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? ANSWER: D-B-A-C-E

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.

ANSWER: 39

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? ANSWER 2.93

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?

Answer: 7

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 4th 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?

Answer: -2 (Lateness)

Answer: 0 (Tardiness)

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. Answer: 12

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)?

Answer: 12 days; $ 1,700.

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? Answer: 11 days; $ 1,850.

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

Answer: B

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?

Answer: 0.9332

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.

Answer: 56.935 (= 57)

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

Answer: (a)

Solution

z= (Due Date – Length of Critical Path)/ standard deviation

The numerator (45 – 50) = -5.

The value of z will be a negative number irrespective of the value of standard deviation.

Therefore, the probability will be less than 50%.