Justifiable Lubricant Cost for a Ring-Rolling Operation: 22-Mec-B4 Integrated Manufacturing Systems, December 2013, Q4

Question 4: Justifiable Lubricant Cost for a Ring-Rolling Operation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ring-rolled ball-bearing races whose inner-surface roughness is specified at 0.10 µm with a bilateral tolerance of ±0.05 µm, produced at a known rate with a known scrap cost, and a candidate emulsion lubricant that shifts the process mean without changing its spread.

Given data
Quantity	Symbol	Value
Nominal roughness	R_a,nom	0.10 µm
Tolerance	±T	±0.05 µm
Lower specification limit	LSL	0.05 µm
Upper specification limit	USL	0.15 µm
Present process mean	μ₀	0.112 µm
Process standard deviation	σ	0.02 µm
Production rate	N	30,000 rings/month
Cost of discarding one defective ring	c	$5.00
Mean after lubricant change, part (a)	μ_a	0.100 µm
Mean after lubricant change, part (b)	μ_b	0.090 µm

Find. The maximum additional monthly expenditure on lubricant that the reduction in scrap would pay for, first when the emulsion centres the process on 0.10 µm and then when it overshoots to 0.09 µm, together with the decision that follows if the emulsion costs nothing extra.

Approach. Model the roughness as normally distributed with the stated mean and standard deviation, compute the fraction falling outside the two-sided specification before and after the lubricant change, convert each fraction to a monthly scrap cost, and take the difference: an expenditure is justified up to the point where it exactly consumes the saving.

Figure 4.1 — Roughness distribution against the specification band. The shaded tails are the present out-of-specification output, dominated by the upper tail because the process mean sits 0.012 µm above nominal. Centring the mean on 0.100 µm makes the two tails symmetrical and much smaller.

Part (a) — locate the specification limits in standard-deviation units for the present process. The specification band runs from $\mathrm{LSL}=0.10-0.05=0.05\ \mu\text{m}$ to $\mathrm{USL}=0.10+0.05=0.15\ \mu\text{m}$. Standardising with $z=(x-\mu)/\sigma$ at $\mu_0=0.112$ and $\sigma=0.02$, $$z_{\mathrm{L}}=\frac{0.05-0.112}{0.02}=-3.10,\qquad z_{\mathrm{U}}=\frac{0.15-0.112}{0.02}=+1.90$$ The process is off centre by 0.012 µm, so the two tails are very unequal.
Convert to a fraction defective. Reading the standard normal table, $$p_0=\Phi(-3.10)+\bigl[1-\Phi(1.90)\bigr]=0.00097+0.02872=0.02968$$ so 2.968 per cent of rings fail. Almost all of the loss is in the upper tail, which is the quantitative statement of the qualitative fact that the process is running rough.
Cost the present scrap. At $N=30{,}000$ rings per month and a discard cost of $5.00 each, $$C_0=p_0Nc=0.02968\times 30{,}000\times \$5.00=890.5\ \text{rings}\times \$5.00 =\$4{,}452.62\ \text{per month}$$
Repeat for the centred process. If the emulsion brings the mean to the nominal 0.100 µm with the spread unchanged, both limits sit $z=\pm 0.05/0.02=\pm 2.50$ from the mean, so $$p_a=2\bigl[1-\Phi(2.50)\bigr]=2(0.00621)=0.012419$$ $$C_a=0.012419\times 30{,}000\times \$5.00=372.6\ \text{rings}\times \$5.00=\$1{,}862.90\ \text{per month}$$
Take the difference to obtain the justifiable expenditure. Any additional lubricant cost below the saving leaves the firm better off, and at exactly the saving the firm is indifferent, so the maximum justifiable additional cost is $$\boxed{\ \Delta C_a=C_0-C_a=\$4{,}452.62-\$1{,}862.90=\$2{,}589.72\ \text{per month}\ }$$ which is about $0.086 per ring, or roughly 518 rings' worth of scrap avoided each month.

Figure 4.2 — Part (b): if the emulsion overshoots to a mean of 0.090 µm the process is off centre again, this time towards the smooth side, and the lower tail becomes the dominant loss. The centred distribution is drawn for comparison.

Part (b) — recompute for an overshoot to 0.09 µm. If the emulsion moves the mean past nominal to $\mu_b=0.090\ \mu\text{m}$, the process is again off centre, this time on the smooth side: $$z_{\mathrm{L}}=\frac{0.05-0.09}{0.02}=-2.00,\qquad z_{\mathrm{U}}=\frac{0.15-0.09}{0.02}=+3.00$$ $$p_b=\Phi(-2.00)+\bigl[1-\Phi(3.00)\bigr]=0.02275+0.00135=0.024100$$ so 2.41 per cent fail — better than the present 2.968 per cent but much worse than the 1.242 per cent available from a centred process.
Cost that outcome and take the difference. $$C_b=0.024100\times 30{,}000\times \$5.00=723.0\ \text{rings}\times \$5.00=\$3{,}615.00\ \text{per month}$$ $$\boxed{\ \Delta C_b=C_0-C_b=\$4{,}452.62-\$3{,}615.00=\$837.62\ \text{per month}\ }$$ The justifiable expenditure collapses to less than a third of the part (a) figure, purely because the mean has been moved to the wrong side of nominal rather than onto it.
Answer the final sub-question: what if the lubricant is free? If the emulsion adds no cost at all, then the 0.09 µm outcome still delivers a net saving of $837.62 per month against doing nothing, so the change should be adopted — but it should not be left there. Adopting it and then re-centring the process on 0.10 µm (by trimming roll pressure, speed or emulsion concentration) recovers the remaining $2,589.72 − $837.62 = $1,752.10 per month at no further material cost. In other words, a free lubricant makes the change unconditionally worthwhile, and makes the centring adjustment the most profitable remaining action.

Check: this analysis assumes that the lubricant change shifts the mean without altering the standard deviation, which is what the question implies by quoting only a new mean, and that every ring outside the specification is scrapped at the full $5.00 with no salvage or rework value and no external failure cost. If out-of-specification rings could be reworked, or if a rough race caused a bearing failure in service, the cost of non-conformance would be larger and the justifiable lubricant spend larger with it. The normal model is also assumed to hold into the tails; with a 2.97 per cent overall reject rate the estimate is dominated by the region within two standard deviations of the mean, where the assumption is safest.

Final results
Case	Mean R_a (µm)	Fraction outside spec	Rejects per month	Scrap cost per month	Justifiable extra lubricant cost
Present process	0.112	0.02968	891	$4,452.62	—
(a) Centred on nominal	0.100	0.01242	373	$1,862.90	$2,589.72 / month
(b) Overshoot	0.090	0.02410	723	$3,615.00	$837.62 / month
(b) If lubricant is free	0.090	0.02410	723	$3,615.00	Adopt: net gain $837.62 / month, then re-centre for a further $1,752.10

Topic. Economic justification of a process improvement from the normal-distribution fraction defective; the cost of an off-centre process.

Key relations. $z=(x-\mu)/\sigma$; $p=\Phi\!\left(\frac{\mathrm{LSL}-\mu}{\sigma}\right)+ \left[1-\Phi\!\left(\frac{\mathrm{USL}-\mu}{\sigma}\right)\right]$; monthly scrap cost $C=pNc$; justifiable expenditure $\Delta C=C_{\text{before}}-C_{\text{after}}$. Capability indices for context: $C_p=(\mathrm{USL}-\mathrm{LSL})/6\sigma=0.833$ here, unchanged by the lubricant, while $C_{pk}=\min(\mu-\mathrm{LSL},\ \mathrm{USL}-\mu)/3\sigma$ rises from 0.633 to 0.833 when the process is centred.

Why this works. The specification is a hard boundary, so the cost of poor quality is proportional to the probability mass lying beyond it, and that probability is completely determined by how many standard deviations the limits sit from the mean. Because the spread here is fixed by the process, the only free variable is the mean, and the tail probability is extremely sensitive to it: moving the mean by 0.012 µm — six tenths of one standard deviation — more than halves the reject rate. That sensitivity is what makes the arithmetic worth doing before spending money on a lubricant, and it is the same sensitivity that makes centring the cheapest quality improvement available in almost every process. The comparison in part (b) makes the point sharply: an emulsion that moves the mean by the same magnitude but past the target captures less than a third of the available benefit.

Common pitfall. Computing only the upper tail because it is obviously the larger one. Here the lower tail contributes just 0.097 per cent before the change but 2.275 per cent after the overshoot in part (b), where ignoring it would understate the reject rate by a factor of eighteen and make the emulsion look far better than it is. A second pitfall is treating the saving as an annual figure: the production rate is quoted per month, so the justified expenditure is per month too.

Source. Kalpakjian & Schmid, Manufacturing Engineering and Technology, 8th ed., Ch. 36 (Quality Assurance, Testing and Inspection) and Ch. 13 (Ring Rolling); Montgomery, Introduction to Statistical Quality Control, 8th ed., Ch. 8 (Process and Measurement System Capability Analysis).

Question 4 of 7: Justifiable Lubricant Cost for a Ring-Rolling Operation

Question 4: Justifiable Lubricant Cost for a Ring-Rolling Operation (20 marks)