MGT 3110: Exam 3 Study Guide
DO NOT RELY ONLY ON THIS STUDY GUIDE ALONETO PREAPRE FOR THE
EXAM
Discussion questions
1. Define independent and dependent demand items.
2. What is Master Production Schedule?
3. What is Bill of Materials?
4. What is Low-Level coding and what how is it used?
5. What are the benefits of MRP?
6. What are the inputs required for MRP?
7. What is “Lot Sizing” in MRP?
8. What are the reasons for using a lot sizing method other than Lot-for-lot?
9. What does the part-period balancing lot-sizing technique attempt to do in deciding the lot
sizes?
10. Explain the terms flowtime and lateness.
11. What are the advantages and disadvantages of shortest processing time (SPT) rule?
12. What is the Critical Ratio? To what jobs does Critical Ratio give priority to?
13. What can be said about the jobs if CR < 1, or =1, or > 1?
14. What is input-output control?
15. Define maintenance.
16. Define reliability.
17. What is FR(%)? Define it.
18. Define FR(N). How is computed?
19. Define MTBF. How is related to FR(N)?
20. What is redundancy?
21. What is the impact on system reliability of adding parts or components in parallel?
22. Increasing the number of parts or components in a product tends to reduce its reliability. Why
is this true only when adding components in series?
23. Explain carefully how redundancy improves product reliability.
24. What is breakdown maintenance?
25. Is there an optimal amount of preventive maintenance? What caution should be exercised
before calculating this optimal amount?
Problems
1.
Consider the following Solver model for an aggregate planning problem.
a.
Determine the excel formula for the following cells:
B18
B19
B20
B23
E23
F23
B24
B31
C31
F31
G31
H31
B32
B37
B38
B39
B40
B41
B42
B43
b.
c.
d.
e.
2.
What is the Solver Target cell?
What are the Solver changing cells?
What are the Solver constraints?
What options of Solver must be checked?
A Bill of Materials is desired for a bracket (A) that is made up of a base (B), two springs
(C) and four clamps (D). The base is assembled from one clamp (D) and two housings (E).
Each clamp has one handle (F) and one casting (G). Each housing has two bearings (H)
and one shaft (I).
a. Develop a product structure tree.
b. The lead time for the parts are given below. Develop a time-phased product structure.
c. The available inventory for each part is given in the table below. Determine the net
requirement quantities of all parts required to assemble 50 units of bracket A.
Item
A
B
C
D
E
F
G
Lead time
1
2
3
2
1
2
1
Available
5
5
10
20
50
150
50
H
I
3.
Lead time
1
3
2
4
On-hand
100
30
0
0
Scheduled receipts
50 in week 1
20 in week 1, 30 in week 2
50 in week 1
30 in week 1, 40 in week 3
For the following item the inventory holding cost is $0.80 per week and the setup cost is
$300. Determine the lot sizes and total cost for this item under (i) Lot-for-Lot, (ii) EOQ,
and (iii) PPB methods
Item
Week:
Gross requirement
Scheduled receipts
Projected on-hand
100
Net Requirement
Planned receipts
Planned order releases
5.
5
0
A product (A) consists of a base (B) and a casting (C). The base consists of a plate (P) and
three fasteners (F). The lead time, current on-hand inventory and scheduled receipts are
given below. All components are lot for lot. The MPS requires start of production of 100
units of product A in week 4 and 150 in week 6. Produce the MRP for the upcoming six
weeks. Produce a list of all planned order releases.
Part
B
C
P
F
4.
1
2
LT =
1
100
1
2
250
3
200
4
150
5
250
6
200
7
200
8
150
Consider the following planned and actual hours of input and output.
Week ending
Planned input
Actual input
Planned output
Actual output
1
2
3
4
5
6
7
8
500
700
650
600
800
700
650
700
700
700
650
800
600
800
650
700
600
600
650
650
800
500
700
500
700
500
700
600
800
800
700
800
Prepare the Input/Output Control chart for this workstation. Assume an initial actual
backlog of 120 hours.
6.
A company wishes to assign a set of jobs to a set of machines. The following table provides
data on the profit margin of each job when performed on a specific machine. Setup an
Excel Solver model to determine the set of assignments that maximizes production value.
a.
Fill out the Excel formulas for the following cells:
G13
G14
G15
G16
C17
D17
E17
F17
G19
b.
c.
d.
e.
What is the Solver Target cell?
What are the Solver changing cells?
What are the Solver constraints?
What options of Solver must be checked?
7.
The following jobs are waiting to be processed on day 250
Job
A
B
C
D
E
F
Date Job received
215
220
225
240
245
250
Production days needed
30
20
40
50
20
35
Date job due
290
415
375
315
420
380
Sequence the jobs in the order of SPT, EDD, and Critical Ratio, and compute (i) Average
flow time, (ii) Average lateness, (iii) Average no. of jobs in the system, and (iv)
Utilization, for each of the three schedule of jobs.
8.
A control rod mechanism is a nuclear reactor has 15 critical components with an average
reliability of 99.5% for each component. Determine the overall reliability of the
mechanism.
9.
Determine the system reliability for the following systems.
(a)
R = 90%
R = 98%
R = 92%
R = 90%
R = 90%
R = 98%
(b)
R = 99%
R = 98%
R = 95%
R = 90%
R = 94%
R = 95%
R = 90%
R = 94%
R = 95%
R = 90%
R = 94%
R = 95%
10.
Fifty components of a safety system were tested for reliability, each for 200 hours of
operation. Of the 50, 2 failed after 50 hours of operation, 3 after 75 hours, 1 after 120
hours, and 1 after 150 hours. The rest of the components did not fail. Determine the
following failure rate statistics.
a.
b.
c.
d.
11.
Percentage of failures.
Number of failures per unit-hour of operation.
Mean time between failures.
What is the expected number of failures over 1000 hours of operation?
In a mining operation an electoral exhaust system is used to pump out the fumes. The
break down data over the past 200 weeks is shown in the table below. Each time the
system breaks down the mining operation must be shut down. The estimated cost to repair
the system and the lost production amounted to $12,000. Routine maintenance service for
the exhaust system may be purchased at a cost of $6,000 per week. With this contract the
number of breakdowns is expected to average only 1.2 per week. Then, the cost to bring
the system back on line is $5,000 in lost production. Determine whether it is economical to
purchase the preventive maintenance contract.
No. of breakdowns No. of weeks that many breakdowns occurred (Frequency)
0
75
1
35
2
50
3
20
4
10
5
10
Answers to discussion questions
1. Define independent and dependent demand items.
Finished products whose demand is independent of production decisions are called
“Independent demand” items. Items for which demand can be directly calculated from
production decisions are called “Dependent demand” items. These are raw-materials and
parts required for the production of the finished goods.
2. What is Master Production Schedule?
Master Production Schedule specifies production quantities of each Independent Demand
item for a planning horizon of 12 to 15 weeks. Total of MPS quantities must be in
accordance with the aggregate production plan.
3. What is Bill of Materials?
Bill of materials is structured list of components, ingredients, and materials needed to make
an end product. Items needed to produce a given part are called components or “children”.
The part into which the components go us called “Parent”. The BOM also gives the number
of units of a child item needed to produce one unit of the parent item.
4. What is Low-Level coding and what how is it used?
A level code starting from zero at the top of the BOM tree and incremented by 1 going down
each level of the BOM tree is assigned. Then, the lowest level at which an item appears is
called Low-Level code. The MRP computations are processed one level at a time, starting
from level zero.
5. What are the benefits of MRP?
 Better response to customer orders
 Faster response to market changes
 Improved utilization of facilities and labor
 Reduced inventory levels
6. What are the inputs required for MRP?
 Master Production Schedule
 Bill of Materials
 Inventory status
7. What is “Lot Sizing” in MRP?
The process of combining net requirements into production lots is called lot sizing.
8. What are the reasons for using a lot sizing method other than Lot-for-lot?
 Lot-for-lot often requires too many lots that may not be economically justifiable
 Sometime lot-for-lot generates absurdly small lots
9. What does the part-period balancing lot-sizing technique attempt to do in deciding the lot
sizes?
It balances the setup and holding costs. PPB uses additional information by changing the lot
size to reflect requirements of the next lot size in the future.
10. 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.
11. 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.
12. 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.
13. 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.
14. 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.
15. Define maintenance.
Maintenance consists of all activities involved in keeping a system's equipment in working
order.
16. Define reliability.
Reliability is the probability that a machine part or product will function properly for a
specified time under stated conditions.
17. What is FR(%)? Define it.
FR(%) refers to product failure rate. It measures the percent of failures among the total
number of products tested.
18. Define FR(N). How is computed?
FR(N) refers to the number of failures during a period of time. It measures the number of
failures over the total unit-hours the products operated without failing. it is the ratio of failed
units to total operating hours.
19. Define MTBF. How is related to FR(N)?
MTBF stands for Mean time between failures. It is the expected time between a repair and
the next failure of a component, machine, process, or product. MTBF is the reciprocal of
FR(N)
20. What is redundancy?
Redundancy is the use of a component in parallel to raise reliabilities.
21. What is the impact on system reliability of adding parts or components in parallel?
This will increase the reliability of the system by introducing redundancy.
22. Increasing the number of parts or components in a product tends to reduce its reliability. Why
is this true only when adding components in series?
Adding parts in series involves an additional multiplication by a value less than one, so that
reliability must fall. Adding parts in parallel (the redundancy concept) increases reliability
because only one part of the parallel system must function.
23. Explain carefully how redundancy improves product reliability.
A redundant part or component is connected in parallel with the primary part or component.
"In parallel" means that either the original part or its backup needs to work, not that both
must work at the same time. Redundancy increases reliability by providing an additional path
(through the redundant part) to provide system reliability.
24. What is breakdown maintenance?
Breakdown maintenance is the remedial maintenance that occurs when equipment fails and
must be repaired on an emergency or priority basis.
25. Is there an optimal amount of preventive maintenance? What caution should be exercised
before calculating this optimal amount?
Too little preventive maintenance causes breakdown costs to rise sharply, adding more to
cost than is saved by less preventive maintenance; too much preventive maintenance reduces
breakdowns, but by an amount insufficient to offset the added cost of preventive
maintenance. Operations managers should assure that all costs of breakdowns have been
properly included in the calculations. There is a history of not including indirect and
subjective breakdown cost elements, which leads to performing too little preventive
maintenance.
Answers to problems
1.
B18
B19
B20
B23
E23
F23
B24
B31
C31
F31
G31
H31
B32
B37
B38
B39
B40
B41
B42
B43
=B14/B13
=B14*B5
=B13*B6
=E13
=B23+C23-D23
=E23*F6*$B$18
=E23
=E14
=F23
=E6
=B31+SUM(C31:E31)-F31
=C31*$B$15
=G31
=SUMPRODUCT(E23:E26,F6:F9)*B19
=D35*B20
=E35*B7
=C27*B9
=D27*B10
=G35*B8
=SUM(B37:B42)
b.
B43
c.
C23:D26, D31:E34
d.
D31:D34 <= H31:H34
G31:G34 >= E15
C23:D26 = Integer (if needed)
D31:E34 = Integer (if needed)
e.
Assume linear model
Assume non-negative
2.
A
B
C2
F
D4
E2
D1
G
H2
I
F
G
F
D
G
B
H
E
I
C
A
F
D
G
1
2
3
4
5
Lead time = 7 weeks
Part
A
B
C
D
E
F
G
H
I
Gross
50
1 x A = 45
2 x A = 2 x 45 = 90
4 x A + 1 x B = 4 x 45 + 40 = 220
2 x B = 80
1 x D = 200
1 x D = 200
2 x E = 2 x 30 = 60
1 x E = 30
Available
5
5
10
20
50
150
50
5
0
Net
50 – 5 = 45
45 – 5 = 40
90 – 10 = 80
220 – 20 = 200
80 – 50 = 30
200 – 150 = 50
200 – 50 = 150
60 – 5 = 55
30 – 0 = 30
6
7
3.
A has releases of 100 in 4, 150 in 7; B has a release of 150 in 6, but on hand inventory accounted
for all other needs; C has releases of 150 in 4, 70 in 1; P has a release of 150 in 5 and F has a release of
450 in 5 (the beginning inventory of B leads to no other gross requirements of P or F).
1
2
3
1
0
50
100
0
0
2
0
3
0
150
0
0
150
0
0
MPS start for A
Item B
Item C
Week:
Gross requirement
Scheduled receipts
Projected on-hand
30
Planned receipts
Planned order releases
Item F
Week
Gross requirement
Scheduled receipts
Projected on-hand
Planned receipts
Planned order releases
5
6
150
4
100
5
0
6
150
150
0
0
50
0
100
50
100
0
Lead time =
Week:
Gross requirement
Scheduled receipts
Projected on-hand
100
Planned receipts
Planned order releases
Item P
Week:
Gross requirement
Scheduled receipts
Projected on-hand
Planned receipts
Planned order releases
4
100
0
0
1
0
20
30
0
20
1
0
50
0
0
0
1
0
30
0
0
230
2
0
30
50
0
0
Lead time =
3
0
4
100
5
0
6
150
80
0
150
80
20
0
0
0
0
0
150
0
4
0
5
100
6
0
50
0
0
50
50
0
0
0
0
4
0
5
300
6
0
70
0
0
70
230
0
0
0
0
Lead time =
3
0
2
0
50
0
0
50
0
50
Lead time =
3
0
40
30
30
0
0
0
0
2
0
1
3
2
4
4.
(i) L-4-L
Item
Week:
Gross requirement
Scheduled receipts
Projected on-hand
Net Requirement
Planned receipts
Planned order releases
LT =
1
100
1
2
250
3
200
4
150
5
250
6
200
7
200
8
150
100
0
0
250
0
250
250
200
0
200
200
150
0
150
150
250
0
250
250
200
0
200
200
200
0
200
200
150
0
150
150
0
5
250
0
150
100
375
0
6
200
0
275
0
0
375
7
200
0
75
125
375
0
8
150
0
250
0
0
0
100
No. of setup =
Carrying cost =
Setup cost = 7 x $300 =
Total cost =
7
0
2100
2100
(ii) EOQ:
Total demand for 8 weeks = 1500
D (1 year) = (1500/8) x 52 weeks/year = 9750
H (for 52 weeks) = $0.80/week x 52 weeks = $41.60
S = 300
Q=√
2(9750)300
41.6
= 375
Item
Week:
Gross requirement
Scheduled receipts
Projected on-hand
Net Requirement
Planned receipts
Planned order releases
100
LT =
1
100
0
100
0
0
375
1
2
250
0
0
250
375
375
3
200
0
125
75
375
0
4
150
0
300
0
0
375
Annual setup cost = (D/Q) S = (9750/375) x 300 =
Annual holding cost = (Q/2)H per year = (375/2) x 41.60 =
Annual cost = Annual setup cost + Annual holding cost = 7800 + 7800 =
Weekly cost = Annual cost/52 = 15600/52 =
Cost for 8 weeks = Cost per week x 8 weeks = 300 x 8 =
(iii) PPB
EPP = 300/0.80 = 375
7800
7800
15600
300
2400
Item
Week:
Gross requirement
Scheduled receipts
Projected on-hand
Net Requirement
Planned receipts
Planned order releases
100
LT =
1
100
0
100
0
0
600
Periods
Quantities
combined
combined
Lot #1 Receipt in week #2
1
2
250
0
0
250
600
3
200
0
350
4
150
0
150
5
250
0
0
250
450
6
200
0
200
7
200
0
0
200
8
150
0
450
Periods brought
forward – last qty
New Partperiods
Total combined partperiods
250
250 + 200 = 450
0
1
0
1x 200 = 200
2, 3, 4
450 + 150 = 550
Lot #2 Receipt in week #5
5
250
2
2 x 150 = 300
0
0 + 200 = 200
200 + 300 = 500
(Close to 375)
0
0
5, 6
250 + 200 = 450
5, 6, 7
450 + 200 = 650
Lot #3 Receipt in week #7
7
200
7, 8
200 + 150 = 350
1
2
1 x 200 = 200
2 x 200 = 400
0
0 + 200 = 200
(Close to 375)
200 + 400 = 600
0
1
0
1x 150 = 150
0
0 + 150 = 150
2
2, 3
Summary:
Lot # in Week
Lot PP Setup cost = 3 setups x $300 = $900
Total Lot PP = 850
500
Total holding cost = 850 x $0.80 = $680
200
Total cost = 900 + 680 = $1,580
150
Lot size
Lot #1 Receipt in week #2
550
Lot #2 Receipt in week #5
450
Lot #3 Receipt in week #7
350
Total cost = 700 + 460 + 420 = $1,580
#5.
Week ending
Planned input
Actual input
Cumulative deviation
Planned output
Actual output
Cumulative deviation
Backlog
120
1
2
3
4
5
6
7
8
500
700
200
650
600
-50
220
800
700
100
650
700
0
220
700
700
100
650
800
150
120
600
800
300
650
700
200
220
600
600
300
650
650
200
170
800
500
0
700
500
0
170
700
500
-200
700
600
-100
70
800
800
-200
700
800
0
70
#6.
G13
=SUM(C13:F13)
G14
=SUM(C14:F14)
G15
=SUM(C15:F15)
G16
=SUM(C16:F16)
C17
=SUM(C13:C16)
D17
=SUM(D13:D16)
E17
=SUM(E13:E16)
F17
=SUM(F13:F16)
G19
=SUMPRODUCT(C5:F8,C13:F16)
Solver parameters
Set Target Cell
Options:
G19 MAX
Changing cells
C13:F16
Constraints
G13:G17 <= 1
C17:F17 = 1
□ Assume linear model
□ Assume non-negative
#7. SPT
Job
B
E
A
F
C
D
Processing
time (Days)
20
20
30
35
40
50
195
Average flow time =
Average lateness =
Average WIP =
Utilization =
95.833
30
2.949
0.339
Days till due
date
165
170
40
130
125
65
Completion
time (Flowtime)
20
40
70
105
145
195
Lateness
0
0
30
0
20
130
575
180
EDD
Job
A
D
C
F
B
E
Processing
time (Days)
30
50
40
35
20
20
195
Days till due
date
40
65
125
130
165
170
Average flow time =
Average lateness =
Average no. of jobs in the system =
Utilization =
Completion
time (Flowtime)
30
80
120
155
175
195
755
Lateness
0
15
0
25
10
25
75
125.83
12.50
3.87
0.26
CR
Job
D
A
C
F
B
E
Processing
time (Days)
50
30
40
35
20
20
195
Days till due
date
65
40
125
130
165
170
Average flow time =
Average lateness =
Average no. of jobs in the system =
Utilization =
Completion
time (Flowtime)
50
80
120
155
175
195
775
Lateness
0
40
0
25
10
25
100
CR
1.30
1.33
3.13
3.71
8.25
8.50
129.17
16.67
3.97
0.25
#8. 0.99515 = 0.9257569 or 92.7569%
#9.
(a)
RRow 1= 0.98 x 0.92 x {1 – (1-.9) (1-.9) (1-.9)} = 0.900698
RRow 2 = 0.98 x 0.92 x 0.98 x 0.95 = 0.903256
RS = 1 – (1 - RRow 1)( 1 - RRow 2) = 1 – (1 – 0.900698)(1 – 0.903256) = 0.990393, or 99.0393%
(b)
R1 = 1 – (1 – 0.95) (1 – 0.95) (1 – 0.95) = 0.999875
R2 = 1 – (1 – 0.90) (1 – 0.90) (1 – 0.90) = 0.999
R3 = 1 – (1 – 0.94) (1 – 0.94) (1 – 0.94) = 0.998659
Rs = R1 x R2 x R3 = .998659 = 99.8659%
#10.
(a)
Failed =
FR(%) =
Total tested unit-hours =
Failed
2
3
1
1
(b)
(c)
(d)
7
14.0%
10000
Operating Non-operating
time
time/unit
50
150
75
125
120
80
150
50
Total non-operating time =
Operating time = 10000 – 805 =
Non-operating
time
300
375
80
50
805
9195
7/9195 = 0.000761283
1/0.000761283 = 1313.57
1000* 0.000761283 = 0.761283306
#11.
No. of breakdowns
0
1
2
3
4
5
No. of weeks that many
breakdowns occurred
(Frequency)
75
35
50
20
10
10
200
Expected number of failures/week =
Average breakdown with routine maintenance =
Repair cost/breakdown without routine maintenance =
Routine maintenance cost/week =
Repair cost/breakdown with routine maintenance =
Cost without routine maintenance/week = 1.425 x $12,000 =
Cost with routine maintenance/week = 6000 + 5000 x 1.2 =
Relative
frequency
Expected
breakdowns
0.375
0.000
0.175
0.175
0.250
0.500
0.100
0.300
0.050
0.200
0.050
1.000
0.250
1.425
1.425
1.2
12000
6000
5000
$17,100
$13,200
Total cost/week with routine maintenance is less expensive compared to without it. Therefore,
purchase routine maintenance contract.