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3. It is important to view a batch process operation as a network of interconnected
queues because each machine or work center is linked to other machines and
work centers in a network fashion. The flows between work centers are clarified
when the flow of materials through the production process is viewed as a network
of queues.
4. Scheduling of patients in a doctor's clinic is similar to scheduling jobs in a factory
in that patients are comparable to jobs, and waiting rooms, examining rooms and
consulting rooms are comparable to various queues and work centers. To
facilitate the "flow" of patients, dispatching rules might be used. The clinic is
different from a factory because people instead of jobs are waiting. Some people
may not appreciate waiting longer than others in an effort to improve the clinic's
5. Gantt charting and FCS schedule jobs one at a time according to priorities on the
resources available. FCS differs from Gantt charting in that each work center
may have multiple machines or resources.
6. Lead time, which is a function of capacity and priority decisions, can be managed
by controlling input, output, capacity and priorities.
Lead time cannot be constant because of job priorities, variability in demand,
processing and supply that lead to changing input, load factors, processing time,
and starting dates.
7. The m x n scheduling algorithms utilize a highly restrictive set of assumptions;
constant processing time, no passing of jobs, no lot splitting, etc. Real sequencing
problems have a great deal of variability in processing times as well as multiple
objectives. Optimal rules should be more widely used provided they can better
model real sequencing problems.
8. The purpose of shop floor control systems is to implement the schedule and then
correct it if necessary to meet the plan.
Scheduling needs monitoring and feedback to be effective, i.e. to meet changing
needs. Therefore, controls are required to provide the necessary information.
Effective scheduling cannot be done without shop floor control systems.
9. The main goal of theory of constraints is to make as much money as possible
from operations. This is done by identifying the bottlenecks in the process and
focusing on eliminating these constraints, one by one, by subordinating all other
parts/elements in the system to eliminating these bottlenecks.
10. A bottleneck is any resource whose capacity is less than the demand placed on it,
and less than the capacity of all other resources. A bottleneck will therefore
constrain the output of the entire plant.
11. All non-bottleneck resources should be scheduled to ensure that the bottleneck is
not starved for materials and can keep busy processing orders needed for
sale. A queue should be formed at the bottleneck to insure that it stays
busy. Non-bottleneck recourses do not need to operate at full capacity,
provided they process enough to keep the bottleneck busy. Thus non-
bottleneck work centers may have idle time in their schedules. Non-
12. Measures that can be taken to provide more capacity at the bottleneck work center
include: reduction of setup time, use of bottleneck resources 24 hours a
day without breaks, and addition of resources to the bottleneck via
additional labor or machines.
Answers to Problems
a. Makespan = 18 hours
1 24 9
2 16 7
3 32 18
4 24 14
d. Job Job Idle Time
1 0
2 0
3 6
4 _5
11 hours total
e. Sequence: 3, 2, 4, 1 has a makespan of 14 hours.
0 3 7 8 9 10 12
14
A ¦____[1]___¦____[2]_______¦_[3] ¦ [4] ¦_ _¦____
__¦_______¦
B
¦_______[3]__________¦__[1]____¦________¦__[4]___¦_
______¦
0 2 6 8 9 12
14
Machine idle time:
Machine A - 5 hours
Machine B - 4 hours
Machine C - 3 hours
1 24 14 5
2 16 7 0
3 32 12 0
4 24 12 3
8 hrs.
The makespan is reduced from 18 hours to 14 hours, machine idle time
from 24 hours to 12 hours, and job idle time from 11 hours to 8 hours.
4. a. Finite capacity schedule by means of a Gantt Chart
0 2 3 4 5 6 7 8 9
A1 |___[1]______| |_[4]_|
A2 |______[2]________|_[3]_|
B1 |___[1]___| |__[4]___|
B2 |__________[3]____________|
C1 |__[2]___| |__[1]___|
0 2 3 4 5 6 7 8 9 12
b. Make Span = 12 hours Job Due Date Delivery Job
Idle
Work Center Idle Time: (hr) Time (hr) Time
(hr)
42 hours 4 24 9 0
Work Center Idle Time Percentage = 58%
The FCS has a shorter makespan but a larger idle time percentage than the
Gantt chart of problem 3. FCS also leads to shorter delivery time for jobs
3 and 4, and no idle time for all jobs. There is also no bottleneck work
center.
5. a. Jobs are sequenced in order 1, 2, 3, 4, 5, 6
0 6 10 13 16 18 22 23 27 35
A1|___[1]__|___[4]______|
A2|_[2]|___[3]___|[5] |___[6]_____|
B1 |______[2]_______|______[5]_____ __|_______[6]______ |
B2 |___[1]__ |____[3]_______|__[4]__|
16 18 23 27 35
Machine center idle (min)
A1 – 17
A2 – 13
B1 – 6
B2 – 18
Job
Job waiting time (min)
Delivery time (min)
1
0
16
2
0
18
3
9
23
4
15
27
5
15
27
6
21
35
b. In comparison to the Gantt Chart from problem 2, which requires 56 min
(1, 2, 3, 4, 5, 6 sequence) to complete the task, the FCS requires only 35 min.
Also the waiting time of the jobs is less for FCS in comparison to Gantt Chart,
however machine utilization is better for the Gantt Chart, when we compare
machine center idle time for both cases.
Job
Job waiting time (min)
Delivery time (min)
1
0
16
2
10
28
3
21
35
4
27
39
5
36
48
6
42
56
6. a. Priority used: 1 - 2 - 3 - 4 – 5 (assumes 8 hour move/wait time or 1 day
between departments)
Hr: 0 3 5 6 8 9 11 17 20 28
I | 1 | 2 | 3 | 4 | 5 |____________________________
U | idle | 1 | 3 |____4___|
7. a. Gantt Chart: Priority Order for Samples is Earliest Due Date (1, 3, 2, 5, 4)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
A | 1 | | 3 | 5 | 4 | | 2 |
B | | 1 | 2 | | 5 | | 4 |
C | 3 | 5 | 1 | 3 | 2 | 4 |
D | 5 | | 3 | 1 | | 4 | | 2 |
0 2 4 6 8 10
A1 | 1 | | 3 | 2 |
A2 | 4 | | 5 |
B1 | 2 | | 5 |
B2 | | 1 | | 4 |
C1 | 3 | 2 | 4 |
Samples
1
2
3
4
5
Completion Date (hr.)
7
10
8
9
9
Due Date (hr.)
6
10
8
14
12
c. Considering the solutions in part (a), samples 1 and 3 are processed with
no waiting because they are the first two priorities. Sample 2 waits for station B for 3
hours and station C for 3 hours. Sample 5 waits for station A for 2 hours and sample 4
8. a. D C E A B
b. D C B A E
c. B C E A D (Note: CR values A:20/12 B: 19/15 C: 16/11 D: 10/5 E:
18/11)
9. a. Job 1 (1 hour processing time for Job 1 versus 2 hours for Job 4)
b. Job 1: 10/7 = 1.43
Job 4: 12/8 = 1.50
Choose Job 1
10. a. For problem 2, machine B is the bottleneck since jobs 3, 4, 5 and 6 all must
wait for machine B to become available. Another machine of type B (or more
hours of capacity) would reduce the make span and thus increase shop
throughput.
b. For problem 3, machine B in clearly the bottleneck since job 3 has to wait 6
11. a. For problem 2, the theory of constraints could be used in two ways. First, we
could find a better sequence of jobs than 1, 2, 3, 4, 5 and thus reduce job interference.
Second, we could make more capacity available at the bottleneck and thus further
improve the make span.
b. For problem 3, the theory of constraints could be used in much the same way as
problem 2, find a better sequence and then add capacity to the bottleneck.
c. For problem 6, the same approach holds true.
3. It is important to view a batch process operation as a network of interconnected
queues because each machine or work center is linked to other machines and
work centers in a network fashion. The flows between work centers are clarified
when the flow of materials through the production process is viewed as a network
of queues.
4. Scheduling of patients in a doctor's clinic is similar to scheduling jobs in a factory
in that patients are comparable to jobs, and waiting rooms, examining rooms and
consulting rooms are comparable to various queues and work centers. To
facilitate the "flow" of patients, dispatching rules might be used. The clinic is
different from a factory because people instead of jobs are waiting. Some people
may not appreciate waiting longer than others in an effort to improve the clinic's
5. Gantt charting and FCS schedule jobs one at a time according to priorities on the
resources available. FCS differs from Gantt charting in that each work center
may have multiple machines or resources.
6. Lead time, which is a function of capacity and priority decisions, can be managed
by controlling input, output, capacity and priorities.
Lead time cannot be constant because of job priorities, variability in demand,
processing and supply that lead to changing input, load factors, processing time,
and starting dates.
7. The m x n scheduling algorithms utilize a highly restrictive set of assumptions;
constant processing time, no passing of jobs, no lot splitting, etc. Real sequencing
problems have a great deal of variability in processing times as well as multiple
objectives. Optimal rules should be more widely used provided they can better
model real sequencing problems.
8. The purpose of shop floor control systems is to implement the schedule and then
correct it if necessary to meet the plan.
Scheduling needs monitoring and feedback to be effective, i.e. to meet changing
needs. Therefore, controls are required to provide the necessary information.
Effective scheduling cannot be done without shop floor control systems.
9. The main goal of theory of constraints is to make as much money as possible
from operations. This is done by identifying the bottlenecks in the process and
focusing on eliminating these constraints, one by one, by subordinating all other
parts/elements in the system to eliminating these bottlenecks.
10. A bottleneck is any resource whose capacity is less than the demand placed on it,
and less than the capacity of all other resources. A bottleneck will therefore
constrain the output of the entire plant.
11. All non-bottleneck resources should be scheduled to ensure that the bottleneck is
not starved for materials and can keep busy processing orders needed for
sale. A queue should be formed at the bottleneck to insure that it stays
busy. Non-bottleneck recourses do not need to operate at full capacity,
provided they process enough to keep the bottleneck busy. Thus non-
bottleneck work centers may have idle time in their schedules. Non-
12. Measures that can be taken to provide more capacity at the bottleneck work center
include: reduction of setup time, use of bottleneck resources 24 hours a
day without breaks, and addition of resources to the bottleneck via
additional labor or machines.
Answers to Problems
a. Makespan = 18 hours
1 24 9
2 16 7
3 32 18
4 24 14
d. Job Job Idle Time
1 0
2 0
3 6
4 _5
11 hours total
e. Sequence: 3, 2, 4, 1 has a makespan of 14 hours.
0 3 7 8 9 10 12
14
A ¦____[1]___¦____[2]_______¦_[3] ¦ [4] ¦_ _¦____
__¦_______¦
B
¦_______[3]__________¦__[1]____¦________¦__[4]___¦_
______¦
0 2 6 8 9 12
14
Machine idle time:
Machine A - 5 hours
Machine B - 4 hours
Machine C - 3 hours
1 24 14 5
2 16 7 0
3 32 12 0
4 24 12 3
8 hrs.
The makespan is reduced from 18 hours to 14 hours, machine idle time
from 24 hours to 12 hours, and job idle time from 11 hours to 8 hours.
4. a. Finite capacity schedule by means of a Gantt Chart
0 2 3 4 5 6 7 8 9
A1 |___[1]______| |_[4]_|
A2 |______[2]________|_[3]_|
B1 |___[1]___| |__[4]___|
B2 |__________[3]____________|
C1 |__[2]___| |__[1]___|
0 2 3 4 5 6 7 8 9 12
b. Make Span = 12 hours Job Due Date Delivery Job
Idle
Work Center Idle Time: (hr) Time (hr) Time
(hr)
42 hours 4 24 9 0
Work Center Idle Time Percentage = 58%
The FCS has a shorter makespan but a larger idle time percentage than the
Gantt chart of problem 3. FCS also leads to shorter delivery time for jobs
3 and 4, and no idle time for all jobs. There is also no bottleneck work
center.
5. a. Jobs are sequenced in order 1, 2, 3, 4, 5, 6
0 6 10 13 16 18 22 23 27 35
A1|___[1]__|___[4]______|
A2|_[2]|___[3]___|[5] |___[6]_____|
B1 |______[2]_______|______[5]_____ __|_______[6]______ |
B2 |___[1]__ |____[3]_______|__[4]__|
16 18 23 27 35
Machine center idle (min)
A1 – 17
A2 – 13
B1 – 6
B2 – 18
Job
Job waiting time (min)
Delivery time (min)
1
0
16
2
0
18
3
9
23
4
15
27
5
15
27
6
21
35
b. In comparison to the Gantt Chart from problem 2, which requires 56 min
(1, 2, 3, 4, 5, 6 sequence) to complete the task, the FCS requires only 35 min.
Also the waiting time of the jobs is less for FCS in comparison to Gantt Chart,
however machine utilization is better for the Gantt Chart, when we compare
machine center idle time for both cases.
Job
Job waiting time (min)
Delivery time (min)
1
0
16
2
10
28
3
21
35
4
27
39
5
36
48
6
42
56
6. a. Priority used: 1 - 2 - 3 - 4 – 5 (assumes 8 hour move/wait time or 1 day
between departments)
Hr: 0 3 5 6 8 9 11 17 20 28
I | 1 | 2 | 3 | 4 | 5 |____________________________
U | idle | 1 | 3 |____4___|
7. a. Gantt Chart: Priority Order for Samples is Earliest Due Date (1, 3, 2, 5, 4)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
A | 1 | | 3 | 5 | 4 | | 2 |
B | | 1 | 2 | | 5 | | 4 |
C | 3 | 5 | 1 | 3 | 2 | 4 |
D | 5 | | 3 | 1 | | 4 | | 2 |
0 2 4 6 8 10
A1 | 1 | | 3 | 2 |
A2 | 4 | | 5 |
B1 | 2 | | 5 |
B2 | | 1 | | 4 |
C1 | 3 | 2 | 4 |
Samples
1
2
3
4
5
Completion Date (hr.)
7
10
8
9
9
Due Date (hr.)
6
10
8
14
12
c. Considering the solutions in part (a), samples 1 and 3 are processed with
no waiting because they are the first two priorities. Sample 2 waits for station B for 3
hours and station C for 3 hours. Sample 5 waits for station A for 2 hours and sample 4
8. a. D C E A B
b. D C B A E
c. B C E A D (Note: CR values A:20/12 B: 19/15 C: 16/11 D: 10/5 E:
18/11)
9. a. Job 1 (1 hour processing time for Job 1 versus 2 hours for Job 4)
b. Job 1: 10/7 = 1.43
Job 4: 12/8 = 1.50
Choose Job 1
10. a. For problem 2, machine B is the bottleneck since jobs 3, 4, 5 and 6 all must
wait for machine B to become available. Another machine of type B (or more
hours of capacity) would reduce the make span and thus increase shop
throughput.
b. For problem 3, machine B in clearly the bottleneck since job 3 has to wait 6
11. a. For problem 2, the theory of constraints could be used in two ways. First, we
could find a better sequence of jobs than 1, 2, 3, 4, 5 and thus reduce job interference.
Second, we could make more capacity available at the bottleneck and thus further
improve the make span.
b. For problem 3, the theory of constraints could be used in much the same way as
problem 2, find a better sequence and then add capacity to the bottleneck.
c. For problem 6, the same approach holds true.
