Effect of High Variability in Downtime
Evaluate the line efficiencies and production rates for the two-stage line in Examples 16A.1 and 16A.2 using the geometric repair distribution instead of the constant downtime distribution.

Question

Accepted Answer

For (a) and (b), values of E0 and E will be the same as in Example 16A.2.
(a) [latex]E_{0} = 0.60[/latex] and [latex]R_{p} = 30 pc/hr[/latex]

(b) [latex]E_{∞} = 0.75[/latex] and [latex]R_{p}[/latex] = 37.5 pc/hr

(c) For b = 10, all of the parameters in Equation (16A.5) remain the same
except h(b).

[latex]E_{b} = E_{0} + D^{′}_{1}h(b)E_{2}[/latex]

Using Equation (16A.12),

Case [latex]1: r = 1.0. h(b) = \frac{b \frac{T_{c}}{T_{d}}}{2 + (b - 1) \frac{T_{c}}{T_{d}}}[/latex]

[latex]h(b) = h(10) = \frac{10(1.2/8.0)}{2 + (10 - 1)(1.2/8.0)}= 0.4478[/latex]

Now using Equations (16A.5) and (16A.15),

[latex]E_{10} = 0.600 + 0.20(0.4478)(0.75) = 0.6672[/latex] and

[latex]R_{p} = \frac{E}{T_{c}}[/latex]

[latex]R_{p} = 0.6672(60) /1.2 = 33.36 pc/hr[/latex]

(d) For b = 100, again the only change is in h(b).

[latex]h(b) = h(100) = \frac{10(1.2/8.0)}{2 + (100 - 1)(1.2/8.0)}= 0.8902[/latex]

[latex]E_{100} = 0.600 + 0.20(0.8902)(0.75) = 0.7333[/latex] and

[latex]R_{p} = 0.7333(60) /1.2 = 36.67 pc/hr[/latex]

Question 16A.3: Effect of High Variability in Downtime Evaluate the line eff...

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