A Proposed Capital Investment to Improve Process Yield Many engineering projects aim at improving facility utilization and process yields. This case study illustrates an engineering economy analysis related to the redesign of a major component in the manufacture of semiconductors. Semiconductor man

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A Proposed Capital Investment to Improve Process Yield

Many engineering projects aim at improving facility utilization and process yields. This case study illustrates an engineering economy analysis related to the redesign of a major component in the manufacture of semiconductors.

Semiconductor manufacturing involves taking a flat disc of silicon, called a wafer, and depositing many layers of material on top of it. Each layer has a pattern on it that, upon completion, defines the electrical circuits of the finished microprocessor. Each 8-inch wafer has up to 100 microprocessors on it. However, the typical average yield of the production line is [latex]75 \%[/latex] good microprocessors per wafer.

At one local company, the process engineers responsible for the chemicalvapor-deposition (CVD) tool (i.e., process equipment) that deposits one of the many layers have an idea for improving overall yield. They propose to improve this tool's vacuum with a redesign of one of its major components. The engineers believe the project will result in a [latex]2 \%[/latex] increase in the average production yield of nondefective microprocessors per wafer.

This company has only one CVD tool, and it can process 10 wafers per hour. The process engineers have determined that the CVD tool has an average utilization rate (i.e., "time running") of [latex]80 \%[/latex]. A wafer costs [latex]\$ 5,000[/latex] to manufacture, and a good microprocessor can be sold for [latex]\$ 100[/latex]. These semiconductor fabrication plants ("fabs") operate 168 hours per week, and all good microprocessors produced can be sold.

The capital investment required for the project is [latex]\$ 250,000[/latex], and maintenance and support expenses are expected to be [latex]\$ 25,000[/latex] per month. The lifetime of the modified tool will be five years, and the company uses a [latex]12 \%[/latex] MARR per year (compounded monthly) as its "hurdle rate."

Before implementing the proposed engineering solution, top management has posed the following questions to you (hired as an independent consultant) to evaluate the merits of the proposal:

(a) Based on the PW method, should the project be approved?

(b) If the achievable improvement in production yield has been overestimated by the process engineers, at what percent yield improvement would the project breakeven?

Accepted Answer

You start your economic evaluation of the engineering proposal by first calculating the production rate of wafers. The average number of wafers per week is

[latex](10[/latex] wafers / hour [latex]) imes(168[/latex] hours / week [latex]) imes(0.80)=1,344[/latex].

Since the cost per wafer is [latex]\$ 5,000[/latex] and good microprocessors can be sold for [latex]\$ 100[/latex] each, you determine that profit is earned on each microprocessor produced and sold after the 50th microprocessor on each wafer. Thus, the [latex]2 \%[/latex] increase in production yield is all profit (i.e., from 75 good microprocessors per wafer on the average to [latex]76.5[/latex] ). The corresponding additional profit per wafer is [latex]\$ 150[/latex]. The added profit per month, assuming a month is (52 weeks/year [latex]\div 12[/latex] months per year) = [latex]4.333[/latex] weeks, is

[latex](1,344 ext { wafers/week)(4.333 weeks/month })(\$ 150 / ext { wafer })=\$ 873,533 .[/latex]

Therefore, the PW of the project is

[latex]\begin{aligned}\operatorname{PW}(1 \%)=&-\$ 250,000-\$ 25,000(P / A, 1 \% ext { per month, } 60 ext { months }) \&+\$ 873,533(P / A, 1 \%, 60) \ =& \$ 37,898,813.\end{aligned}[/latex]

You advise company management that, based on PW, the project should be undertaken.

It is known that at the breakeven point, profit equals zero. That is, the [latex]P W[/latex] of the project is equal to zero, or PW of costs [latex]=\mathrm{PW}[/latex] of revenues. In other words,

[latex]\begin{aligned}\$ 1,373,875=&(1,344 ext { wafers } / ext { week }) imes(4.333 ext { weeks } / ext { month }) imes(\$ X / ext { wafer }) \& imes(P / A, 1 \%, 60),\end{aligned}[/latex]

where [latex]X=\$ 100[/latex] times the number of extra microprocessors per wafer. Then,

[latex]\frac{\$ 1,373,875}{(1,344)(4.333)(44.955)}=X, ext { or } X \cong \$ 5.25 ext { per wafer. }[/latex]

Thus, [latex](\$ 5.25 / \$ 100)=0.0525[/latex] extra microprocessors per wafer (total of [latex]75.0525[/latex] ) equates [latex]P W[/latex] of costs to PW of revenues. This corresponds to a breakeven increase in yield of

[latex]\frac{1.5 ext { die per wafer }}{0.0525 ext { die per wafer }}=\frac{2.0 \% ext { increase }}{ ext { breakeven increase }} ext {, }[/latex]

or breakeven increase in yield [latex]=0.07 \%[/latex].

You advise management that an increase of only [latex]0.07 \%[/latex] in process yield would enable the project to breakeven. Thus, although management may believe that the process engineers have overestimated projected process yield improvements in the past, there is quite a bit of "economic safety margin" provided by the engineers in their current projection of process yield improvement as long as the other assumptions concerning the average utilization rate of the CVD tool, the wafer production rate, and the plant operating hours are valid.

Question 5.C-S.9: A Proposed Capital Investment to Improve Process Yield Many ...

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