MGT 3110: Exam 3 Study Guide
Discussion questions
1.
What is safety stock, and what is its purpose?
2.
Under what circumstances would the amount of safety stock held be (a) Large? (b) Small?
(c) Zero?
3.
What is meant by the term service level? Generally speaking, how is service level related to
the amount of safety stock held?
4.
What is the single-period model, and under what circumstances is it appropriate‘?
5.
Can the optimal stocking level in the single-period model ever be less than expected
demand? Explain briefly.
6.
Why is scheduling fairly simple for repetitive systems but fairly complex for job shops?
7.
What are the main decision areas of job-shop scheduling?
8.
What are Gantt charts‘? How are they used in scheduling? What are the advantages of
using Gantt charts?
9.
What are the basic assumptions of the assignment method of linear programming?
10. Briefly describe each of these priority rules: a. FCFS b. SPT c. EDD d. S/O e. Rush
11. Why are priority rules needed?
12. Explain forward and backward scheduling and each one’s advantage.
13. How are scheduling and productivity related?
14. What factors would you take into account in deciding whether to split a job?
15. Explain the term makespan.
16. Explain the terms flowtime and lateness.
17. What are the advantages and disadvantages of shortest processing time (SPT) rule?
18. What is the Critical Ratio? To what jobs does Critical Ratio give priority to?
19. What can be said about the jobs if CR < 1, or =1, or > 1?
20. What can be said about the jobs negative values for S/O?
21. What is input-output control?
22. Identify the term being described for each of the following:
a. A sequence of activities in a project.
b. The longest time sequence of activities in a project.
c. when both earliest and latest times are the same.
d. The difference in time length of any path and the critical path.
e. The statistical distribution used to describe variability of an activity time.
f. The statistical distribution used to describe path variability.
g. Shortening an activity by allocating additional resources.
23.
24.
25.
26.
27.
Why might a probabilistic estimate of a project’s completion time based solely on the
variance of the critical path be misleading? Under what circumstances would it be
acceptable?
What is AOA convention? Describe its salient aspects. What is main disadvantage of this
convention?
What is AON convention? Describe its salient aspects.
How is slack time of an activity calculated?
What are the three activity times used in a PERT project? How are they used?
Problems
1.
The Winfield Distributing Company has maintained an 80% service level policy for
inventory of string trimmers. Mean demand during the reorder period is 130 trimmers, and
the standard deviation is 80 trimmers. What is the value of ROP and SS?
2.
The new office supply discounter, Paper Clips, Etc. (PCE), sells a certain type of
ergonomically correct office chair which costs $300. The annual holding cost rate is 40%,
annual demand is 600, and the order cost is $20 per order. The store is open 300 days per
year and PCE has decided to establish a customer service level of 90%.
a.
Suppose that the lead time is a constant 4 days and the demand is variable with a
standard deviation of 2.4 chairs per day. What is the safety stock and reorder point?
b. Suppose that the lead time is a variable with an average of 4 days and standard
deviation of 3 days. Further suppose that the demand is constant. What is the safety
stock and reorder point?
c.
Suppose that the lead time is a variable with an average of 4 days and standard
deviation of 3 days. Further suppose that the demand is also variable with a standard
deviation of 2.4 chairs per day. What is the safety stock and reorder point?
3.
An oyster bar buys fresh oysters for $3 per pound and sells them for $10 per pound.
Unsold oyster at the end of the day is sold to a grocery store for $1.20 per pound.
Determine the pounds of oysters that must be ordered each day under each of the following
conditions.
a. The daily demand follows normal distribution with mean of 150 pounds and standard
deviation of 12 pounds.
b. The daily demand follows uniform distribution with between 100 and 200 pounds.
c. The daily demand follows a discrete distribution with the following probabilities.
Demand during lead time
100
120
140
160
180
200
4.
Probability
0.05
0.10
0.10
0.25
0.30
0.20
Consider the following planned and actual hours of input and output.
Week ending
Planned input
Actual input
Planned output
Actual output
1
500
700
650
600
2
800
700
650
700
3
700
700
650
800
4
600
800
650
700
5
600
600
650
650
6
800
500
700
500
Prepare the Input/Output Control chart for this workstation. Assume an initial actual
backlog of 120 hours and zero for the two cumulative deviations.
5.
The following jobs are waiting to be processed at a work center.
Job
A
B
C
D
E
F
Production days needed
30
20
40
50
20
35
Date job due
Remaining operations
2
4
3
1
5
2
40
165
125
65
170
130
Sequence the jobs in the order of SPT, EDD, S/O, and Critical Ratio, and compute (i) Average
flow time, (ii) Average lateness, and (iii) Average no. of jobs in the system, for each of the three
schedule of jobs.
6.
An antique restoration operation uses a two-step sequence that all jobs in a certain
category follow. For the group of jobs listed.
a. Find the sequence that will minimize total completion time.
b. Determine the amount of idle time for workstation 102.
c. What jobs are candidates for splitting‘? Why? If they were split, how much would idle
time and makespan time be reduced? Assume the very first job can be split 50% and
the remaining jobs can be spit 60%.
JOB TIMES (minutes)
Workstation 101
Workstation 102
7.
A
27
45
B
18
33
C
70
30
D
26
24
E
15
10
Consider the tasks, durations in weeks, and predecessor relationships in the following
network.
Immediate
Activity Description Predecessor(s)
A
--B
A
C
A
D
B
E
D, C
F
C
G
F
H
F
I
E, G
J
I
a.
to
4
2
8
1
6
2
2
6
4
1
Find the expected time te for all the activities.
tm
7
8
12
2
8
3
2
8
8
2
tp
10
20
16
3
22
4
2
10
12
3
b.
c.
d.
e.
f.
g.
h.
Draw the AON network
Determine all the paths unique paths and find the expected project completion time
using the te values found in (a).
Determine the earliest and latest start and finish times, slacks, and all the critical
paths.
Determine the path variance for the critical path and the next longest path.
What is the probability of completion of the project before week 42?
What is the probability of completion of the project before week 35?
With 99% confidence what is your estimate for the project completion time.
Answers to discussion questions
1.
What is safety stock, and what is its purpose?
Safety stock is inventory held in excess of expected demand to reduce the risk of stockout
presented by variability in either lead time or demand rates.
2.
Under what circumstances would the amount of safety stock held be (a) Large? (b) Small?
(c) Zero?
Safety stock is large when large variations in lead time and/or usage are present.
Conversely, small variations in usage or lead time require small safety stock. Safety stock
is zero when usage and lead time are constant, or when the service level is 50 percent (and
hence, z = 0).
3.
What is meant by the term service level? Generally speaking, how is service level related to
the amount of safety stock held?
Service level can be defined in at least two different ways – (a) as the probability that
demand will be met with available stock on hand, and (b) the percentage of annual demand
satisfied from inventory.
Increasing the service level requires increasing the amount of safety stock.
4.
What is the single-period model, and under what circumstances is it appropriate‘?
The Single-Period Model is used when inventory items have a limited useful life (i.e., items
are not carried over from one period to the next).
5.
Can the optimal stocking level in the single-period model ever be less than expected
demand? Explain briefly.
Yes. When excess costs are high and shortage costs are low, the optimum stocking level is
less than expected demand.
6.
Why is scheduling fairly simple for repetitive systems but fairly complex for job shops?
Job shops are intended to handle a wide range of processing requirements; jobs tend to
follow many different paths through the shop, and often differ significantly with respect to
processing times. In addition, rush orders, changes in specs, and short planning horizons
must be contended with. Consequently, scheduling can be exceedingly complex.
Conversely, in continuous systems there is a high degree of uniformity which presents
much less difficulty in terms of scheduling.
7.
What are the main decision areas of job-shop scheduling?
The main decision areas of job shop scheduling concern loading and sequencing.
8.
What are Gantt charts‘? How are they used in scheduling? What are the advantages of
using Gantt charts?
Gantt charts are visual aids used by managers to plan and adjust facility and equipment
loading. Advantages of Gantt charts include ease of manipulation, the fact that they provide
a visual model of loading, and the ability to assist in trial and error changes. Disadvantages
include the need to repeatedly update the chart, the inability to reveal costs associated with
different alternatives and the inability to include other details (e.g., processing items which
are dependent on the equipment being used rather than the same regardless of the
equipment being used).
9.
What are the basic assumptions of the assignment method of linear programming?
The assignment model assumes a one-for-one matching is possible, that costs for each
combination are known and fixed, and that in general, each machine is capable of handling
each job. (The last assumption is not strict, however. For example, certain combinations
may be “undesirable” because certain jobs cannot be processed on certain pieces of
equipment.)
10.
Briefly describe each of these priority rules: a. FCFS b. SPT c. EDD d. S/O e. Rush
a. FCFS: process jobs in order of arrival.
b. SPT: process jobs according to processing times, shortest ones first.
c. DD: process jobs by due date, earliest due dates first.
d. S/O: takes into account remaining processing time on all remaining operations for
each job. Jobs with lowest slack per remaining operations are scheduled first.
e. Rush: Emergency or preferred customers first.
11.
Why are priority rules needed?
Priority rules allow for the fact that jobs are not equally important: different processing
sequences will have different consequences for the organization. Priority rules enable an
organization to emphasize those measures of effectiveness deemed most important.
12.
Explain forward and backward scheduling and each one’s advantage.
Forward scheduling is when scheduling starts with a specific date and moves forward;
backward scheduling starts at a specific end date and moves backward. Forward
scheduling shows the earliest finish date; backward scheduling shows the latest start date.
13.
How are scheduling and productivity related?
To the extent that scheduling efforts can achieve a balance in facility or equipment loading,
the amount of output for a given amount of input will be higher than it otherwise would,
and hence, productivity will be higher than otherwise. Conversely, poor scheduling will
adversely affect productivity because it will not enable maximum use of resources.
14.
What factors would you take into account in deciding whether to split a job?
Will throughput time decrease? Is it technically feasible to split a job? How disruptive will
it be? What additional costs will be involved (e.g., setup, increased paperwork)?
15.
Explain the term makespan.
Makespan is the total time needed to complete a group of jobs from the beginning of the
first job to the completion of the last job.
16.
Explain the terms flowtime and lateness.
Flow time is the length of time a job is in the system; lateness is completion time minus
due date.
17.
What are the advantages and disadvantages of shortest processing time (SPT) rule?
SPT minimizes the average flow time, average lateness, and average number of jobs in the
system. It maximizes the number of jobs completed at any point. The disadvantage is that
long jobs are pushed back in the schedule.
18.
What is the Critical Ratio? To what jobs does Critical Ratio give priority to?
The Critical Ratio (CR) is an index number computed by diving the time until due date by
the working time remaining. The CR gives priority to jobs that must be done to keep
shipping on schedule.
19.
What can be said about the jobs if CR < 1, or =1, or > 1?
If CR < 1, then the job has fallen behind, the work remaining exceeds the time until due
date. If CR = 1, then the job is on schedule, the work remaining exactly equals the time
until due date. If CR > 1, then there is slack, the time until due date exceeds the work
remaining.
20.
What can be said about the jobs negative values for S/O?
Negative S/O means the job has more processing time needed than the remaining time till
due date. The more negative S/O value, the higher the priority for the job.
21.
What is input-output control?
Input/output control keeps track of planned versus actual inputs and outputs, highlighting
deviations and indicating bottlenecks using cumulative backlog.
22.
Identify the term being described for each of the following:
a. A path
b. The critical path
c. Critical activity
d. The path slack
e. Beta distribution
f. Normal distribution
g. Shortening an activity by allocating additional resources.
23.
Why might a probabilistic estimate of a project’s completion time based solely on the
variance of the critical path be misleading? Under what circumstances would it be
acceptable?
Near-critical paths can have larger path variance than the critical path itself, which may
result in a lower probability of timely completion than that based solely on the time
distribution of the critical path. Probabilistic estimates based solely on the critical path are
acceptable if the variance of the second longest path is smaller.
24.
What is AOA convention? Describe its salient aspects. What is main disadvantage of this
convention?
AOA stands for Activity On Arrow convention. Under this convention, activities are
represented by arrows in the network. The nodes represent beginning and/or ending of a
given activity and the beginning and/or ending of the next/previous activity. The main
disadvantage is, AOA convention requires the use of “dummy activities” to properly
represent some precedence relationships.
25.
What is AON convention? Describe its salient aspects.
AON stands for Activity On Node convention. Under this convention, activities are
represented by nodes. Arrows used to connect the nodes that represent the precedence
relationships.
26.
How is slack time of an activity calculated?
Slack time may be calculated as the difference between the latest and earliest start times of
an activity, or the latest and earliest finish times.
27.
What are the three activity times used in a PERT project?
They are optimistic time (to), pessimistic time (tp), and most likely time (tm).
Answers to problems
1.
Given dL = 130, dLT = 80, and for 80% service level, Z = 0.84
ROP = 130 + 0.84 x 80 = 197.2, or round up to 198 for at least 80% service level
2.
d = D/No. of days per year = 600/300 = 2 per day, Z for 90% service level = 1.285
a.
Given: L = 4 days Constant, d = 2.4 per day, therefore dLT = 2.4 √4 = 4.8
Safety stock = Z dLT = 1.285 x 4.8 = 6.2 or 7 (round up for at least 90% service level)
ROP = dL + SS = (2 chairs/day * 4) + 7 = 15
b.
Given: L = 4 days with L = 3 and demand is constant, dLT = 2 (3) = 6
Safety stock = Z dLT = 1.285 x 6 = 7.7 or 8 (round up for at least 90% service level)
ROP = dL + SS = (2 chairs/day * 4) + 8 = 16
c.
Given: L = 4 days with L = 3 , and d = 2.4 per day,
therefore dLT = √4(2.4)2 + 22 32 = 7.684
Safety stock = Z dLT = 1.285 x 7.684 = 9.9 or 10 (round up for at least 90% service level)
ROP = dL + SS = (2 chairs/day * 4) + 10 = 18
3.
Cs = Lost profit = Selling price per unit – Cost per unit = 10 – 3 = $7
Co = Cost/unit – salvage value/unit = 3 – 1.20 = $1.80
Optimum service level = 7/(7 + 1.80) = 0.795 = 79.5%
a. From normal table, for 79.5% service level, Z = 0.83
Stock =  + Z  = 150 + 0.825 (12) = 159.9 or 160
b. Stock = a + (b – a) SL = 100 + (200 – 100) 0.795 = 179.5
c. Stock = 180
Demand during lead time
100
120
140
160
180
200
Probability
0.05
0.10
0.10
0.25
0.30
0.20
Cum. Probability
0.05
0.15
0.25
0.50
0.80
1.00
#4.
Week ending
Planned input
Actual input
Cumulative deviation
Planned output
Actual output
Cumulative deviation
Backlog
1
500
700
200
650
600
-50
220
120
2
800
700
100
650
700
0
220
3
700
700
100
650
800
150
120
4
600
800
300
650
700
200
220
5
600
600
300
650
650
200
170
6
800
500
0
700
500
0
170
#5. SPT
Job
B
E
A
F
C
D
Processing
time (Days)
20
20
30
35
40
50
195
Average flow time =
Average tardiness =
Average no. of jobs =
Completion time
(Flowtime)
20
40
70
105
145
195
575
Due date
165
170
40
130
125
65
Tardiness
0
0
30
0
20
130
180
95.833
30
2.949
EDD
Job
A
D
C
F
B
E
Processing
time (Days)
30
50
40
35
20
20
195
Average flow time =
Average tardiness =
Average no. of jobs =
Due date
40
65
125
130
165
170
Completion time
(Flowtime)
30
80
120
155
175
195
755
Tardiness
0
15
0
25
10
25
75
125.83
12.50
3.872
S/O
Job
Processing
time (Days)
Due date
Slack
S/O
Completion
time
(Flowtime) Tardiness
A
D
C
E
B
F
30
50
40
20
20
35
195
40
65
125
170
165
130
Average flow time =
Average tardiness =
Average no. of jobs =
CR
Job
D
A
C
F
B
E
10
145
85
15
150
95
5.00
36.25
28.33
15.00
30.00
47.50
30
80
120
140
160
195
725
0
15
0
0
0
65
80
120.83
13.33
3.718
Processing time
50
30
40
35
20
20
195
Due date
65
40
125
130
165
170
Flow time
50
80
120
155
175
195
775
Job
A
B
C
D
E
F
Processing time
30
20
40
50
20
35
Due date
40
165
125
65
170
130
CR
1.333
8.25
3.125
1.3 D
8.5
3.714
Job
A
B
C
E
F
Processing time
30
20
40
20
35
Due date
-10
115
75
120
80
CR
-0.333 A
5.75
1.875
6
2.285
Job
B
C
E
F
Processing time
20
40
20
35
Due date
85
45
90
50
CR
4.25
1.125 C
4.5
1.43
Tardiness
0
40
0
25
10
25
100
Job
B
E
F
Processing time
20
20
35
Due date
45
50
10
CR
2.25
2.5
0.29 F
Job
B
E
Processing time
20
20
Due date
10
15
CR
0.5 B
0.75
Average flow time =
Average tardiness =
Average no. of jobs =
129.167
16.667
3.974
SPT
95.833
30
2.949
Average flow time =
Average tardiness =
Average no. of jobs =
#6.
EDD
125.83
12.50
3.872
S/O
CR
120.83 129.167
13.33 16.667
3.718
3.974
JOB TIMES (minutes)
A
27
45
Workstation 101
Workstation 102
B
18
33
C
70
30
D
26
24
E
15
10
a. Schedule: B – A – C – D – E
Workstation 101
Processing time
Finish time
18
18
27
45
70
115
26
141
15
156
Job
B
A
C
D
E
Processing time
33
45
30
24
10
Workstation 102
Start time
Finish time
18
51
51
96
115
145
145
169
169
179
Idle time
18
0
19
0
0
37
b. Total idle time for workstation 102 is 37 minutes
B
A
C
18
45
Idle
20
A
Idle
51
30
40
50
E
115
B
18
10
D
96
60
70
80
90
100
141
D
C
115
110
156
120
145
130
140
150
160
E
169
179
170
180
c. Job B and C are candidates for splitting. Starting with 50% splitting for B the
schedule will be:
Workstation 101
Processing time
Finish time
9
9
27
36
70
106
Job
B
A
C
Processing time
33
45
30
Workstation 102
Start time
Finish time
9
42
42
87
115
145
Idle time
9
0
At this point, since WS 102 is ready for C at 87, but C won’t be finished in WS 101 till 115, C
is candidate for splitting. You really need to split C such that it gets finished in WS 101 at 87,
no earlier, i.e. a splitting of 106 – 87 = 19. 19 hours is within the 60% splitting allowed. So,
using a processing time of 70 – 19 = 51, redo C as below:
C
D
E
51
26
15
87
113
128
30
24
10
87
117
141
117
141
151
0
0
0
9
New idle time = 9 hours.
I have now given the correct answer to (c), it turned out to be more complicated than I intended.
So, if you don’t follow the logic you have two options (i) skip it, there will not be a problem like
this in the test, or (ii) see me in Google Hangouts and I will explain it.
#7. (a)
Task
A
B
C
D
E
F
G
H
I
J
to
4
2
8
1
6
2
2
6
4
1
tm
7
8
12
2
8
3
2
8
8
2
tp
10
20
16
3
22
4
2
10
12
3
t
7
9
12
2
10
3
2
8
8
2
Variance
1
64/36
256/36
64/36
4/36
b.
B
D
E
Start
A
I
G
C
F
H
J
Fin
ish
c.
Path
P1
P2
P3
P4
Paths
A-B-D-E-I-J
A-C-E-I-J
A-C-F-G-I-J
A-C-F-H
Path duration
38
39
34
30
Critical path: A-C-E-I-J
d.
Task
Start
A
B
C
D
E
F
G
H
I
J
Mean
ES
7
9
12
2
10
3
2
8
8
2
0
7
7
16
19
19
22
22
29
37
EF
0
7
16
19
18
29
22
24
30
37
39
LS
0
8
7
17
19
24
27
31
29
37
LF
0
7
17
19
19
29
27
29
39
37
39
2
Slack
0
1
0
1
0
5
5
9
0
0
Critical
Critical
Critical
Critical
Critical
36/36
324/36
64/36
4/36
256/36
4/36
0/36
16/36
64/36
4/36
Critical path: A-C-E-I-J
c.
2P = 11.778
2P1 = 19.111
d. P(T<=42):
Probability =
Path:
TE =
2

z=
e. P(T<=35):
Probability =
z=
f. Confidence = 99%
Z = 2.325
Critical path:
A-C-E-I-J
39
11.778
3.4319
(42 – 39)/3.4319 = 0.87
Normal table = 0.8078
Second longest path: A-B-DE-I-J
38
19.111
4.3716
(42 – 38)/4.3716 = 0.91
Normal table = 0.8186
(35 – 39)/3.4319 = -1.17
1- 0.8790 = 0.1210
(35 - 38)/4.3716 = -0.69
= 1 – 0.7549 = 0.2451
0.1210
T = 39 + 2.325 (3.4319)
= 46.98
T = 38 + 2.325 (4.3716) =
48.16
T = 48.16
Answer
0.8078