Quality Control and Process Analysis in Mechanical Engineering

This assignment solution addresses quality control and process analysis in a manufacturing context. It begins with calculations related to sample size determination, considering factors like allowance, observation time, and acceptable error. The core of the solution focuses on quality control charts, specifically X-bar, R, and C charts. The solution provides step-by-step calculations for determining control limits (UCL and LCL) for both X-bar and R charts, utilizing data on coil spring lengths. It includes the construction and interpretation of these charts, determining whether the process is under control. Furthermore, the solution incorporates run tests and median tests to analyze data patterns. The second part of the assignment delves into the application of c-charts for analyzing defects, calculating control limits and determining whether the process is under control. Finally, the solution calculates production costs and the impact of rework on the overall cost per unit, providing a comprehensive analysis of the manufacturing process and quality control measures.

PART-1
ANSWER - A
Given Details
(PR) 1.1
Allowance Percentage 15% of job.
Observation Time (minutes)
1 2.07
2 2.08
3 2.11
4 2.12
5 2.06
6 1.98
7 2.04
8 2.02
9 2.05
10 2.03
11 2.09
Calculations: Formula
Average Observation Time
(OT) 2.059

Normal Time (NT) 2.265 Avg OT * PR Factor
Standard Time (ST) 2.60475 NT (1+ Allowance Factor)
ANSWER - B
Given Details
Times busy Times idle Total observations
1 8 1 9
2 9 2 11
3 10 3 13
4 8 0 8
5 9 2 11
6 11 4 15
7 9 3 12
Calculations:
Sum 64 15 79
Estimated % of Idle Time 0.1899 18.98734177
ANSWER - C
Given Details:
n = required sample size
z = 2.33 for a 98% confidence level
p = estimate of idle proportion 0.1899

h = acceptable error of 5% 0.08
Calculation:
N = (Z^2P(1-P))/H^2
n 130
Excess observation Required 51
ANSWER - D
Given Details:
n = required sample size
z = 2.33 for a 98% confidence level
Observations 100.0000
p = estimate of idle proportion 0.1500
h = acceptable error of 5% 0.05
Calculation:
N = (Z^2P(1-P))/H^2
n 277
Excess observation Required 177
PART-II
Q5A. Quality Control Chart

Answer- A
Observations (Coil spring length, cm), n
Samp
le k 1 2 3 4 5 X-BAR R
1 10.2 10 9.94 9.99 9.96 9.982 0.08
2 10.01 10.02 10.07 9.95 9.96 10.002 0.12
3 9.99 10 9.93 9.92 9.99 9.966 0.08
4 10.03 9.91 10 9.98 9.89 9.962 0.14
5 9.95 9.92 10.03 10.05 10.01 9.992 0.13
6 9.97 10.06 10.05 9.96 10.02 10.012 0.1
7 10.05 10.01 10.1 8.96 9.98 9.82 1.14
8 10.09 10.1 10 9.99 10.08 10.052 0.11
9 10.14 10.1 9.99 10.08 10.09 10.08 0.15
10 10.01 9.98 10.08 10.07 9.99 10.026 0.1
99.894 2.15
9.9894 0.215
Observations (Coil spring length, cm), n
Samp
le k 1 2 3 4 5 X-BAR R
1 10.2 10 9.94 9.99 9.96 =(SUM(B3:F3))/
5 =B3-D3
2 10.01 10.02 10.07 9.95 9.96 =(SUM(B4:F4))/
5 =D4-E4
3 9.99 10 9.93 9.92 9.99 =(SUM(B5:F5))/
5 =C5-E5
4 10.03 9.91 10 9.98 9.89 =(SUM(B6:F6))/
5 =B6-F6
5 9.95 9.92 10.03 10.05 10.01 =(SUM(B7:F7))/
5 =E7-C7
6 9.97 10.06 10.05 9.96 10.02 =(SUM(B8:F8))/
5 =C8-E8

7 10.05 10.01 10.1 8.96 9.98 =(SUM(B8:F8))/
5 =D9-E9
8 10.09 10.1 10 9.99 10.08 =(SUM(B9:F9))/
5 C10-E10
9 10.14 10.1 9.99 10.08 10.09 =(SUM(B10:F10
))/5 B11-D11
10 10.01 9.98 10.08 10.07 9.99 =(SUM(B11:F11
))/5 D12-C12
=SUM(G3:G12) =SUM(H3:H
12)
=G13/10 =H3/10
For Sample
n=5; A=0.58; B=2.11 & C=0
X̿̿̿̿̿ = 99.89/10
= 9.98
R̅ = 2.15/10
= 0.215
Determination of UCL & LCL
X-BAR
UCL= X̿̿̿̿̿ + BR̅
= 9.98+ (0.58x0.21)
UCL= 10.101
LCL= X̿̿̿̿̿ + AR̅
= 9.98- (0.58x0.21)

LCL=9.8582
R-CHART
UCL= BR̅
= 2.11x0.21
UCL= 0.4431
LCL= CR̅
= 0x0.21
LCL=0
CL for x-bar chart = x_double_bar = 9.989
UCL for x-bar chart = x_double_bar + A2*R_bar = 9.989 + 0.577*0.215 = 10.114
LCL for x-bar chart = x_double_bar - A2*R_bar = 9.989- 0.577*0.215 = 9.865
CL for R chart = R_bar = 0.215
LCL for R chart = D3*R_bar = 0*0.215 = 0
UCL for R chart = D4*R_bar = 2.114*0.215 = 0.454
Answer-b
X-BAR CHART

Inference:
From the graph it is found that the process is under control, at one point the process went beyond
control but after that point the process is under control.
R-Chart
Inference:
From the graph it can be seen that the process is beyond control at some regions but after that the
process is made under control.
Answer - C

Obs Above/ below median Run
10.02 above 1
10.00 below 2
9.94 below
9.99 below
9.96 below
10.01 above 3
10.02 above
10.07 above
9.95 below 4
9.96 below
9.99 below
10.00 below
9.93 below
9.92 below
9.99 below
10.03 above 5
9.91 below 6

10.00 below
9.98 below
9.89 below
9.95 below
9.92 below
10.03 above 7
10.05 above
10.01 above
9.97 below 8
10.06 above 9
10.05 above
9.96 below 10
10.02 above 11
10.05 above
10.01 above
10.10 above
8.96 below 12
9.98 below
10.09 above 13

10.10 above
10.00 below 14
9.99 below
10.08 above 15
10.14 above
10.10 above
9.99 below 16
10.08 above 17
10.09 above
10.01 above
9.98 below 18
10.08 above 19
10.07 above
9.99 below 20
N=50
Observed runs = 20
Expected number of runs = (N/2) + 1 = (50/2) + 1 = 26
Variance = (N/2)*((N/2) - 1)/(N - 1) = = (50/2)*((50/2) - 1)/(50 - 1) = 12.245
Zmedian = (Observed runs - Expected runs) / Std. Dev = (20 - 26)/SQRT(12.245) = -1.714

Obs Up/ down Run
10.02 down 1
10.00 down
9.94 down
9.99 Up 2
9.96 down 3
10.01 Up 4
10.02 Up

10.07 Up
9.95 down 5
9.96 Up 6
9.99 Up
10.00 Up
9.93 down 7
9.92 down
9.99 Up 8
10.03 Up
9.91 down 9
10.00 Up 10
9.98 down 11
9.89 down
9.95 Up 12
9.92 down 13
10.03 Up 14
10.05 Up
10.01 down 15
9.97 down

10.06 Up 16
10.05 down 17
9.96 down
10.02 Up 18
10.05 Up
10.01 down 19
10.10 Up 20
8.96 down 21
9.98 Up 22
10.09 Up
10.10 Up
10.00 down 23
9.99 down
10.08 Up 24
10.14 Up
10.10 down 25
9.99 down
10.08 Up 26
10.09 Up

10.01 down 27
9.98 down
10.08 Up 28
10.07 down 29
9.99 down
N=50
Observed runs = 29
Expected number of runs = (2N - 1)/3 = (2*50 - 1) /3 = 33
Variance = (16N - 29)/90 = (16*50 - 29) / 90 = 8.5667
ZU/D = (Observed runs - Expected runs) / Std. Dev = (29 - 33)/SQRT(8.5667) = -1.367
Answers to Q5B.
Answer- A
We need to use c chart for this case because:
● This concerns a sample and not all products
● This data corresponds to defects per sample and not defectives
Answer - B
The process is not under control as according to the formulas for UCL = average +
3*(sqrt(average)) and LCL = average - 3*(sqrt(average))
Sample Defects Center LCL UCL
1 3 4.925 0 11.5827

2 2 4.925 0 11.5827
3 4 4.925 0 11.5827
4 5 4.925 0 11.5827
5 1 4.925 0 11.5827
6 2 4.925 0 11.5827
7 4 4.925 0 11.5827
8 1 4.925 0 11.5827
9 2 4.925 0 11.5827
10 7 4.925 0 11.5827
11 8 4.925 0 11.5827
12 4 4.925 0 11.5827
13 6 4.925 0 11.5827
14 7 4.925 0 11.5827
15 13 4.925 0 11.5827

16 3 4.925 0 11.5827
17 5 4.925 0 11.5827
18 6 4.925 0 11.5827
19 7 4.925 0 11.5827
20 2 4.925 0 11.5827
21 1 4.925 0 11.5827
22 2 4.925 0 11.5827
23 8 4.925 0 11.5827
24 3 4.925 0 11.5827
25 4 4.925 0 11.5827
26 6 4.925 0 11.5827
27 9 4.925 0 11.5827
28 5 4.925 0 11.5827
29 6 4.925 0 11.5827

30 1 4.925 0 11.5827
31 12 4.925 0 11.5827
32 9 4.925 0 11.5827
33 7 4.925 0 11.5827
34 5 4.925 0 11.5827
35 3 4.925 0 11.5827
36 5 4.925 0 11.5827
37 6 4.925 0 11.5827
38 7 4.925 0 11.5827
39 4 4.925 0 11.5827
40 2 4.925 0 11.5827
Average c bar 4.925
sigma c bar 2.219234
z 3

UCL = 11.5827
LCL = -1.7327 but as defects cannot be negative, hence, taking this as 0
As seen from the data, the samples shoot out of the control limits as they have more defects than
11 in some cases. Hence, the process is out of control.
Answer - C
No.of defects in Coil Spring
Daily good quality product yield = 616 parts
Cost per spring = $450
Total cost of the batch = $450*640 = 288000
Cost of rework = $45*136 = 6120
Total cost of the batch = 294120
Cost per unit = 294120/616 = 477.4675
Restaurant cost per spring = $477.46
Answer - D
Nos.

a Springs produced per hr 80
b No. of hrs per day 8
c Total springs produced 640
d Perfect springs (95% of c) 608
e Defectives (5% of c) 32
f Reworked (70% of e) 22.4
g Total effective springs (d+f) 630.4
Total cost of the batch = $450*640 = 288000
Cost of rework = $45*136 = 1008
Total cost of the batch = 289008
Cost per unit = 294120/616 = 458.4518
Restaurant cost per spring = $458.45