### Practice Examination Questions

Practice Examination 3 – CMBA Spring 2012 Page 1 of 10
Practice Examination Questions
QUESTIONS ON PRODUCTION SCHEDULING
SCH 1: What is the order of processing the jobs using Johnson’s rule for the data given in the
following table?

 PROCESSING TIME (DAYS) JOB MACHINE 1 MACHINE 2 A 8 12 B 4 9 C 11 7 D 2 6 E 10 5

SCH 2: Draw a Gantt chart and find the value of make-span
(time to complete all jobs) using the sequence A-B-CD-E for the following problem.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
M1 A A A A A A A A B B B B B B B B B C C C D D E E E E E E
M2 A A A A A A B B B B C C C C C C C D D D D D D E E E E E
Time Days

 Processing Time (Days) Job M1 M2 A 8 6 B 9 4 C 3 7 D 2 6 E 6 5

Practice Examination 3 – CMBA Spring 2012 Page 2 of 10
DATA FOR QUESTIONS SCH 3 to SCH 6

 JOB PROCESSING TIME (days) DUE DATE (days) A 18 46 B 13 33 C 21 25 D 19 52 E 24 48

SCH 3: What is the average number of jobs in the system using the Earliest Due Date (EDD)
rule?
SCH 4: What is the average tardiness using the First Come First Served (FCFS) rule? ANSWER:
18.6
SCH 5: What is the number of tardy jobs using Shortest Processing Time rule? ANSWER: 2
SCH 6: What is the average completion time using First Come First Served rule? ANSWER: 53.4
Job Time Due Date Completion
Time Tardiness Job Time Due Date Completion Time Tardiness
C 21 25 21 0 B 13 33 13 0
B 13 33 34 1 A 18 46 31 0
A 18 46 52 6 D 19 52 50 0
E 24 48 76 28 C 21 25 71 46 tardy
D 19 52 95 43 E 24 48 95 47 tardy
Total 278 Number of Tardy Jobs
2
Average Number of Jobs 278/95 2.9263158
Job Time Due Date Completion
Time Tardiness Job Time Due Date Completion Time
A 18 46 18 0 A 18 46 18
B 13 33 31 0 B 13 33 31
C 21 25 52 27 C 21 25 52
D 19 52 71 19 D 19 52 71
E 24 48 95 47 E 24 48 95
Average
18.6 Average Completion Time 53.4
SCH 4 – FCFS
SCH 3 – EDD SCH 5 – SPT
SCH 6 – FCFS

Practice Examination 3 – CMBA Spring 2012 Page 3 of 10
SCH 7: Suppose SPT rule is being used in a “dynamic” scheduling problem. There are four jobs
A, B, C, and D ready to be processed at the present time. The processing times for the five jobs
are 3, 7, 6 and 2 days respectively. A new job E will arrive on the 4th day. The processing time
for job E is 2 days.
On which day (from present time) job E will be completed?
The sequencing order using SPT is D-A-C-B (D = 2 days, A = 3 days, C = 6 days and B = 7 days).
When job E comes on 4
th day, job A is being processed and will be completed at time 5. After
job A is completed, there are three jobs waiting to be processed B (7 days), C (6 days) and E (2
days). Using SPT, job E will be picked up for processing. Job E will then be completed at time 7.
SCH 8: The due date for a job is day 20. It is finished on day 18. What is the lateness of the job?
What is the tardiness of the job?
Lateness is defined as completion time – due date = 18 – 20 = -2. The lateness can be positive
(job is delayed), or 0 (on time job), or negative (job is completed early).
The tardiness refers only to the positive lateness. For this problem, the tardiness is 0 (zero)
because the job is not late. Mathematically, tardiness = maximum (0, lateness). For this
problem, tardiness = maximum (0, -2) = 0.

Practice Examination 3 – CMBA Spring 2012 Page 4 of 10
Practice Examination Questions
QUESTIONS ON PROJECT MANAGEMENT
Consider the following AON network and the data given in the following table to answer the
next two questions.

 Activity Time (days) A 5 B 6 C 7 D 4 E 9 F 3 G 4

PROJ 1: Find the earliest completion time of the project. Answer: 28
Solution
List all paths and identify the longest path.
A-B-D-E-G: 28 days (longest path, critical path, gives the project duration)
A-B-D-F-G: 22 days
A-C-F-G: 19 days
A
B
C
D
E F
G

Practice Examination 3 – CMBA Spring 2012 Page 5 of 10
PROJ 2: What is the latest start time of activity D if the project completion time (due date) is
set at 29 days.
Solution: Find the LST and LFT for each activity. These are shown in the following table.

 Activity Time (days) EST EFT LST LFT A 5 0 5 1 6 B 6 5 11 6 12 C 7 5 12 15 22 D 4 11 15 12 16 E 9 15 24 16 25 F 3 15 18 22 25 G 4 24 28 25 29

Note: EST and EFT are also shown in the table although these are not required to solve the
problem.
A
B C
D
E F
G

Practice Examination 3 – CMBA Spring 2012 Page 6 of 10
PROJ 3: Below are the data for a Time-Cost CPM Scheduling model analysis. The time is in days.
Immediate Normal Crash Normal Crash
Activity Predecessor Time Time Cost Cost
A None 3 2 \$200 \$400
B A 4 3 \$300 \$600
C A 1 1 \$200 \$200
D B and C 3 2 \$500 \$550
E D 2 1 \$500 \$900
What is the time to complete this project and the total normal cost (before crashing any of the
activities)?
Solution
There are two paths in the network shown below for this problem that include ABDE (12 days)
and ACDE (9 days). Therefore, project duration is 12 days.
The total of normal cost is \$ 17,000 (200+300+200+500+500).
A
B C
D E

Practice Examination 3 – CMBA Spring 2012 Page 7 of 10
PROJ 4: Below are the data for a Time-Cost CPM Scheduling model analysis. The time is in days.
Immediate Normal Crash Normal Crash
Activity Predecessor Time Time Cost Cost
A None 3 2 \$200 \$400
B A 4 3 \$300 \$600
C A 1 1 \$300 \$300
D B and C 3 2 \$500 \$550
E D 2 1 \$500 \$900
If you crash this project to reduce the total time by one day what is the time to complete the
project and total cost?
Solution
There are two paths in the network shown below for this problem that include ABDE (12 days)
and ACDE (9 days).
One of the activities has to be crashed to reduce the project duration by one day. Find the cost
of crashing/day for each activity. These costs are A (\$ 200); B (\$ 300), C (can not be crashed), D
(\$ 50) and E (\$ 400). Pick up the activity that belongs to the critical path (ABDE) and has the
minimum cost of crashing/day. The activity selected is D at a cost of \$ 50.
After crashing D, the length of path ABDE will become 11 days and the cost to complete the
project will be \$ 1,850 (1,800 + 50). The current normal cost, without crashing is \$ 1,800.
Also note that the length of path ACDE will become 8 days since activity D belongs to this path
also.
A
B C
D E

Practice Examination 3 – CMBA Spring 2012 Page 8 of 10
PROJ 5: You have collected the data for a Time-Cost CPM Scheduling model analysis. The time is
in days and the project “direct costs” are given below.

 Activity Immediate Predecessor Normal Time (days) Crash Time (days) Normal Cost (Direct) \$ Crash Cost (Direct) \$ A None 3 2 300 400 B A 3 3 100 100 C A 1 1 200 200 D B and C 3 2 400 550 E D 2 1 500 900 F E 3 3 200 200 G F 2 2 100 100

The indirect costs for the project are determined on a daily duration basis and are given below:
Total indirect costs for 16 days of project duration: \$ 400
Total indirect costs for 15 days of project duration: \$ 250
Total indirect costs for 14 days of project duration: \$ 200
Total indirect costs for 13 days of project duration: \$ 100
At what project duration do we achieve the lowest total project cost (i.e., direct plus indirect
costs)?
A) 16 days
B) 15 days
C) 14 days
D) 13 days
E) 12 days
The network for this problem is given below. The costs of crashing activities are given in the last
column in the following table.

 Activity Immediate Predecessor Normal Time (days) Crash Time (days) Normal Cost (Direct) \$ Crash Cost (Direct) \$ Cost of Crashing/day A None 3 2 300 400 100 B A 3 3 100 100 No crashing C A 1 1 200 200 No crashing D B and C 3 2 400 550 150 E D 2 1 500 900 400 F E 3 3 200 200 No crashing G F 2 2 100 100 No crashing

Practice Examination 3 – CMBA Spring 2012 Page 9 of 10
There are two paths in the network. The sequence of crashing the activities and the costs are
given in the following table. The minimum cost is for project duration 15 days.

 Schedule 1 Schedule 2 Schedule 3 Schedule 4 Paths No crashing Crash A (1 day) Crash (D 1 day) Crash E ( 1 day) ABDEFG 16 DAYS 15 14 13 ACDEFG 14 DAYS 13 12 11 Crashing Cost 0 100 150 400 Activity Cost 1800 1900 (1800+100) 2050 (1900+150) 2450 (2050+400) Indirect Cost 400 250 200 100 Total Cost 220 2150 2250 2550

A
B C
D E F G

Practice Examination 3 – CMBA Spring 2012 Page 10 of 10
PROJ 6: Suppose the length of critical path in a project is 38 days. The critical path consists of
four activities whose variances are 1.5, 1.2, 1.0 and 0.3. What is the probability of completing
the project within 41 days?
Solution
Find the variance of the critical path, V = 1.5 + 1.2 + 1.0 + 0.3 = 4.0
Standard Deviation of the critical path =
σ = 4.00 = 2.0
Find z = (Due Date – Length of Critical Path)/
σ = (41 – 38)/2 = 1.5
Probability for z = 1.5 is 0.9332 (use table or the Excel function).
PROJ 7: Suppose the length of critical path in a project is 52 days and the standard deviation is 3
days. What due date should be set for the project so that the probability of completing the
project is 95%? Round the answer to the nearest higher integer.
Solution
Find the value of z for the given probability (0.95).
Value of z = 1.645
z= (Due Date – Length of Critical Path)/ standard deviation
Therefore, 1.645 = (Due Date – 52)/3; Solving this equation will give D = 56.935 = 57.
PROJ 8: Suppose the length of critical path in a project is 50 days. The probability of completing
the project within 45days is:
(a) Less than 50%
(b) Greater than 50%
(c) Equal to 50%
(d) Can not be determined from the above data