, SOLUTION MANUAL FOR Manufacturing System
Throughput Excellence Analysis Improvement and
Design by Herman Tang
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
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k
1
Manufacturing System Throughput Excellence – Analysis,
Improvement, and Design
Chapter 1: Throughput Concepts
1.1 A shift has 7.5 hours of planned production time. The actual production time in a shift was
6.8 hours. The shift experienced 0.15 hour of starved time, 0.25 hour of blocked time, 0.2 hour
of equipment downtime, and 0.1 hour of work delay time. Calculate the operational availabil-
ity and Toyota’s formula availability for this shift.
Solution:
Actual production time 6.8
● A= Planned production time
= 7.5
= 90.7%
Planned production time−Starved time−Equipment time−Work delay time
● Atoyota = Planned production time
7.5−0.15−0.2−0.1 6.9
= = = 94.0%
k 7.5 7.5 k
1.2 A manufacturing system employs 320 people. Over one week, it utilized 13,400 labor hours
and produced 4800 units. Calculate the productivity per employee and per hour for this week.
Solution:
Produced units 4800
●
Number of employees
= 320
= 15.0 units∕employee
Produced units 4800
●
Production hours
= 13,400
= 0.358 units∕hour
1.3 In one week, a shop incurred a direct labor cost of $100 k, material cost of $150 k, equipment
cost of $45 k, utility cost of $7 k and overhead cost of $13 k. Calculate the percentage of direct
labor cost in the total cost.
Solution:
Clabor 100
●
Clabor +Cmaterial +Cequipment +Cutility +Cfacility +Coverhead
= 315
= 31.7%
1.4 An operation has a known throughput time of 4.5 hours and a throughput rate of 50 jobs per
hour. Calculate the average WIP level for this operation in the long run based on Little’s Law.
Solution:
● WIP = TT × TR = 4.5 × 50 = 225 (units)
Manufacturing System Throughput Excellence: Analysis, Improvement, and Design, First Edition. Herman Tang.
© 2024 John Wiley & Sons, Inc. Published 2024 by John Wiley & Sons, Inc.
Companion website: www.wiley.com/go/Tang/ManufacturingSystem
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2 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
1.5 A production line has eight workstations with cycle times of 58, 56, 57, 60, 59, 56, 55, and 59
seconds, respectively. Estimate the long-run average WIP for this production line.
Solution:
● TT = 58 + 56 + 57 + 60 + 59 + 56 + 55 + 59 = 460 (seconds) ≈ 0.128 (hour)
3600 3600
● TR = max (58,56,57,60,59,56,55,59)
= 60
= 60 (JPH)
● WIP = TT × TR = 0.128 × 60 = 7.7 (units)
1.6 If a process has a cycle time of 48 seconds, estimate the corresponding theoretical throughput
rate in JPH.
Solution:
3600 3600
● TR
gross = CT (seconds) = 48 = 75.0 (JPH)
1.7 Based on the data from Exercise 1.1 and known production output of 480 units, calculate the
standalone availability (Asa ) and standalone throughput rate (TRsa ).
Solution:
Actual production time 6.8
● A
sa = Planned production time−Starved time−Blocked time = 7.5−0.15−.025 = 95.8%
480
● TRgross = 6.8
= 70.59 (JPH)
Asa 0.958
● TRsa = A
× TR = 0.907
× 70.59 = 74.56 (JPH)
Chapter 2: System Performance Metrics
k 2.1 A shop had planned to produce 450 units during a standard eight-hour shift. However, it k
ended up producing 451 units in 8.5 hours. Calculate the shop’s volume attainment (VA) and
schedule attainment (SA) performance. Comment on the results.
Solution:
Actual production time
● A= Planned production time
= 8.58
= 106.25%
Actual produced units 451
● VA = Planned output units = 450 = 100.22%
Planned production time
● SA = VA × Actual production time = VA × A1 = 100.22% 1
106.25%
= 94.3%
2.2 A plant operates two shifts, six days a week. Each shift is 8.5 hours long, inclusive of
a half-hour lunch break and three 15-minute breaks for nonproduction activities. The
plant is designed to have a throughput rate of 60 units per hour. Over the week, the plant
manufactured 4440 units, out of which 90 required repairs. Also, there were events leading
to 12 hours of unplanned downtime. Calculate the plant’s speed performance of the week.
Solution:
Actual units produced 4440
● P(unit based) = = ((8.5−0.5−3×0.25)×6×2−12)×60 = 98.7(%)
Planned units in uptime
3600
● Desing CT = TR
= 3600
60
= 60 (seconds)
((8.5−0.5−3×0.25)×6×2−12)×3600
● Actual CT = 4440
= 60.81 (seconds)
Design CT 60
● P(cycle time based) = Actual CT = 60.81 = 98.7(%)
2.3 A production line is designed to have a cycle time of 60 seconds, but it operates at 61.2 seconds.
Estimate the throughput rate of this production line. If the production runs for 7.5 hours per
shift, determine the number of units lost a shift due to the speed performance rate.
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Manufacturing System Throughput Excellence – Analysis, Improvement, and Design 3
Solution:
design CT 60
● P(cycle time based) = = = 98.04(%)
actual CT 61.2
3600
● Lost TR = 60 − 61.2
= 1.18 (JPH)
● Shift lost = 1.18 × 7.5 = 8.8 (units)
2.4 Using the data provided in Exercise 2.2, calculate the OEE for the week.
Solution:
(8.5−0.5−3×0.25)×6×2−12
● A = = 86.21%
(8.5−0.5−3×0.25)×6×2
4440−90
● Q= 4440
= 97.97%
● OEE = A × P × Q = 86.21% × 98.67% × 97.97% = 83.3%
2.5 A production system operated for 80 hours across five working days, with an OEE of 85%.
Calculate its TEEP.
Solution:
80
● TEEP = OEE × U = 85% × = 56.7%
5×24
2.6 A quality improvement project can enhance the quality rate from 98.0% to 98.5%. It can also
lead to a 0.1% improvement (92.0–92.1%) in operational availability and a 0.2% improvement
(94.3–94.5%) in the speed performance. Compare the estimated OEE and calculated OEE
(refer to subsection 2.2.3).
Solution:
k ● old OEE = A × P × Q = 92% × 94.3% × 98% = 85.02%
1 k
● new OEE = A × P × Q = 92.1% × 94.5% × 98.5% = 85.73%
2
● ΔOEE = OEE − OEE = 0.71%
2 1
● ΔOEE ≈ ΔA + ΔP + ΔQ = 0.1% + 0.2% + 0.5% = 0.8%
2.7 For an operation, the significance of the three elements of OEE are assigned values of 7, 4, and
5, respectively. Over a week, these three elements were measured to be 91%, 99%, and 95%,
respectively. Calculate the original OEE and the weighted OEE using the method introduced
in subsection 2.3.1.2.
Solution:
● OEE = A × P × Q = 85.59%
3×7 3×4 3×5
● w
A = 7+4+5 = 1.31, wP = 7+4+5 = 0.75, and wQ = 7+4+5 = 0.94
● OEEw = (AwA ) × (PwP ) × (QwQ ) = 83.58%
2.8 Continuing from Exercise 2.4, assume a standalone availability of 96%. What would be the
simplified standalone OEE? Compare this with the results obtained in Exercise 2.2.
Solution:
● OEE
sa − simplified = Asa × P × Q = 96% × 98.67% × 97.97% = 92.8%
2.9 If a system’s throughput is 0.5 JPH less than its target, and the unit profit is $1400, calculate
the monetary loss for one week of production, assuming 75 working hours.
Solution:
● −0.5 × 75 × $1400 = − $52,500
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4 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
Chapter 3: Bottleneck Identification and Buffer Analysis
3.1 A system comprises five workstations, each with cycle times of 52, 55, 58, 50, and 51 seconds,
respectively. Estimate the throughput rates of the system and identify which workstation
serves as the throughput bottleneck.
Solution:
3600
● Station 3: TR =
CT (seconds)
= 3600
58
= 62.07 (JPH) is the slowest workstation.
3.2 Four manufacturing subsystems, arranged in series, have active periods accounting for 95%,
89%, 91%, and 85% of the production time in a week, respectively. Determine which subsystem
acts as the throughput bottleneck?
Solution:
● Subsystem 1, as it has the highest active time.
3.3 A system consists of seven operations. Their starved and blocked times (due to various
reasons) over a week of production are listed in Table 3.3. Using the concept of a “turning
point,” identify the operation that is the bottleneck and provide a rationale for your choice.
Table 3.3
Operation 1 2 3 4 5 6 7
Starved time (minute) 50 70 20 40 30 40 50
k Blocked time (minute) 40 30 50 70 60 60 50 k
Solution:
● Operation 3, as it has the lowest combined starved and blocked time.
3.4 A conveyor, which has a transfer time of one minute between two systems with a cycle time
of 55 seconds, is in operation. What would be the recommended minimum quantity of WIP
units to support system throughput?
Solution:
ttransfer
● Min =
CT
+ 1 = 60
55
+ 1 = 2.1 (units)
3.5 A system that includes four subsystems arranged in series operates in a continuous flow.
The five conveyors associated with these subsystems hold WIP units of 40, 65, 10, 20 and 32,
respectively (refer to Figure 3.22). Identify the bottleneck in the system’s throughput.
Solution:
● Subsystem 2, as its upstream conveyor has the highest WIP units.
3.6 A conveyor, with a capacity equivalent to 15 minutes of production time and typically filled
to two-thirds of its capacity, is in operation. Determine the duration for which the conveyor
can compensate for the downtime of its upstream and downstream systems.
Solution:
● (2/3) Cover up to ten minutes of the upstream system until the conveyor becomes
empty.
● (1/3) Cover up five minutes of the downstream system until the conveyor becomes full.
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Manufacturing System Throughput Excellence – Analysis, Improvement, and Design 5
3.7 A production line functions at a rate of 50 units per minute, and the WIP change alarm is set
to activate at three-fourths of the production rate. If the number of WIP units on the conveyor
drops by 120 units within three minutes, would the WIP change alarm be triggered?
Solution:
3
● Alarm: × 50 = 37.5 units∕min
4
ΔWIP 120
●
Δt
= 3
= 40 (units∕min ) > 37.5
Chapter 4: Quality Management and Throughput
4.1 The implementation of an automated in-process inspection necessitates a new investment
of $40,000, aimed at reducing the cost of internal failures. Given an estimated weekly cost
saving of $1500 from internal failures and assuming no other changes in costs, calculate the
break-even point for this investment.
Solution:
Total cost
● Break–even =
Total gains or savings
= 40,000
1500
= 26.67 (weeks)
4.2 An improvement project requires an investment of $15,000 and is projected to yield a cost
saving of $3500 per quarter due to improved product quality over the next two years. The
company’s minimum acceptable rate of return (MARR) is set at 14%. Determine whether
this project is worth undertaking.
Solution:
● Using Excel’s rate() function: “=rate(2*4,3500,–15000)”
k ● Rate = 16.4% > MARR
k
4.3 An improvement project requires an investment of $15,000, and the company’s MARR is 14%.
Calculate the minimum expected cost saving per quarter from improved product quality over
two years, which would make this project worthwhile.
Solution:
● Using Excel’s pmt() function: “=pmt(0.14,2*4,–15000)”
● Saving = $3.33.55
4.4 A manufacturing system consists of eight workstations in series, with TPY values of 0.99,
0.98, 0.95, 0.97, 0.98, 0.95, 0.98, and 0.99, respectively. Calculate the RTY of the entire system.
Solution:
● RTY = 0.99 × 0.98 × 0.95 × 0.97 × 0.98 × 0.95 × 0.98 × 0.99 = 0.808
4.5 A system comprises three parallel, identical subsystems, each with mean values of a quality
attribute of 15.2, 14.5, and 13.7, respectively. Estimate the mean value of this quality attribute
for the combined products from all subsystems.
Solution:
15.2+14.5+13.7
● 𝜇= 3
= 14.47
4.6 A system comprises two parallel, identical subsystems, each with similar mean values for
a quality attribute but different variations. The standard deviations for this attribute are 5.5
and 7.3 for the two subsystems, respectively. Estimate the standard deviation value of this
quality attribute for the combined products from both subsystems.
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6 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
Solution:
√ √
𝜎12 + 𝜎22 5.52 + 7.32
● 𝜎 = = = 6.46
2 2
Chapter 5: Maintenance Management and Throughput
5.1 A robot has a failure probability of 0.0002 per hour. Calculate its reliability for eight hours
of work. If the failure probability is reduced to 0.0001 per hour, what would be the improved
reliability over eight work hours?
Solution:
● Using Excel’s exp() function.
● = exp(−0.0002*8) = 99.84%
● = exp(−0.0001*8) = 99.92%
5.2 Over a three-month production period, the four most frequent failure modes (A, B, C, and D)
occurred 10, 13, 9, and 8 times, respectively. The average downtime for these modes was
8, 3, 12, and 7 minutes, respectively. If the consequence rating is considered the same for all
modes, which failure mode should be prioritized for maintenance?
Solution:
● Failure C has the highest severity, as S = F × D × C = 108
3 3 3 3
5.3 An assembly line encountered 92 failures over a three-month production period, amounting
to 900 working hours. Determine the MTBF for this line.
k Solution: k
Total available time 900
● MTBF = Number of failures
= 92
= 9.78 (hours)
5.4 The maintenance department spent 1050 minutes rectifying 92 failures on an assembly line
over a three-month period. What is the MTTR for this line?
Solution:
Total down time 1050
● MTTR = Number of downtime or repair
= 92
= 11.41 (hours)
5.5 If the MTBF of a production line is 15.5 hours for a given period, what would be the corre-
sponding failure rate?
Solution:
Number of failures 1 1
● 𝜆= Total time
= MTBF
= 15.5
= 6.45%
5.6 If the MTBF and MTTR of a production line are 15.5 and 0.25 hours, respectively, for a given
period, calculate the availability of this production line.
Solution:
MTBF 15.5
● A= MTBF+MTTR
= 15.5+0.25
= 98.41%
5.7 During a month’s production, unplanned maintenance for downtime accounted for 125 out
of the total 600 maintenance hours. What would be the PMP and the unscheduled downtime
ratio?
Solution:
Planned maintenance hours
● PMP =
Total maintenance hours
= 600−125
600
= 79.2%
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Manufacturing System Throughput Excellence – Analysis, Improvement, and Design 7
5.8 A maintenance department incurred expenses of $85 k and $100 k on various maintenance
tasks over two consecutive months. The estimated costs for throughput and quality for these
two months were $250 k and $230 k, respectively. Calculate the total maintenance cost for
these two months.
Solution:
● $50 + $250 = $335 (k)
● $100 + $230 = $330 (k)
5.9 A maintenance department adheres to the three elements of OEE to guide its maintenance
activities. During a shift, three fundamental issues are reported, as listed Table 5.10. Due
to resource constraints, the maintenance department can immediately address only two of
these issues. Which two should be prioritized for immediate attention? (refer to Table 5.8
for the ratings)
Table 5.10
Issue A P Q
1 Downtime = 4 minutes Slow by 0.6 minute None
2 None Slow by 2.5 minutes Defect = 1.9%
3 Downtime = 8 minutes None Defect = 0.3%
Solution:
k k
Issue RA RP RQ PI
1 8 3 1 24
2 1 5 4 20
3 9 1 2 18
Chapter 6: Throughput Enhancement Methodology
6.1 The TR data for a month exhibits a standard deviation of 2.7 and a mean of 56. Calculate the
coefficient of variation (CV) for this data.
Solution:
Standard deviation
● CV =
Mean
= 2.7
56
= 4.82%
6.2 The table (Table 6.6) lists the downtime durations and corresponding frequencies of a system
for one week. Utilize MS Excel or similar software to perform data curve fitting and analyze
the overall characteristics of the system.
Table 6.6
Downtime (<minute) 1 2 3 4 5 10 15 20
Frequency (times) 35 20 7 4 2 1 1 1
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8 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
Solution:
● Using Excel’s trendline function:
6.3 A serial production line consists of five workstations, each with cycle times of 55, 58, 54, 57,
and 52 seconds, respectively. Estimate the cycle time for this line.
Solution:
● CT
serial ≈ Max {CTop.1 , CTop.2 , …CT op.n } = Max {55, 58, 54, 57, 52} = 58 (seconds)
k k
6.4 A parallel manufacturing system comprises three identical legs, each with an equal produc-
tion volume. The cycle times for these three legs are 85, 88, and 87 seconds, respectively.
Determine the cycle time for this system.
Solution:
Average{CTseg.1 ,CTseg.2 ,…,CTseg.n } Average{85,88,87}
● CT = 28.9 (seconds)
parallel ≈ n
= 3
6.5 Three improvement project proposals have been rated (without weights) as shown in Table
6.7. Based on the three factors, determine which proposal should be prioritized for support.
Provide a rationale for your choice.
Table 6.7
Proposal (1) Benefits (2) Technology (3) Resources
A 3 4 2
B 3.5 3.5 3
C 4 2.5 3.5
Solution:
● Proposal B has the highest priority with an overall score of 36.75.
6.6 An internal survey was conducted to assess the levels of importance and agreement for var-
ious improvement tasks, as listed in Table 6.8. Based on the survey results, identify the top
three tasks that should be prioritized.
k
Throughput Excellence Analysis Improvement and
Design by Herman Tang
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
, Trim Size: 7in x 10in Single Column Tang190324 both02.tex V1 - 06/24/2024 5:16pm Page 1
k
1
Manufacturing System Throughput Excellence – Analysis,
Improvement, and Design
Chapter 1: Throughput Concepts
1.1 A shift has 7.5 hours of planned production time. The actual production time in a shift was
6.8 hours. The shift experienced 0.15 hour of starved time, 0.25 hour of blocked time, 0.2 hour
of equipment downtime, and 0.1 hour of work delay time. Calculate the operational availabil-
ity and Toyota’s formula availability for this shift.
Solution:
Actual production time 6.8
● A= Planned production time
= 7.5
= 90.7%
Planned production time−Starved time−Equipment time−Work delay time
● Atoyota = Planned production time
7.5−0.15−0.2−0.1 6.9
= = = 94.0%
k 7.5 7.5 k
1.2 A manufacturing system employs 320 people. Over one week, it utilized 13,400 labor hours
and produced 4800 units. Calculate the productivity per employee and per hour for this week.
Solution:
Produced units 4800
●
Number of employees
= 320
= 15.0 units∕employee
Produced units 4800
●
Production hours
= 13,400
= 0.358 units∕hour
1.3 In one week, a shop incurred a direct labor cost of $100 k, material cost of $150 k, equipment
cost of $45 k, utility cost of $7 k and overhead cost of $13 k. Calculate the percentage of direct
labor cost in the total cost.
Solution:
Clabor 100
●
Clabor +Cmaterial +Cequipment +Cutility +Cfacility +Coverhead
= 315
= 31.7%
1.4 An operation has a known throughput time of 4.5 hours and a throughput rate of 50 jobs per
hour. Calculate the average WIP level for this operation in the long run based on Little’s Law.
Solution:
● WIP = TT × TR = 4.5 × 50 = 225 (units)
Manufacturing System Throughput Excellence: Analysis, Improvement, and Design, First Edition. Herman Tang.
© 2024 John Wiley & Sons, Inc. Published 2024 by John Wiley & Sons, Inc.
Companion website: www.wiley.com/go/Tang/ManufacturingSystem
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2 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
1.5 A production line has eight workstations with cycle times of 58, 56, 57, 60, 59, 56, 55, and 59
seconds, respectively. Estimate the long-run average WIP for this production line.
Solution:
● TT = 58 + 56 + 57 + 60 + 59 + 56 + 55 + 59 = 460 (seconds) ≈ 0.128 (hour)
3600 3600
● TR = max (58,56,57,60,59,56,55,59)
= 60
= 60 (JPH)
● WIP = TT × TR = 0.128 × 60 = 7.7 (units)
1.6 If a process has a cycle time of 48 seconds, estimate the corresponding theoretical throughput
rate in JPH.
Solution:
3600 3600
● TR
gross = CT (seconds) = 48 = 75.0 (JPH)
1.7 Based on the data from Exercise 1.1 and known production output of 480 units, calculate the
standalone availability (Asa ) and standalone throughput rate (TRsa ).
Solution:
Actual production time 6.8
● A
sa = Planned production time−Starved time−Blocked time = 7.5−0.15−.025 = 95.8%
480
● TRgross = 6.8
= 70.59 (JPH)
Asa 0.958
● TRsa = A
× TR = 0.907
× 70.59 = 74.56 (JPH)
Chapter 2: System Performance Metrics
k 2.1 A shop had planned to produce 450 units during a standard eight-hour shift. However, it k
ended up producing 451 units in 8.5 hours. Calculate the shop’s volume attainment (VA) and
schedule attainment (SA) performance. Comment on the results.
Solution:
Actual production time
● A= Planned production time
= 8.58
= 106.25%
Actual produced units 451
● VA = Planned output units = 450 = 100.22%
Planned production time
● SA = VA × Actual production time = VA × A1 = 100.22% 1
106.25%
= 94.3%
2.2 A plant operates two shifts, six days a week. Each shift is 8.5 hours long, inclusive of
a half-hour lunch break and three 15-minute breaks for nonproduction activities. The
plant is designed to have a throughput rate of 60 units per hour. Over the week, the plant
manufactured 4440 units, out of which 90 required repairs. Also, there were events leading
to 12 hours of unplanned downtime. Calculate the plant’s speed performance of the week.
Solution:
Actual units produced 4440
● P(unit based) = = ((8.5−0.5−3×0.25)×6×2−12)×60 = 98.7(%)
Planned units in uptime
3600
● Desing CT = TR
= 3600
60
= 60 (seconds)
((8.5−0.5−3×0.25)×6×2−12)×3600
● Actual CT = 4440
= 60.81 (seconds)
Design CT 60
● P(cycle time based) = Actual CT = 60.81 = 98.7(%)
2.3 A production line is designed to have a cycle time of 60 seconds, but it operates at 61.2 seconds.
Estimate the throughput rate of this production line. If the production runs for 7.5 hours per
shift, determine the number of units lost a shift due to the speed performance rate.
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Manufacturing System Throughput Excellence – Analysis, Improvement, and Design 3
Solution:
design CT 60
● P(cycle time based) = = = 98.04(%)
actual CT 61.2
3600
● Lost TR = 60 − 61.2
= 1.18 (JPH)
● Shift lost = 1.18 × 7.5 = 8.8 (units)
2.4 Using the data provided in Exercise 2.2, calculate the OEE for the week.
Solution:
(8.5−0.5−3×0.25)×6×2−12
● A = = 86.21%
(8.5−0.5−3×0.25)×6×2
4440−90
● Q= 4440
= 97.97%
● OEE = A × P × Q = 86.21% × 98.67% × 97.97% = 83.3%
2.5 A production system operated for 80 hours across five working days, with an OEE of 85%.
Calculate its TEEP.
Solution:
80
● TEEP = OEE × U = 85% × = 56.7%
5×24
2.6 A quality improvement project can enhance the quality rate from 98.0% to 98.5%. It can also
lead to a 0.1% improvement (92.0–92.1%) in operational availability and a 0.2% improvement
(94.3–94.5%) in the speed performance. Compare the estimated OEE and calculated OEE
(refer to subsection 2.2.3).
Solution:
k ● old OEE = A × P × Q = 92% × 94.3% × 98% = 85.02%
1 k
● new OEE = A × P × Q = 92.1% × 94.5% × 98.5% = 85.73%
2
● ΔOEE = OEE − OEE = 0.71%
2 1
● ΔOEE ≈ ΔA + ΔP + ΔQ = 0.1% + 0.2% + 0.5% = 0.8%
2.7 For an operation, the significance of the three elements of OEE are assigned values of 7, 4, and
5, respectively. Over a week, these three elements were measured to be 91%, 99%, and 95%,
respectively. Calculate the original OEE and the weighted OEE using the method introduced
in subsection 2.3.1.2.
Solution:
● OEE = A × P × Q = 85.59%
3×7 3×4 3×5
● w
A = 7+4+5 = 1.31, wP = 7+4+5 = 0.75, and wQ = 7+4+5 = 0.94
● OEEw = (AwA ) × (PwP ) × (QwQ ) = 83.58%
2.8 Continuing from Exercise 2.4, assume a standalone availability of 96%. What would be the
simplified standalone OEE? Compare this with the results obtained in Exercise 2.2.
Solution:
● OEE
sa − simplified = Asa × P × Q = 96% × 98.67% × 97.97% = 92.8%
2.9 If a system’s throughput is 0.5 JPH less than its target, and the unit profit is $1400, calculate
the monetary loss for one week of production, assuming 75 working hours.
Solution:
● −0.5 × 75 × $1400 = − $52,500
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4 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
Chapter 3: Bottleneck Identification and Buffer Analysis
3.1 A system comprises five workstations, each with cycle times of 52, 55, 58, 50, and 51 seconds,
respectively. Estimate the throughput rates of the system and identify which workstation
serves as the throughput bottleneck.
Solution:
3600
● Station 3: TR =
CT (seconds)
= 3600
58
= 62.07 (JPH) is the slowest workstation.
3.2 Four manufacturing subsystems, arranged in series, have active periods accounting for 95%,
89%, 91%, and 85% of the production time in a week, respectively. Determine which subsystem
acts as the throughput bottleneck?
Solution:
● Subsystem 1, as it has the highest active time.
3.3 A system consists of seven operations. Their starved and blocked times (due to various
reasons) over a week of production are listed in Table 3.3. Using the concept of a “turning
point,” identify the operation that is the bottleneck and provide a rationale for your choice.
Table 3.3
Operation 1 2 3 4 5 6 7
Starved time (minute) 50 70 20 40 30 40 50
k Blocked time (minute) 40 30 50 70 60 60 50 k
Solution:
● Operation 3, as it has the lowest combined starved and blocked time.
3.4 A conveyor, which has a transfer time of one minute between two systems with a cycle time
of 55 seconds, is in operation. What would be the recommended minimum quantity of WIP
units to support system throughput?
Solution:
ttransfer
● Min =
CT
+ 1 = 60
55
+ 1 = 2.1 (units)
3.5 A system that includes four subsystems arranged in series operates in a continuous flow.
The five conveyors associated with these subsystems hold WIP units of 40, 65, 10, 20 and 32,
respectively (refer to Figure 3.22). Identify the bottleneck in the system’s throughput.
Solution:
● Subsystem 2, as its upstream conveyor has the highest WIP units.
3.6 A conveyor, with a capacity equivalent to 15 minutes of production time and typically filled
to two-thirds of its capacity, is in operation. Determine the duration for which the conveyor
can compensate for the downtime of its upstream and downstream systems.
Solution:
● (2/3) Cover up to ten minutes of the upstream system until the conveyor becomes
empty.
● (1/3) Cover up five minutes of the downstream system until the conveyor becomes full.
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Manufacturing System Throughput Excellence – Analysis, Improvement, and Design 5
3.7 A production line functions at a rate of 50 units per minute, and the WIP change alarm is set
to activate at three-fourths of the production rate. If the number of WIP units on the conveyor
drops by 120 units within three minutes, would the WIP change alarm be triggered?
Solution:
3
● Alarm: × 50 = 37.5 units∕min
4
ΔWIP 120
●
Δt
= 3
= 40 (units∕min ) > 37.5
Chapter 4: Quality Management and Throughput
4.1 The implementation of an automated in-process inspection necessitates a new investment
of $40,000, aimed at reducing the cost of internal failures. Given an estimated weekly cost
saving of $1500 from internal failures and assuming no other changes in costs, calculate the
break-even point for this investment.
Solution:
Total cost
● Break–even =
Total gains or savings
= 40,000
1500
= 26.67 (weeks)
4.2 An improvement project requires an investment of $15,000 and is projected to yield a cost
saving of $3500 per quarter due to improved product quality over the next two years. The
company’s minimum acceptable rate of return (MARR) is set at 14%. Determine whether
this project is worth undertaking.
Solution:
● Using Excel’s rate() function: “=rate(2*4,3500,–15000)”
k ● Rate = 16.4% > MARR
k
4.3 An improvement project requires an investment of $15,000, and the company’s MARR is 14%.
Calculate the minimum expected cost saving per quarter from improved product quality over
two years, which would make this project worthwhile.
Solution:
● Using Excel’s pmt() function: “=pmt(0.14,2*4,–15000)”
● Saving = $3.33.55
4.4 A manufacturing system consists of eight workstations in series, with TPY values of 0.99,
0.98, 0.95, 0.97, 0.98, 0.95, 0.98, and 0.99, respectively. Calculate the RTY of the entire system.
Solution:
● RTY = 0.99 × 0.98 × 0.95 × 0.97 × 0.98 × 0.95 × 0.98 × 0.99 = 0.808
4.5 A system comprises three parallel, identical subsystems, each with mean values of a quality
attribute of 15.2, 14.5, and 13.7, respectively. Estimate the mean value of this quality attribute
for the combined products from all subsystems.
Solution:
15.2+14.5+13.7
● 𝜇= 3
= 14.47
4.6 A system comprises two parallel, identical subsystems, each with similar mean values for
a quality attribute but different variations. The standard deviations for this attribute are 5.5
and 7.3 for the two subsystems, respectively. Estimate the standard deviation value of this
quality attribute for the combined products from both subsystems.
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6 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
Solution:
√ √
𝜎12 + 𝜎22 5.52 + 7.32
● 𝜎 = = = 6.46
2 2
Chapter 5: Maintenance Management and Throughput
5.1 A robot has a failure probability of 0.0002 per hour. Calculate its reliability for eight hours
of work. If the failure probability is reduced to 0.0001 per hour, what would be the improved
reliability over eight work hours?
Solution:
● Using Excel’s exp() function.
● = exp(−0.0002*8) = 99.84%
● = exp(−0.0001*8) = 99.92%
5.2 Over a three-month production period, the four most frequent failure modes (A, B, C, and D)
occurred 10, 13, 9, and 8 times, respectively. The average downtime for these modes was
8, 3, 12, and 7 minutes, respectively. If the consequence rating is considered the same for all
modes, which failure mode should be prioritized for maintenance?
Solution:
● Failure C has the highest severity, as S = F × D × C = 108
3 3 3 3
5.3 An assembly line encountered 92 failures over a three-month production period, amounting
to 900 working hours. Determine the MTBF for this line.
k Solution: k
Total available time 900
● MTBF = Number of failures
= 92
= 9.78 (hours)
5.4 The maintenance department spent 1050 minutes rectifying 92 failures on an assembly line
over a three-month period. What is the MTTR for this line?
Solution:
Total down time 1050
● MTTR = Number of downtime or repair
= 92
= 11.41 (hours)
5.5 If the MTBF of a production line is 15.5 hours for a given period, what would be the corre-
sponding failure rate?
Solution:
Number of failures 1 1
● 𝜆= Total time
= MTBF
= 15.5
= 6.45%
5.6 If the MTBF and MTTR of a production line are 15.5 and 0.25 hours, respectively, for a given
period, calculate the availability of this production line.
Solution:
MTBF 15.5
● A= MTBF+MTTR
= 15.5+0.25
= 98.41%
5.7 During a month’s production, unplanned maintenance for downtime accounted for 125 out
of the total 600 maintenance hours. What would be the PMP and the unscheduled downtime
ratio?
Solution:
Planned maintenance hours
● PMP =
Total maintenance hours
= 600−125
600
= 79.2%
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Manufacturing System Throughput Excellence – Analysis, Improvement, and Design 7
5.8 A maintenance department incurred expenses of $85 k and $100 k on various maintenance
tasks over two consecutive months. The estimated costs for throughput and quality for these
two months were $250 k and $230 k, respectively. Calculate the total maintenance cost for
these two months.
Solution:
● $50 + $250 = $335 (k)
● $100 + $230 = $330 (k)
5.9 A maintenance department adheres to the three elements of OEE to guide its maintenance
activities. During a shift, three fundamental issues are reported, as listed Table 5.10. Due
to resource constraints, the maintenance department can immediately address only two of
these issues. Which two should be prioritized for immediate attention? (refer to Table 5.8
for the ratings)
Table 5.10
Issue A P Q
1 Downtime = 4 minutes Slow by 0.6 minute None
2 None Slow by 2.5 minutes Defect = 1.9%
3 Downtime = 8 minutes None Defect = 0.3%
Solution:
k k
Issue RA RP RQ PI
1 8 3 1 24
2 1 5 4 20
3 9 1 2 18
Chapter 6: Throughput Enhancement Methodology
6.1 The TR data for a month exhibits a standard deviation of 2.7 and a mean of 56. Calculate the
coefficient of variation (CV) for this data.
Solution:
Standard deviation
● CV =
Mean
= 2.7
56
= 4.82%
6.2 The table (Table 6.6) lists the downtime durations and corresponding frequencies of a system
for one week. Utilize MS Excel or similar software to perform data curve fitting and analyze
the overall characteristics of the system.
Table 6.6
Downtime (<minute) 1 2 3 4 5 10 15 20
Frequency (times) 35 20 7 4 2 1 1 1
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8 Manufacturing System Throughput Excellence – Analysis, Improvement, and Design
Solution:
● Using Excel’s trendline function:
6.3 A serial production line consists of five workstations, each with cycle times of 55, 58, 54, 57,
and 52 seconds, respectively. Estimate the cycle time for this line.
Solution:
● CT
serial ≈ Max {CTop.1 , CTop.2 , …CT op.n } = Max {55, 58, 54, 57, 52} = 58 (seconds)
k k
6.4 A parallel manufacturing system comprises three identical legs, each with an equal produc-
tion volume. The cycle times for these three legs are 85, 88, and 87 seconds, respectively.
Determine the cycle time for this system.
Solution:
Average{CTseg.1 ,CTseg.2 ,…,CTseg.n } Average{85,88,87}
● CT = 28.9 (seconds)
parallel ≈ n
= 3
6.5 Three improvement project proposals have been rated (without weights) as shown in Table
6.7. Based on the three factors, determine which proposal should be prioritized for support.
Provide a rationale for your choice.
Table 6.7
Proposal (1) Benefits (2) Technology (3) Resources
A 3 4 2
B 3.5 3.5 3
C 4 2.5 3.5
Solution:
● Proposal B has the highest priority with an overall score of 36.75.
6.6 An internal survey was conducted to assess the levels of importance and agreement for var-
ious improvement tasks, as listed in Table 6.8. Based on the survey results, identify the top
three tasks that should be prioritized.
k
