Question

A part has a mean of 75.51 mm with an LSL of 75.29 and a USL of 75.71 and a standard deviation of 0.08mm. Any part above the USL is reground for a cost of $3.10. Every rejected part (below LSL) costs $4.65 for parts and labor. A modification costs $6200. The modification brings the standard deviation to 0.04 mm. How many parts of production (including good and bad parts) does it take for this modification to pay for itself?

Answer 1

Solution

Let X = measurement of the part under question.

We assume X ~ N(µ, σ²), where given µ = 75.51, and σ = 0.08 …………………..........................................………………… (1)

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] ...……...…(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables ………….… (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) …...................................................(2b)

Now to work out the solution,

Also given LSL = 75.29, USL = 75.71 …………………………......…………………………….. (3)

Proportion of oversize (above USL)

= P(X > 75.71) [vide (3)]

= P[Z > {(75.71 – 75.51)/0.08}] [vide (2) and (1)]

= P(Z > 2.5)

= 0.0062 [vide (2b)] …………………………………………………………....…….....…………… (4)

Proportion of undersize (below LSL)

= P(X < 75.29) [vide (3)]

= P[Z > {(75.29 – 75.51)/0.08}] [vide (2) and (1)]

= P(Z > - 2.75)

= 0.0030 [vide (2b)] …………………………………………………………......……………...……. (5)

Just for ease in presentation and explanation, let the production quantity be 10000

(4) and (5) =>

number of oversize parts = 10000 x 0.0062 = 62 and

number of undersize parts = 10000 x 0.0030 = 30 ………………………………………...……… (6)

Given undersize parts cost $4.65 and oversize parts are reworked at a cost of $3.10, (6) =>

Total cost per 10000 production = (62 x 4.65) + (30 x 3.10) = 381.3 ……………………………. (7)

Now, with modification, σ = 0.04 and reworking as above,

Proportion of oversize (above USL)

= P(X > 75.71) [vide (3)]

= P[Z > {(75.71 – 75.51)/0.04}] [vide (2) and (1)]

= P(Z > 5)

= 0.00000029 [vide (2b)] ……………………………………………………………....…………… (4a)

Proportion of undersize (below LSL)

= P(X < 75.29) [vide (3)]

= P[Z > {(75.29 – 75.51)/0.04}] [vide (2) and (1)]

= P(Z > - 5.5)

= 0.000000019 [vide (2b)] ………………………………....…………………………...……………. (5a)

So, for a production quantity of 10000

(4a) and (5a) =>

number of oversize parts = 10000 x 0.00000029 = 0.0029 and

number of undersize parts = 10000 x 0.000000019 = 0.00019 ………………..…….…………… (6a)

Total cost per 10000 production = (0.0029 x 4.65) + (0.00019 x 3.10) = 0.0194 ………………. (7a)

(7) and (7a) => cost saving per 10000 production = 381.30 – 0.0194 = 381.2816 ………………. (8)

Given the cost of modification = 6200, quantity of production required to pay the modification cost

= 10000 x (6200/381.2816)

= 162609 Answer

DONE