Question

samples of n=4 items each are taken from a manufacturing process at regular intervals. a quality characteristic is measured, and x-bar and r values are calculated for each sample. after 25 samples, we have: x-bar = 107.5 r = 12.5 assume that the quality characteristic is normally distributed.

a)compute control limits for the x-bar and r control charts

b)estimate the process mean and standard deviation

c)assuming that the process is in control, what are the natural tolerance limits of the process?

d)if the specifications limits are 4.4+- 0.2. what is the process capability? is the process capable of meeting the specifications?

e)assuming that if any item exceeds the upper specification limit it can be reworked, and if it is below the lower specification limit, it must be scrapped, what percent scrap and rework is the process producing?

f)if the unit cost of scrap and rework are $2.4 and $0.75, respectively, find the total daily cost of scrap and rework. g)if a process average shifts to 4.5 mm, what is the impact on the proportion of scrap and rework produced?

Answer 1

Solution

Let xij = quality characteristic value of the jth unit in the ith sample, j = 1 to 4, i = 24

Part (a)

Control limits for the x-bar and r control charts

The Xbar part of Xbar-R Chart has:

Central line: CL = Xdouble bar = 107.5/25 = 4.3

Lower Control Limit: LCL = Xdouble bar – A₂Rbar = 4.3 – (0.729 x 0.5) = 3.93

Upper Control Limit: UCL = Xdouble bar + A₂Rbar = 4.3 + (0.729 x 0.5) = 4.66

The R part of Xbar-R Chart has:

Central line: CL = Rbar = 12.5/25 = 0.5

Lower Control Limit: LCL = D₃Rbar = 0 x 0.5 = 0

Upper Control Limit: UCL = D₄Rbar = 2.282 x 0.5 = 1.141

Answer 1

[Detils

Xdouble bar = (1/k)∑(i = 1, k)x_ibar, x_ibar = (1/n)∑(i = 1, n)x_ij

Rbar = (1/k)∑(i = 1, k)R_i, R_i = Max(xij) - min(xij)

A₂, D₃, D₄ , d2 are constants which can be directly obtained from standard Control Chart Constants Table

Part (b)

Estimate the process mean = Xdouble bar = 4.3 Answer 2

Estimate of process standard deviation, σ_hat = Rbar/d2 = 0.5/2.059 = 0.2428 Answer 3

Part (c)

Natural tolerance limits of the process

= Xdouble bar ± 3σ_hat

= 4.3 ± (3 x 0.2428)

= 4.3 ± 0.73 Answer 4

Part (d)

Process capability

= 6σ_hat

= (6 x 0.2428)

= 1.46 Answer 5

Given the specification limits as 4.4 + 0.2, tolerance band = 0.4. Since 6σ_hat > 0.4,

process is not capable of meeting the specifications. Answer 6

Part (e)

Given that any item exceeds the upper specification limit it can be reworked, and if it is below the lower specification limit, it must be scrapped,

Percent scrap

= Percent (Proportion) Defectives below LSL

= 100 x P[Z < {(LSL – Xdoublebar)/σ_hat)],Z ~ N(0, 1).

= 100 x P[Z < {(4.2 – 4.3)/0.5)]

= 100 x P[Z < - 0.2]

= 100 x 0.4207 [Using Excel Function: Statistical NORMSDIST]

= 42.07% Answer 7

Percent rewok

= Percent (Proportion) Defectives above USL

= 100 x P[Z > {(USL – Xdoublebar)/σ_hat)],Z ~ N(0, 1).

= 100 x P[Z > {(4.6 – 4.3)/0.5)]

= 100 x P[Z > 0.6]

= 100 x 0.2743 [Using Excel Function: Statistical NORMSDIST]

= 27.43% Answer 8

Part (f)

Given the unit cost of scrap and rework are $2.4 and $0.75, respectively, the total daily cost of scrap and rework when daily production is N units

= N x {(2.4 x 0.4207) + (0.75 x 0.2743)}

= 1.2154N Answer 9

Part (g)

If a process average shifts to 4.5 mm,

Percent scrap

= Percent (Proportion) Defectives below LSL

= 100 x P[Z < {(LSL – process average)/σ_hat)],Z ~ N(0, 1).

= 100 x P[Z < {(4.2 – 4.5)/0.5)]

= 100 x P[Z < - 0.6]

= 100 x 0.2742 [Using Excel Function: Statistical NORMSDIST]

= 27.42% Answer 10

Percent rewok

= Percent (Proportion) Defectives above USL

= 100 x P[Z > {(USL – process average)/σ_hat)],Z ~ N(0, 1).

= 100 x P[Z > {(4.6 – 4.5)/0.5)]

= 100 x P[Z > 0.2]

= 100 x 0.4208 [Using Excel Function: Statistical NORMSDIST]

= 42.08% Answer 11

Thus, the impact on the proportion of scrap and rework produced is that

While rework increases, scrap decreases Answer 12

DONE