, 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:
, 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.
k
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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
k
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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.
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
k
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k
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.
k
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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
k
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k
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.
k
