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ch17, File: Chapter 17: Scheduling
True/False
1. Scheduling specifies when labor, equipment and facilities are needed to produce a
product.
Difficulty: Easy
Feedback: Objectives in Scheduling
2. Scheduling is considered the last stage of the planning process.
Difficulty: Easy
Feedback: Objectives in Scheduling
3. The wide variety of jobs in a job shop makes scheduling difficult.
Difficulty: Easy
Feedback: Objectives in Scheduling
4. Managers typically use multiple objective when constructing a schedule.
Difficulty: Easy
Feedback: Objectives in Scheduling
5. Minimizing overtime is one of several objectives that could be considered when
constructing a schedule.
Difficulty: Easy
Feedback: Objectives in Scheduling
6. Job shop scheduling is usually performed by skilled machine operators.
Difficulty: Easy
Feedback: Objectives in Scheduling
7. The process of assigning work to limited resources is known as scheduling.
Difficulty: Moderate
Feedback: Loading
8. Sequencing involves determining the order in which jobs are released to the shop
floor.
Difficulty: Hard
Feedback: Sequencing
9. The sequence in which jobs should be processed is determined using linear
programming.
Difficulty: Moderate
Feedback: Sequencing
10. Determining the optimal allocation of jobs to machines or workers can be solved
using the assignment method of linear programming.
Difficulty: Moderate
Feedback: Sequencing
11. The process of prioritizing jobs that have been assigned to a resource is called
sequencing.
Difficulty: Easy
Feedback: Sequencing
12. The time required for a job to move through the system is referred to as flow time.
Difficulty: Moderate
Feedback: Sequencing
13. All sequencing rules strive to optimize both processing efficiency and due date
performance.
Difficulty: Moderate
Feedback: Sequencing
14. Johnsons rule gives an optimal sequence for jobs processed serially through a two
step system.
Difficulty: Moderate
Feedback: Sequencing
15. The shortest processing time (SPT) rule is most useful when the job shop is not
congested.
Difficulty: Moderate
Feedback: Sequencing
16. The longest processing time (LPT) rule tends to complete larger jobs in-house when
subcontracting is anticipated.
Difficulty: Moderate
Feedback: Sequencing
17. The first-come, first-served (FCFS) sequencing rule should be only used when
operating at high levels of capacity.
Difficulty: Moderate
Feedback: Sequencing
18. Gantt charts show the planned and yet to be completed activities against total
activities.
Difficulty: Moderate
Feedback: Objectives in Scheduling
19. Increasing the capacity of a work center that is a bottleneck increases output.
Difficulty: Moderate
Feedback: Theory of Constraints
20. Finite scheduling will not load a resource beyond its fixed capacity.
Difficulty: Moderate
Feedback: Advanced Planning and Scheduling Systems
21. The theory of constraints (TOC) approach to scheduling concentrates on scheduling
the bottleneck resources.
Difficulty: Moderate
Feedback: Theory of Constraints
22. With drum-buffer-rope (DBR) the bottleneck resource is always the buffer.
Difficulty: Moderate
Feedback: Theory of Constraints
23. Scheduling using the theory of constraints (TOC) requires that process batch sizes
and transfer batch sizes always match.
Difficulty: Moderate
Feedback: Theory of Constraints
24. Typically, there are more options available when scheduling employees because labor
is a very flexible resource.
Difficulty: Moderate
Feedback: Employee Scheduling
25. The scheduling function for mass production and job shop production are basically
the same.
Difficulty: Moderate
Feedback: Objectives in Scheduling
26. Job shop scheduling is also known as shop floor control.
Difficulty: Moderate
Feedback: Objectives in Scheduling
27. The dispatch list contains the order in which work is assigned to a productive
resource.
Difficulty: Moderate
Feedback: Sequencing
28. In a job shop environment the process of prioritizing jobs assigned to a productive
resource is called sequencing.
Difficulty: Moderate
Feedback: Sequencing
29. Tardiness is defined as the difference between a jobs flow time and makespan.
Difficulty: Moderate
Feedback: Sequencing
30. Input/output control provides the information necessary to regulate work flow in a job
shop environment.
Difficulty: Moderate
Feedback: Monitoring
31. The assignment method of linear programming can be used for both loading and
employee scheduling.
Difficulty: Moderate
Feedback: Employee Scheduling
Multiple Choice
32. Which of the following is not typically considered an objective of scheduling?
a. minimizing job lateness
b. maximizing time in the system
c. minimizing overtime
d. maximizing machine utilization
Difficulty: Easy
Feedback: Objectives of Scheduling
33. The process of assigning work to limited resources is known as
a. loading.
b. sequencing.
c. monitoring.
d. dispatching
Difficulty: Easy
Feedback: Loading
34. The process of prioritizing jobs that have been assigned to a resource is called
a. loading.
b. monitoring.
c. sequencing.
d. input/output analysis.
Difficulty: Easy
Feedback: Sequencing
35. If a job is ahead of schedule, then its critical ratio (CR) will be
a. less than one.
b. greater than one.
c. equal to one.
d. none of the above.
Difficulty: Moderate
Feedback: Sequencing
36. If the work remaining is greater than the time remaining, the critical ratio (CR) will
be
a. less than one.
b. greater than one.
c. equal to one.
d. none of the above.
Difficulty: Moderate
Feedback: Sequencing
37. The time required to complete a group of jobs is referred to as
a. flow time.
b. makespan.
c. completion time.
d. none of the above.
Difficulty: Easy
Feedback: Sequencing
38. The sequencing rule that will minimize average job completion time for a set number
of jobs to be processed on one machine is
a. first-come-first served (FCFS).
b. longest processing time (LPT).
c. shortest process time (SPT).
d. due date (DDATE).
Difficulty: Moderate
Feedback: Sequencing
39. The sequencing rule that minimizes average tardiness for a set of jobs to be
processed on one machine is
a. first-come-first served (FCFS).
b. longest processing time (LPT)
c. shortest process time (SPT)
d. due date (DDATE).
Difficulty: Moderate
Feedback: Sequencing
40. If the following jobs are sequenced according to the first-come-first-served (FCFS)
rule then the mean completion time in days for all jobs is (assume zero for todays
date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
33 days.
22 days.
6.6 days.
4.4 days.
Difficulty: Moderate
Feedback: Sequencing
41. If the following jobs are sequenced according to the FCFS rule then the mean
tardiness (in days) for all jobs would be (assume zero for todays date)
Job
Processing
Time (days)
A
B
8
6
Due
Date
12
15
C
D
5
E
a.
b.
c.
d.
11
3
17
10
8
5 days.
8.33 days.
10.6 days.
25 days
Difficulty: Moderate
Feedback: Sequencing
42. If the following jobs are sequenced according to the FCFS rule then the maximum job
tardiness (in days) is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
0 days.
8 days.
20 days.
25 days.
Difficulty: Moderate
Feedback: Sequencing
43. If the following jobs are sequenced according to the FCFS rule then the total number
of jobs that would be late is (assume zero for todays date)
Job
Processing
Time (days)
Due
Date
A
B
C
11
D
5
E
a.
b.
c.
d.
6
3
12
8
15
17
10
8
4.
3.
2.
1.
Difficulty: Moderate
Feedback: Sequencing
44. If the following jobs are sequenced according to the FCFS rule then job C is
completed on day (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
11.
17.
25.
30
Difficulty: Moderate
Feedback: Sequencing
45. If the following jobs are sequenced according to the SPT rule then the mean
completion time (in days) for all jobs is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
6.6 days.
16 days.
22 days.
33 days.
Difficulty: Moderate
Feedback: Sequencing
46. If the following jobs are sequenced according to the SPT rule then the mean tardiness
(in days) for all jobs is (assume zero for todays date)
Job
A
5
E
Difficulty: Moderate
Feedback: Sequencing
11
D
16 days.
13 days.
5.2 days.
3.2 days
6
C
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
47. If the following jobs are sequenced according to the SPT rule then the maximum job
tardiness (in days) is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
26 days
16 days
10 days
5.2 days
Difficulty: Moderate
Feedback: Sequencing
48. If the following jobs are sequenced according to the SPT rule then the total number of
jobs that would be late is (assume zero for todays date)
Job
A
5
E
Difficulty: Moderate
Feedback: Sequencing
11
D
0 jobs.
1 job.
2 jobs.
3 jobs.
6
C
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
49. If the following jobs are sequenced according to the SPT rule then job B is completed
on day (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
6.
14.
15.
22
Difficulty: Moderate
Feedback: Sequencing
50. If the following jobs are sequenced according to the LPT rule then the mean
completion time (in days) for all jobs is (assume zero for todays date)
Job
A
5
E
Difficulty: Moderate
11
D
118 days.
6.6 days.
33 days.
23.6 days
6
C
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
Feedback: Sequencing
51. If the following jobs are sequenced according to the LPT rule then the mean
tardiness (in days) for all jobs is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
62 days.
12.4 days.
15.5 days.
25 days
Difficulty: Moderate
Feedback: Sequencing
52. If the following jobs are sequenced according to the LPT rule then the maximum job
tardiness (in days) is (assume zero for todays date)
Job
A
11
D
5
E
7 days.
10 days.
20 days.
25 days.
6
C
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
Difficulty: Moderate
Feedback: Sequencing
53. If the following jobs are sequenced according to the LPT rule then the total number of
jobs that would be late is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
5 jobs.
4 jobs.
3 jobs.
2 jobs.
Difficulty: Moderate
Feedback: Sequencing
54. If the following jobs are sequenced according to the LPT rule then job D would be
completed on day (assume zero for todays date )
Job
A
6
C
11
D
5
E
a. 5.
b. 25.
c. 30.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
d. 33.
Difficulty: Moderate
Feedback: Sequencing
55. If the following jobs are sequenced according to the SLACK rule then the mean
completion time (in days) for all jobs is (assume zero for todays date)
Job
A
6
C
11
D
7
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
35 days.
20.8 days.
18.4 days.
7 days.
Difficulty: Hard
Feedback: Sequencing
56. If the following jobs are sequenced according to the SLACK rule then the mean
tardiness (in days) for all jobs is (assume zero for todays 11.25 date)
Job
A
6
C
11
D
7
E
a. days.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
b. 9 days.
c. 20 days.
d. 12.5 days.
Difficulty: Hard
Feedback: Sequencing
57. If the following jobs are sequenced according to the SLACK rule then the maximum
job tardiness (in days) is (assume zero for todays date)
Job
A
6
C
11
D
7
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
20 days.
12 days.
10 days.
7 days
Difficulty: Hard
Feedback: Sequencing
58. If the following jobs are sequenced according to the SLACK rule then the total
number of jobs that would be late is (assume zero for todays date)
Job
Processing
Time (days)
A
8
B
6
C
11
D
7
Due
Date
12
15
17
10
E
a.
b.
c.
d.
3
8
1 jobs.
2 jobs.
3 jobs
4 jobs
Difficulty: Hard
Feedback: Sequencing
59. If the following jobs are sequenced according to the SLACK rule then job A would
be completed on day (assume zero for todays date)
Job
A
6
C
11
D
7
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
8.
7.
15.
12.
Difficulty: Hard
Feedback: Sequencing
60. If the following jobs are sequenced according to the DDATE rule then the mean
completion time (in days) for all jobs is (assume zero for todays date)
Job
Processing
Time (days)
A
8
B
C
6
11
Due
Date
12
15
17
D
E
a.
b.
c.
d.
5
3
10
8
16.4 days.
22.6 days.
28.7 days.
33.0 days
Difficulty: Moderate
Feedback: Sequencing
61. If the following jobs are sequenced according to the DDATE rule then the mean
tardiness (in days) for all jobs is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
9.0 days.
5.4 days.
3.2 days.
2.8 days
Difficulty: Moderate
Feedback: Sequencing
62. If the following jobs are sequenced according to the DDATE rule then the maximum
job tardiness (in days) is (assume zero for todays date)
Job
A
Processing
Time (days)
8
Due
Date
12
B
C
11
D
5
E
a.
b.
c.
d.
6
3
15
17
10
8
12 days.
15 days.
16 days.
27 days.
Ans:: c
Difficulty: Moderate
Feedback: Sequencing
63. If the following jobs are sequenced according to the DDATE rule then the total
number of jobs that would be late is (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
5 jobs.
4 jobs.
3 jobs.
2 jobs.
Difficulty: Moderate
Feedback: Sequencing
64. If the following jobs are sequenced according to the DDATE rule then job A would
be completed on day (assume zero for todays date)
Job
A
6
C
11
D
5
E
a.
b.
c.
d.
8
B
3
Due
Date
Processing
Time (days)
12
15
17
10
8
8.
12.
16.
22
Ans c
Difficulty: Moderate
Feedback: Sequencing
65. Four products (1, 2, 3, and 4) must be processed on one of four machines (A, B, C,
and D). The times required in minutes for each product on each machine are shown
below.
Machine
Product
A
B
C
D
1
10
9
16
12
2
8
14
17
5
3
19
20
11
7
4
8
18
5
10
If management wishes to assign products to machines so that the total time to
complete all the products is minimized, then Product 1 is assigned to
a.
b.
c.
d.
machine A
machine B
machine C
machine D
Difficulty: Hard
Feedback: Sequencing
66. Four products (1, 2, 3, and 4) must be processed on one of four machines (A, B, C,
and D). The times required in minutes for each product on each machine are shown
below.
Machine
Product
A
B
C
D
1
10
9
16
12
2
8
14
17
5
3
19
20
11
7
4
8
18
5
10
If management wishes to assign products to machines so that the total time to
complete all the products is minimized, then Product 3 is assigned to
a. machine A.
b. machine B.
c. machine C.
d. machine D
Difficulty: Hard
Feedback: Sequencing
67. Four products (1, 2, 3, and 4) must be processed on one of four machines (A, B, C,
and D). The times required in minutes for each product on each machine are shown
below.
Machine
Product
A
B
C
D
1
10
9
16
12
2
8
14
17
5
3
19
20
11
7
4
8
18
5
10
If management wishes to assign products to machines so that the total time to
complete all the products is minimized, then Product 4 is assigned to
a.
b.
c.
d.
machine A.
machine B.
machine C.
machine D.
Difficulty: Hard
Feedback: Sequencing
68. Four products (1, 2, 3, and 4) must be processed on one of four machines (A, B, C,
and D). The times required in minutes for each product on each machine are shown
below.
Machine
Product
A
B
C
D
1
10
9
16
12
2
8
14
17
5
3
19
20
11
7
4
8
18
5
10
If management assigns products to machines so that the total time to complete all jobs
is minimized, then the time to complete Product 1 is
a. 10 minutes.
b. 9 minutes.
c. 16 minutes.
d. 12 minutes
Difficulty: Hard
Feedback: Sequencing
69. Four products (1, 2, 3, and 4) must be processed on one of four machines (A, B, C,
and D). The times required in minutes for each product on each machine are shown
below.
Machine
Product
A
B
C
D
1
10
9
16
12
2
8
14
17
5
3
19
20
11
7
4
8
18
5
10
The minimum time required to complete all the products is
a. 29 minutes.
b. 27 minutes.
c. 33 minutes.
d. 36 minutes.
Difficulty: Hard
Feedback: Sequencing
70. The scheduling rule that minimizes the makespan time for a set of jobs that must be
processed through a two-step system where every job follows the same sequence
through the two processes is
a. the shortest processing time (SPT).
b. Johnsons Rule.
c. the assignment method.
d. the minimal slack rule.
Difficulty: Easy
Feedback: Sequencing
71. The following set of jobs must be processed serially through a two-step system. If
Johnsons Rule is used to sequence the jobs then the order in which the jobs would be
performed is
Job
Process
1
A
12
B
8
C
7
D
10
E
5
Process
2
9
11
6
14
8
a.
b.
c.
d.
A-B-C-D-E
C-B-E-D-A
E-B-D-A-C
D-E-B-A-C
Difficulty: Hard
Feedback: Sequencing
72. The following set of jobs must be processed serially through a two-step system. The
times at each process are in hours. If Johnsons Rule is used to sequence the jobs
then the makespan time (in hours) for all jobs is
Job
A
8
C
7
D
10
E
5
Process
2
12
B
a.
b.
c.
d.
Process
1
9
11
6
14
8
42 hours.
53 hours.
90 hours.
95 hours.
Difficulty: Hard
Feedback: Sequencing
73. The following set of jobs must be processed serially through a two-step system. The
times at each process are in hours. If Johnsons Rule is used to sequence the jobs
then Job C would start processing on operation 2 at
Job
Process
1
A
12
B
C
8
7
Process
2
9
11
6
D
E
a.
b.
c.
d.
10
5
14
8
hour 13.
hour 26.
hour 47.
hour 53
Difficulty: Hard
Feedback: Sequencing
74. The following set of jobs must be processed serially through a two-step system. The
times at each process are in hours. If Johnsons Rule is used to sequence the jobs
then Job A would complete processing on operation 2 at
Job
A
8
C
7
D
10
E
5
Process
2
12
B
a.
b.
c.
d.
Process
1
9
11
6
14
8
hour 21.
hour 35.
hour 38.
hour 47.
Difficulty: Hard
Feedback: Sequencing
75. A ________ chart is used to show both planned and completed activities against a
time scale.
a. Gantt
b. Dispatch
c. Sequencing
d. None of the above.
Difficulty: Easy
Feedback: Monitoring
Short Answer Questions
76. What makes scheduling so difficult in a job shop?
Ans: Job shop scheduling is difficult because of the variety of jobs (or customers) that are
processed, each with distinctive routing and processing requirements. In addition,
although the volume of each customer order may be small, there are probably a great
number of different orders in the shop at any one time. This necessitates planning for the
production of each job as it arrives, scheduling its use of limited resources, and
monitoring its progress through the system.
Difficulty: Easy
Feedback: Objectives in Scheduling
77. List some of the objectives in scheduling.
Ans: There are many different possible objectives in constructing a schedule, including:
meeting customer due dates, minimizing job lateness, minimizing response time,
minimizing completion time, minimizing time in the system, minimizing overtime,
maximizing machine or labor utilization, minimizing idle time, and minimizing work-inprocess inventory.
Difficulty: Moderate
Feedback: Objectives of Scheduling
78. Why is monitoring an important component of job shop scheduling?
Ans: In a job shop environment, where jobs follow different paths through the shop, visit
many different machine centers, and compete for similar resources, it is not always easy
to keep track of the status of a job. When jobs are first released to the shop, it is relatively
easy to observe the queue that they join and predict when their initial operations might be
completed. As the job progresses, however, or the shop becomes more congested, it
becomes increasingly difficult to follow the job through the system. Competition for
resources (resulting in ling queues), machine breakdowns, quality programs, and setup
requirements are just a few of the things that can delay a jobs progress. Monitoring
develops progress reports on each job until it is completed and helps maintain reliable
schedules in the system.
Difficulty: Moderate
Feedback: Monitoring
79. What is the difference between infinite scheduling and finite scheduling?
Ans: Infinite scheduling loads work without regard to capacity and then levels the load
and sequences the jobs. Thus, leveling and sequencing decisions are made after overloads
and underloads have been identified. This iterative process is very time consuming and
not efficient. Finite scheduling assumes a fixed maximum capacity and will not load the
resource beyond its capacity. Loading and sequencing decisions are made at the same
time, so that the first jobs loaded onto a work center are of highest priority. Any jobs
remaining after the capacity of the work center or resource has been reached are of lower
priority and are scheduled for later time periods. This approach is easier than the infinite
scheduling approach, but it will be successful only if the criteria for choosing the work to
be performed, as well as capacity limitations, can be expressed accurately and concisely.
Difficulty: Moderate
Feedback: Advanced Planning and Scheduling Systems
80. Who developed the theory of constraints (TOC) and what is its approach to
scheduling?
Ans: The theory of constraints (TOC) was developed by an Israeli physicist named
Eliyahu Goldratt. Because he did not have a business or manufacturing background, Dr.
Goldratt took a commonsense, intuitive approach to scheduling. He developed a software
system that used mathematical programming and simulation to create a schedule that
realistically considered the constraints of the manufacturing system. Dr. Goldratts first
insight into the scheduling problem led him to simplify the number of variables
considered. Instead of trying to balance the capacity of the manufacturing system, he
thought it best to balance the flow of work through the system. He identified resources as
bottleneck or non-bottleneck and noted that the systems flow is controlled by the
bottleneck resources. These resources should always have material to work on, should
spend little time on nonproductive activities, should be fully staffed, and should be the
focus of continuous improvement activities. Goldratt pointed out that an hours worth of
production lost at a bottleneck reduces the output of the system by the same amount of
time, whereas an hour lost at a non-bottleneck resource may have no effect at all.
Consequently, Goldratt simplified the scheduling problem by initially scheduling the
bottleneck resources and then the non-bottleneck ones. In this way production could be
synchronized with the needs of the bottleneck and the system as a whole.
Difficulty: Hard
Feedback: Theory of Constraints

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UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSExamination 1Posted Saturday, 16 October 2010.Due no later than 5pm on Wednesday, 20 October 2010.Instructions:1. Time limit: 4 uninterrupted hours.2. There are four questions worth a total of 100 points, plus a 5-poin

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSExamination 1Posted Friday, 17 October 2008.Due no later than 5pm on Tuesday, 21 October 2008.Instructions:1. Time limit: 4 uninterrupted hours.2. There are three questions worth a total of 105 points.(Any scores abov

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONS IN SCIENCE AND ENGINEERING Examination 1Posted 20 February 2008, 10:00 AM. Due 12 noon on Wednesday, 27 February 2008.Instructions: 1. Time limit: 4 uninterrupted hours. 2. There are ve questions worth a total of 100 poin

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSExamination 2Posted Monday, 8 December 2008.Due no later than 5pm on Wednesday, 17 December 2008.Instructions:1. Time limit: 4 uninterrupted hours.2. There are three questions worth a total of 100 points, plus a 10 poi

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 1Posted Wednesday 25 August 2010. Due Wednesday 1 September 2010, 5pm.1. [18 points]For each of the following equations, specify whether each is (a) an ODE or a PDE; (b) determine itsorder; (c) specify wheth

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 1Distributed Wednesday January 10. Due Wednesday January 17, 2007 in class. 1. [30 points] Determine whether each of the following dierential equations is an ODE or a PDE, determine its order, and specify wheth

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 2Posted Wednesday 1 September 2010. (Corrected 6 Sept. 2010.) Due Wednesday 8 September 2010, 5pm.1. [27 points]Consider the following sets of functions. Demonstrate whether or not each is a vector space(wit

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 2Posted January 17. Due Wednesday January 24, 2007 in class.1. [35 points]Consider the temperature functionu(x, t) = e2t/(c)sin(x)for constant , , c, and .(a) Show that this function u(x, t) is a soluti

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 3Posted Wednesday 8 September 2010. Corrected 14 September. Due Wednesday 15 September 2010, 5pm.1. [20 points]Determine whether each of the following functions (, ) determines an inner product on the vector

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 3Posted Wednesday 24 January 2007. Due Wednesday 31 January 2007 in class.1. [20 points]Consider the following sets. Demonstrate whether or not each is a vector space(with addition and scalar multiplication

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 4Posted Wednesday 15 September 2010. Due Wednesday 22 September 2010, 5pm.1. [20 points]The equation x1 + x2 + x3 = 0 denes a plane in(a) Find two linearly independent vectors inR3 that passes through the o

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 5Posted Thursday 23 September 2010. Corrected 28 Sept 2010. Due Wednesday 29 September 2010, 5pm.All of the problems on this set use the inner product1(u, v ) =u(x)v (x) dx.01. [30 points]2Consider the

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 6Posted Wednesday 29 September 2010. Due Wednesday 6 October 2010, 5pm.General advice: You may compute any integrals you encounter using symbolic mathematics tools such asWolframAlpha, Mathematica, or the Sym

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 7Posted Wednesday 6 October 2010. Due Wednesday 13 October 2010, 5pm.General advice: You may compute any integrals you encounter using symbolic mathematics tools such asWolframAlpha, Mathematica, or the Symbo

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 8Posted Thursday 21 October 2010. Due Wednesday 27 October 2010, 5pm.1. [50 points]Consider the following three matrices:(i) A =0110(ii) A =50494950(iii) A =011.0(a) For each of the matrices

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 9Posted Thursday 28 October 2010. Due Wednesday 3 November 2010, 5pm.1. [40 points] 1, x [0, 1/3];0, x (1/3, 2/3);(a) Consider the function u0 (x) =1, x [2/3, 1].2Recall that the eigenvalues of the opera

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 9Posted Friday 14 March 2008. Due Friday 21 March 2008 in class.1. [70 points]Use the Fourier series method to solve the following initial boundary value problemu 2 u2txu(0, t)xu(1, t)xu(x, 0)= 4x

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 10Posted Wednesday 3 November 2010. Due Wednesday 10 November 2010, 5pm.1. [50 points: 8 points each for (a), (b), (d), (e); 4 points for (c); 14 points for (f)]This problem and the next study the heat equati

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 10Posted Wednesday 19 March 2008. Due Wednesday 26 March 2008 in class. 1. [40 points] For each of the following examples, (1) compute the matrix exponential etA by hand, and (2) explain the behavior of dx/dt =

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 11Posted Thursday 18 November 2010. Due Tuesday 23 November 2010, 5pm.This problem set counts for 75 points.Late problem sets are due by 5pm on Wednesday 24 November 2010.1. [30 points: 7 points each for (a)

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 11Posted Wednesday 26 March 2008. Due Wednesday 2 April 2008 in class. 1. [50 points] Consider the wave equation 2u 2u = c2 2 t2 x for 0 x 1 and t 0 subject to the mixed boundary conditions u(0, t) = 0, for all

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 12Posted Friday 26 November 2010. Due Friday 3 December 2010, 5pm.This problem set counts for 100 points, plus a 20 point bonus.1. [50 points]On Problem Set 10, you solved the heat equation on a two-dimensio

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 12Due Wednesday 9 April 2008 in class.1. [60 points]We wish to approximate the solution to the heat equationu 2 u 2 = 100 tx,tx0 x 1, t 0with homogeneous Dirichlet boundary conditionsu(0, t) = u(1, t)

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 13Posted Friday 11 April, 2008. Due Friday 18 April, 2006 in class.1. [50 points]We wish to approximate the solution to the homogenous wave equation2u2u c2 2 = 0 ,t2x0 x 1, t 0with homogeneous Dirichle

UVA - CAAM - 330

CAAM 336DIFFERENTIAL EQUATIONS IN SCIENCE AND ENGINEERINGExamination 1SOLUTIONS1. [20 points]Vector spaces(a) Determine if the following sets are vector spaces or not- justify your answer. (Deneaddition and scalar multiplication in the obvious way.

UVA - CAAM - 330

UVA - CAAM - 330

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 8Posted Wednesday 27 February 2008. Due Friday 7 March 2008 in class.1. [50 points]Use the nite element method to solve the dierential equationddu(1 + x2 )dxdx= 2x,0<x<1subject to the inhomogeneous bo

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 1 SolutionsPosted Wednesday 25 August 2010. Due Wednesday 1 September 2010, 5pm.1. [18 points: 3 points per part]For each of the following equations, specify whether each is (a) an ODE or a PDE; (b) determine

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 2 SolutionsPosted Wednesday 1 September 2010. (Corrected 6 Sept. 2010.) Due Wednesday 8 September 2010, 5pm.1. [27 points: 4.5 points per problem]Consider the following sets of functions. Demonstrate whether

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 3 SolutionsPosted Wednesday 8 September 2010. Corrected 14 September. Due Wednesday 15 September 2010, 5pm.1. [20 points: 5 points per part]Determine whether each of the following functions (, ) determines an

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 4 SolutionsPosted Wednesday 15 September 2010. Due Wednesday 22 September 2010, 5pm.1. [20 points]The equation x1 + x2 + x3 = 0 denes a plane in(a) Find two linearly independent vectors inR3 that passes thr

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 5 SolutionsPosted Thursday 23 September 2010. Due Wednesday 29 September 2010, 5pm.All of the problems on this set use the inner product1(u, v ) =0u(x)v (x) dx.1. [30 points: 6 points each for (a) and (b)

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 6 SolutionsPosted Wednesday 29 September 2010. Due Wednesday 6 October 2010, 5pm.General advice: You may compute any integrals you encounter using symbolic mathematics tools such asWolframAlpha, Mathematica,

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 7 SolutionsPosted Wednesday 6 October 2010. Due Wednesday 13 October 2010, 5pm.General advice: You may compute any integrals you encounter using symbolic mathematics tools such asWolframAlpha, Mathematica, or

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 8 SolutionsPosted Thursday 21 October 2010. Due Wednesday 27 October 2010, 5pm.1. [50 points: 18 points for (a); 12 points for (b); 10 points each for (c) and (d)]Consider the following three matrices:(i) A

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 9 SolutionsPosted Thursday 28 October 2010. Due Wednesday 3 November 2010, 5pm.1. [40 points: 12 points each for (a) and (b); 6 points for (c); 10 points for (d)] 1, x [0, 1/3];0, x (1/3, 2/3);(a) Consider

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 10 SolutionsPosted Wednesday 3 November 2010. Due Wednesday 10 November 2010, 5pm.1. [50 points: 8 points each for (a), (b), (d), (e); 4 points for (c); 14 points for (f)]This problem and the next study the h

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 11 SolutionsPosted Thursday 18 November 2010. Due Tuesday 23 November 2010, 5pm.This problem set counts for 75 points.Late problem sets are due by 5pm on Wednesday 24 November 2010.1. [30 points: 7 points ea

UVA - CAAM - 330

CAAM 336 DIFFERENTIAL EQUATIONSProblem Set 12 SolutionsPosted Friday 26 November 2010. Due Friday 3 December 2010, 5pm.This problem set counts for 100 points, plus a 20 point bonus.1. [50 points: 15 points for (a); 10 points for (b); 5 points for (c);

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Technology and Health Care 18 (2010) 137144DOI 10.3233/THC-2010-0576IOS Press137A text message-based intervention to bridgethe healthcare communication gap in therural developing worldNadim Mahmuda,b, Joce Rodriguezb,c and Josh Nesbitba StanfordU

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Project MasilulekeA Breakthrough Initiative to Combat HIV/AIDSUtilizing Mobile Technology & HIV Self-Testing in South AfricaA signature program of the PopTech Accelerator, Project Masiluleke is a breakthrough crosssector collaboration that employs mobi

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8Current GIS Research in HealthCare and Disease Analyses8.1Geography of HealthThe objective of spatial epidemiology is to identify the causes of diseasesby correlating or relating spatial disease patterns to geographic variation inhealth risks, as

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International Journal of HealthGeographicsBioMed CentralOpen AccessResearchOpen source GIS for HIV/AIDS managementBas Vanmeulebrouk*1, Ulrike Rivett2, Adam Ricketts3 and Melissa Loudon2Address: 1Centre for Geo-Information, Wageningen University and

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Robertson and Nelson International Journal of Health Geographics 2010, 9:16http:/www.ij-healthgeographics.com/content/9/1/16INTERNATIONAL JOURNALOF HEALTH GEOGRAPHICSREVIEWOpen AccessReview of software for space-time diseasesurveillanceColin Rober

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REVIEW ARTICLEPharmacoeconomics 2011; 29 (7): 579-5991170-7690/11/0007-0579/$49.95/0 2011 Adis Data Information BV. All rights reserved.Unit Costs for Delivery of AntiretroviralTreatment and Prevention of Mother-toChild Transmission of HIVA Systemat

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Barriers and Gaps Affecting mHealthin Low and Middle Income CountriesA Policy White PaperMarch 2010Researched and written by:The Earth Institute, Columbia UniversityCommissioned by:The mHealth AllianceAuthorCenter for Global Health and Economic D

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