 8 = 13


16-24



A–B–C 2.4 + 1.8 + 1.4 = 5.6

A–C–B 2.4 + 2.2 + 1.3 = 5.9

B–A–C 3.2 + 0.8 + 2.2 = 6.2

B–C–A 3.2 + 1.4 + 2.6 = 7.2

C–A–B 2.0 + 2.6 + 1.8 = 6.4

C–B–A 2.0 + 1.3 + 0.8 = 4.1 (best)


A–B–C–D 2 + 5 + 3 + 2 = 12

A–B–D–C 2 + 5 + 2 + 6 = 15

A–D–B–C 2 + 4 + 3 + 3 = 12

A–D–C–B 2 + 4 + 6 + 2 = 14

B–A–D–C 1 + 7 + 4 + 6 = 18

B–C–D–A 1 + 3 + 2 + 4 = 10 (best)

C–B–A–D 3 + 2 + 7 + 4 = 16

C–B–D–A 3 + 2 + 2 + 4 = 11

C–D–A–B 3 + 2 + 4 + 5 = 14

C–D–B–A 3 + 2 + 3 + 7 = 15

D–A–B–C 2 + 4 + 5 + 3 = 14

D–C–B–A 2 + 6 + 2 + 7 = 17

22. Each period’s backlog is equal to actual input – actual output. That amount is added to (or

subtracted from) the previous backlog to obtain the current (shown) backlog for the period.

Period

Input

1 2 3 4 5

Planned 24 24 24 24 20

Actual 25 27 20 22 24

Output Planned 24 24 24 24 23

Actual 24 22 23 24 24

Backlog 12 13 18 15 13 13


16-25


23. Period

1 2 3 4 5 6

Input Planned 200 200 180 190 190 200

Actual 210 200 179 195 193 194

Deviation +10 0 -1 +5 +3 -6

Cum. Dev. +10 +10 +9 +14 +17 +11

Period

1 2 3 4 5 6

Output Planned 200 200 180 190 190 200

Actual 205 194 177 195 193 200

Deviation +5 -6 -3 +5 +3 0

Cum. Dev. +5 -1 -4 +1 +4 +4

Backlog 7 12 18 20 20 20 14

24. Day Mon Tue Wed Thu Fri Sat

Staff needed 2 3 1 2 4 3

Worker 1 2 3 1 2 4 3

Worker 2 1 2 1 2 3 2 (tie)

Worker 3 0 2 1 1 2 1

Worker 4 0 1 0 0 1 1 Part-time worker

No. working: 2 3 1 2 4 3

25. Day Mon Tue Wed Thu Fri Sat


Worker 1 3 4 2 3 4 5

Worker 2 2 3 2 3 3 4 (tie)

Worker 3 1 3 2 2 2 3 (tie)

Worker 4 0 2 1 2 2 2

Worker 5 0 2 0 1 1 1 (part-time worker)

Worker 6 0 1 0 1 0 0 (tie) (part-time worker)

No. working: 3 4 2 3 4 5


16-26


26.

Day Mon Tue Wed Thu Fri Sat


Worker 1 4 4 5 6 7 8

Worker 2 4 4 4 5 6 7 (tie)

Worker 3 3 4 4 4 5 6

Worker 4 3 4 3 3 4 5

Worker 5 2 3 3 3 3 4

Worker 6 2 3 2 2 2 3 (tie)

Worker 7 1 2 2 2 1 2 (tie)

Worker 8 0 1 1 1 1 2

Worker 9 0 1 0 0 0 1 (tie) Part-time worker

No. working: 4 4 5 6 7 8

Documents

CHAPTER 16: SCHEDULING - DrJimMirabella.com · · 2010-03-21Chapter 16 - Scheduling 16-1 CHAPTER 16: SCHEDULING Solutions: 1. Job A B C A B C 1 5 8 6 row 1 0 3 1 Worker 2 6 7 9