1. MMAE557 Personal Consulting Project
Analysis of Ball Bearing Manufacturing
Li He A20358122
Xingye Dai A20365915
Instructor: John Cesarone
Apr.30, 2016
2. Section.1 Client Description
Our client is Weili bearing factory in China which we have ever visited. It’s a company located in
Hebei province and produces many models of bearings. It’s a private company so the factory
has not a large scale of manufacturing and their economic situation does not allow them to hire
a large amounts of workers. In that case, to improve their efficiency of manufacturing system
becomes essential.
In this factory, one of the production workshop was investigated. Totally three lines of work
stations are applied to the manufacturing process of bearings. The first line includes 4 steps
(Turning, heat treatment, grinding and testing) of manufacturing the inner and outer rings. The
second line includes 5 steps (shearing & heading, forging, heating treatment, lapping, testing)
of manufacturing the bearing balls. The third line includes 3 steps (blanking & punching,
forming and testing) of manufacturing the cages. All the machining processes are finished by
corresponding machining equipment and the heat treatments are processing in furnace. The
products are annealed at high temperature and quenched in oil for hardening. The inspections
are done by workers and operators. Each line has two operators, one for testing and the other
one for loading and unloading. The inspection for size and surface finish of balls is carried out
on a sample basis by means of microscopes and other precision equipment. The balls are then
cleaned and packed ready for bearing assembly operations. For inspections of cages, blanking,
punching, forming rivet holes and visual inspection is carried for any deformity. For inner and
outer rings, the diameters are measured. The finished parts of bearings are then manually
transport to assembling station after a batch of parts are produced. Then the assembling
station assembles the parts into bearings automatically. The finally products are then tested by
2 inspectors at next station. For each line, because all the steps are continuous and the
transport time between stations is accounted in load and unload time. the products are
transport so the multiple station cells are not applied to the manufacturing system. Two types
of products are recorded for the entire process and their average time for each procedure were
calculated and listed in following tables.
The classifications of manufacturing system of this factory belongs to ‘’Type II-A B. Type II-A
means it works on multiple automated stations and B means that the cell produces bearing
parts in batch production system.
3. Section 2: Client analysis
This factory is capable of manufacturing of large amount of bearings depends on the demands
and orders. In our investigation, processing times of two products are recorded and their cycle
time and production rates are calculated in following steps.
Line 1: Inner/Outer Ring Manufacturing Process and Step Time
Turning (s) Heat Treatment (S) Grinding (s) Testing (s) Load & Unload (s)
Product A 3 2 4 2 2.5
Product B 4 2.5 5 2 2.5
Note: The time of Heat Treatment is the average time of each ring.
Line 2: Ball Manufacturing Process and Step Time
Shearing & Heading
(s)
Forging
(s)
Heat Treatment
(s)
Lapping
(s)
Testing
(s)
Load & Unload
(s)
Product A 0.1 0.1 0.05 0.3 0.3 0.3
Product B 0.1 0.1 0.05 0.3 0.3 0.3
Note: Each of Product A needs 8 balls; Each of Product B needs 10 balls.
Line 3: Cage Manufacturing Process and Step Time
Blanking & Punching (s) Forming (s) Testing (s) Load & Unload (s)
Product A 0.5 2 1 1.5
Product B 0.8 3 1 1.5
Assembly:
Note: the Assembly & Testing time (Cycle Time) of Product A is 10s, the Assembly & Testing time (Cycle
Time) of Product B is 12s.
Plant Layout:
Cage
Rivet
Outer Ring
Inner Ring
Balls
Assembling
(Pillow Wrapping Machine) Testing
4. Cycle Time and Production Rate
Line 1:
2 operators: one for Testing, one for loading & unloading
Turning Heating Grinding Testing
Breakdown Probability 0.005 0.003 0.008 0.007
Down Time (min) 5 10 6 3
𝑇𝐶 = 𝑇 𝑀 + 𝑇𝑆
A: Tc1=3+2+4+2+2.5=13.5S
B: TC1=4+2.5+5+2+2.5=16S
𝑇𝑃 = 𝑇𝐶 + 𝐹 ∗ 𝑇𝑑
F*Td=0.005×5+0.003×10+0.008×6+0.007×3=0.124min=7.44s
A: Tp1=13.5+7.44=20.94s
B: Tp1=16+7.44=23.44s
𝑅 𝑃 =
3600
𝑇 𝑃
(Parts/hr)
A: Rp1=
3600
20.94
=171.92 Parts/hr.
B: Rp1=
3600
23.44
=153.58 Parts/hr.
Line 1 Line 2 Line 3
Assembly
Testing
5. Line 2:
2 operators: one for Testing, one for loading & unloading
Shearing & Heading Forging Heat Treatment Lapping Testing
Breakdown Probability 0.0008 0.0003 0.0003 0.0005 0.0007
Down Time (min) 3 5 10 5 3
𝑇𝐶 = 𝑇 𝑀 + 𝑇𝑆
Ball: Tc=0.1+0.1+0.05+0.2+0.3+0.3=1.05s
𝑇𝑃 = 𝑇𝐶 + 𝐹 ∗ 𝑇𝑑
F*Td=0.0008×3+0.0003×5+0.0003×10+0.0005×5+0.0007×3=0.0115min=0.69s
Ball: Tp=1.05+0.69=1.74s
𝑅 𝑃 =
3600
𝑇 𝑃
(Parts/hr)
Ball: Rp=
3600
20.91.744
=2068.97 Parts/hr.
Tc
A: Tc2=8*Tc=8×1.05=8.4s
B: TC2=10* Tc=10×1.05=10.5s
Rp
A: Rp2=2068.97×
1
8
=258.62 Parts/hr.
B: Rp2=2068.97×
1
10
=106.90 Parts/hr.
Line 3:
2 operators: one for Testing, one for loading & unloading
Blanking & Punching Forming Testing
Breakdown Probability 0.005 0.003 0.004
Down Time (min) 5 8 3
𝑇𝐶 = 𝑇 𝑀 + 𝑇𝑆
A: Tc3=0.5+2+1+1.5=5s
B: TC3=0.8+3+1+1.5=6.3s
7. Sections 3: optimize the system
Notice that the original system is not perfect efficient based on current process of
manufacturing, some improvement changes could be applied into the system like the number
of workstations and workers.
3.1 Optimum Number of Workstation—Assembling Workstation
Product A Product B
Annual Goal (Q) 800000 600000
Cycle Time without testing (TC) 8s 10s
Scrap Time (µ) 2%
Worker Efficiency (Ew) 95%
Machine Running Reliability (RM) 98%
Machine Setup Reliability (RS) 99.5%
Setup Time (TSP) 20min
Batch Size (N) 5000
Original Factory Operation Hours: 8 Hours/day; 5 Days/week; 50 Weeks/year.
Workload:
𝑊𝐿 =
𝑄 ∗ 𝑇𝐶
1 − µ
A: WL(A)=
800000×8
1−2%
=6530612s=1814.06hr.
B: WL(B)=
600000×10
1−2%
=6122449s=1700.68hr.
No. of setup changes:
No.=
𝑄(𝐴)+𝑄(𝐵)
𝑁
=
800000+600000
5000
=280
Setup: WL(SP)=No. ×TSP=280×20=5600min=93.33hr.
Available Time:
AT=8×5×50=2000hr/yr.
8. Optimum number of workstations:
𝑁𝑜. =
𝑊𝐿(𝐴) + 𝑊𝐿(𝐵)
𝐴𝑇 ∗ 𝐸 𝑊 ∗ 𝑅 𝑀
+
𝑊𝐿(𝑆𝑃)
𝐴𝑇 ∗ 𝑅 𝑆
𝑁𝑜. =
1814.06+1700.68
2000×0.95×0.98
+
93.33
2000×0.995
=1.93452
So, 1 workstation 2 workstations OR double the operation hours.
Considering the high price of assembling machines, we choose doubling the operation hours as the
solution.
Then, the cycle time of assembling (with testing) changes from 8+2 to 4+2 for product A; 10+2 to 5+2 for
product B.
Ball Bearing:
Cycle Time:
A: TC=TC1+TCA=13.5+6=19.5s
B: TC=TC1+TCA=16+7=23s
Production Time:
A: TP=TP1+TPA=20.94+(6+0.008×8×60)=30.78s
B: TP=TP1+TPA=23.44+(7+0.008×8×60)=34.28s
Production Rate:
A: RP=
3600
30.78
=116.96 Bearing/hr.
B: RP=
3600
34.28
=105.02 Bearing/hr.
3.2 Machine Cluster Analysis (Determine the number of workers)
Line 1 Line 2 Line 3 Line 4
A (s) B (s) A (s) B (s) A (s) B (s) A (s) B (s)
Machine Time (TM) 6 7 0.5 0.625 1.5 2 6 7.5
Worker Service Time (Ts) 5 5 0.3 0.3 1.5 1.5 4 4.5
Repositioning Time (Tr) 5 5.5 0.15 0.1875 2.3 3.5 5 6
9. The number of machines could be operated by one person (N): 𝑁 =
𝑇 𝑀+𝑇 𝑆
𝑇 𝑆+𝑇𝑟
Line 1:
A: 𝑁 =
6+5
5+5
= 1.1 one worker could operate 1.1 line 1. So, in line 1, employee 1 worker.
B: 𝑁 =
7+5
6+5.5
= 1.14 1 worker
Line 2:
A: 𝑁 =
0.5+0.3
0.3+0.15
= 1.78 1 worker
B: 𝑁 =
0.625+0.3
0.3+0.1875
= 1.90 1 worker
Line 3:
A: 𝑁 =
1.5+1.5
1.5+2.3
= 0.79 2 workers
B: 𝑁 =
2+1.5
1.5+3.5
= 0.7 2 workers
Assembly:
A: 𝑁 =
6+4
4+5
= 1.11 1 worker
B: 𝑁 =
7.5+4.5
4.5+6
= 1.14 1 worker
Conclusion: Theoretically, set 1 worker in Line 1; 1 worker in Line 2; 2 workers in Line 3; 1 worker in
Assembly.
However, as to practice, some of the stuff could not be lifted by one worker, and there are a lot of
emergency situations which could cause the stop of whole factory if there is only one worker in a line,
such as sickness or vacation.
As to Line 2, one worker almost could operate 2 lines. The company could try to train the other guys,
and ask them to do both line 1 & line2, or both line 2 & line 3. Another method, the company could try
to buy standard Balls from other companies, just like buying Rivets.
Cost Analysis (Machine never idle):
Labor Cost ( CL): $2.5/hr. Machine/Line Cost (CM): $20/hr.
Cost/part=(
𝐶 𝐿
𝑛
+ 𝐶 𝑀) ∗ (𝑇 𝑀 + 𝑇𝑆)
Original:
Line 1, Labor 2:
10. A: $=(2.5×2+20) ×13.5÷3600=0.09375
B: $=(2.5×2+20) ×16÷3600=0.1111
Line 2, Labor 2:
A: $=(2.5×2+20) ×8.4÷3600=0.0583
B: $=(2.5×2+20) ×10.5÷3600=0.0729
Line 3, Labor 2:
A: $=(2.5×2+20) ×5÷3600=0.0347
B: $=(2.5×2+20) ×6.3÷3600=0.0438
Assembly, Labor 2:
In order to meet the annual goal, we choose double the operator hours. And the Labor Cost at night is
$3/hr, which means the average Labor Cost is $2.75/hr.
A: $=(2.75×2+20) ×10÷3600=0.0708
B: $=(2.75×2+20) ×12÷3600=0.085
Total:
A: ∑$=0.09375+0.0583+0.0347+0.0708=0.25755 $/bearing
B: ∑$=0.1111+0.0729+0.0438+0.085=0.3128 $/bearing
Optimization (theoretically)
Line 1, Labor 2:
A: $=(2.5+20) ×13.5÷3600=0.0844
B: $=(2.5+20) ×16÷3600=0.1000
Line 2, Labor 2:
A: $=(2.5+20) ×8.4÷3600=0.0525
B: $=(2.5+20) ×10.5÷3600=0.0656
Line 3, Labor 2:
A: $=(2.5×2+20) ×5÷3600=0.0347
B: $=(2.5×2+20) ×6.3÷3600=0.0438
Assembly, Labor 2:
A: $=(2.75+20) ×10÷3600=0.0632
11. B: $=(2.75+20) ×12÷3600=0.0758
Total:
A: ∑$=0.0844+0.0525+0.0347+0.0632=0.2348 $/bearing
B: ∑$=0.1000+0.0656+0.0438+0.0758=0.2852 $/bearing
Conclusion: After optimizing the number of workers, the cost per product A reduces from $0.25755 to
$0.2348, the cost per product B reduces from $0.3128 to $0.2852.
Section 4 Automate the system
Notice that the original layout of the system has different transportation time for each lines whereas
only the maximum Tr could be applied in to the calculation of calculation time. Then we assumed a new
lay out of the system and all the transportation will be carried out by conveyor belt automatically.
Original:
Tr=max{1, 1.5, 1, 1.5}=1.5s
Optimum:
0.5s
1s
1s
Ring Ball Cage
Assembly
Testing
Rivet
12. Same distance between each line to assembly workstation.
Tr=max{1.25, 1.25, 1.25, 1.25}=1.25s
Then optimized layout could save repositioning time from 1.5s to 1.25s
New Tc and Rp
Tc:
A: Tc=19.5-0.25=19.25s
B: Tc=23-0.25=22.75s
Rp:
Due to the Tp of line 1 which is the maximum can save 0.25s, the whole Tp could save 0.25s too.
A: Rp=
3600
30.53
=117.92 Bearing/hr.
B: Rp=
3600
34.03
=105.79 Bearing/hr.
1.25s
1.25s
Ring
Assembly
Ball
Rivet Cage
Testing
1.25s
Half Circle Layout
13. Section 5 Quality assurance
For any factory, quality is the most important specification for their reputations. Therefore, the
inspection procedure becomes essential. The statistical process and quality analysis is given by
the control charts. The mean and range charts are also plotted.
5.1 Product A
Y
direction
Radius Testing Data AVG(x) Range
Y1
r 2.0032 2.0018 1.9958 1.9972 2.0081 2.0026 2.00145 0.0123
R 2.5092 2.5039 2.4909 2.4992 2.5041 2.4955 2.500467 0.0183
Y2
r 2.0071 2.0062 2.0021 1.9988 1.9943 2.0011 2.0016 0.0128
R 2.4989 2.509 2.5021 2.4991 2.4982 2.5033 2.501767 0.0108
Y3
r 1.9978 1.9954 1.9942 1.9933 1.9961 2.0006 1.996233 0.0073
R 2.5032 2.4929 2.4932 2.5077 2.4954 2.5031 2.49925 0.0148
Y4
r 1.9963 2.0085 1.9909 1.9959 2.0085 2.0044 2.00075 0.0176
R 2.5019 2.5055 2.5042 2.4959 2.5044 2.4951 2.501167 0.0104
Y5
r 2.0088 2.0105 2.0049 2.0088 2.0038 1.9992 2.006 0.0113
R 2.4982 2.4966 2.504 2.4921 2.5089 2.4906 2.4984 0.0183
1.988
1.99
1.992
1.994
1.996
1.998
2
2.002
2.004
2.006
2.008
2.01
1 2 3 4 5
DemensionofInnerCircle(inch)
Sample Number,s
Chart of r
UCL
Center
LCL
14. r
Chart Range Chart
CL UCL LCL CL UCL LCL
2.00120667 2.0071282 1.995285 0.01226 0.024569 0
0
0.005
0.01
0.015
0.02
0.025
0.03
1 2 3 4 5
Range
Sample Number, s
Range Chart of r
Center
2.485
2.49
2.495
2.5
2.505
2.51
1 2 3 4 5
DemensionofOuterCircle(inch)
Sample Number, s
Chart of R
UCL
Center
LCL
𝑋
15. 5.2 Product B
Y
direction
Radius Testing Data AVG(x) Range
Y1
r 3.0098 3.0044 3.0051 3.0026 2.9964 2.9956 3.002317 0.0142
R 3.9976 3.9985 4.0021 4.0016 4.0032 3.9925 3.99925 0.0107
Y2
r 2.9984 2.9973 3.0024 3.0036 2.9979 3.0035 3.000517 0.0063
R 4.0035 3.9946 3.9959 4.0031 3.9916 4.0035 3.9987 0.0119
Y3
r 3.0019 3.0026 2.9959 2.9967 3.0016 3.0025 3.0002 0.0067
R 4.0038 3.9958 3.9986 4.0025 4.0016 4.0035 4.000967 0.008
Y4
r 3.0026 3.0053 2.9935 2.9979 3.0019 3.0014 3.000433 0.0118
R 4.0009 3.9959 4.0016 3.9953 3.9929 4.0031 3.998283 0.0102
Y5
r 3.0049 3.0053 2.9959 2.9975 3.0041 3.0029 3.001767 0.0094
R 3.9989 4.0011 4.0021 4.0013 3.9989 4.0019 4.0007 0.0032
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
1 2 3 4 5
Range
Sample Number,s
Range Chart of R
UCL
Center
LCL
R
Chart Range Chart
CL UCL LCL CL UCL LCL
2.50021 2.5072232 2.493197 0.01452 0.029098 0
16. r
Chart Range Chart
CL UCL LCL CL UCL LCL
3.00104667 3.0057221 2.996371 0.00968 0.019399 0
2.99
2.992
2.994
2.996
2.998
3
3.002
3.004
3.006
3.008
1 2 3 4 5
DemensionofInnerCircle(inch)
Sample Number,s
Chart of r
UCL
Center
LCL
0
0.005
0.01
0.015
0.02
0.025
1 2 3 4 5
Range
Sample Number, s
Range Chart of r
Center
17. R
Chart Range Chart
CL UCL LCL CL UCL LCL
3.99958 4.0038304 3.99533 0.0088 0.017635 0
The charts show us the quality is quite well controlled during the process. While consider that the
company are using some low quality machining equipment. The way to make their defect rate better is
to purchase some new equipment with higher capability of controlling the precision and qualities, which
would also reduce the cost significantly. Assume the defect rate reduces from 5% to 3% after using the
new equipment, the quantity of required bearings to meet every 1 million will reduces from 1,052,632
3.99
3.992
3.994
3.996
3.998
4
4.002
4.004
4.006
1 2 3 4 5
DemensionofOuterCircle(inch)
Sample Number, s
Chart of R
UCL
Center
LCL
𝑋
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
1 2 3 4 5
Range
Sample Number,s
Range Chart of R
UCL
Center
LCL
18. to 1,030,928. Cost saved per million is $5,096.14 for A and $6,190.03 for B. Calculated as the following
steps.
Defect Rate=5%:
𝑁 =
1,000,000
1 − 5%
= 1,052,631.58 = 1,052,632
Defect Rate=3%:
𝑁 =
1,000,000
1 − 3%
= 1,030,927.84 = 1,030,928
Cost Saved (Price per Bearing is $0.2348 for A, $0.2852 for B):
CA=0.2348×(1,052,632-1,030,928)=$5,096.14/million
CB=0.2852×(1,052,632-1,030,928)=$6,190.03/million
