Transfer Lines with More Than One Storage Buffer For the same 20-station transfer line considered in the previous examples, compare line efficiencies and production rates for the following cases, assuming an infinite buffer capacity: (a) no storage buffer, (b) one buffer, (c) three buffers, and (d)

Question

Transfer Lines with More Than One Storage Buffer
For the same 20-station transfer line considered in the previous examples, compare line efficiencies and production rates for the following cases, assuming an infinite buffer capacity: (a) no storage buffer, (b) one buffer, (c) three buffers, and (d) 19 buffers. Base the comparison on constant repair times.
Assume in cases (b) and (c) that the buffers are located in the line so as to equalize the downtime frequencies, that is, all [latex]F_{i}[/latex] are equal.

Accepted Answer

The answers for (a) and (b) have already been computed in Example 16A.2.
(a) For the case of no storage buffer, [latex]E_{∞} = 0.60[/latex] and [latex]R_{p} = 0.60(60) /1.2 = 30 pc/hr[/latex]

b) For one storage buffer (a two-stage line), [latex]E_{∞} = 0.75[/latex] and [latex]R_{p} = 0.75(60) /1.2 = 37.5 pc/hr[/latex]

(c) For the case of three storage buffers (a four-stage line),
[latex]F_{1} = F_{2} = F_{3} = F_{4} = 5(.005) = 0.025[/latex] and

[latex]T_{p} = 1.2 + 0.025(8) = 1.4 min/p[/latex]

[latex]E_{∞} = 1.2/1.4 = 0.8571[/latex] and [latex]R_{p} = 0.8571(60) /1.2 = 42.86 pc/hr[/latex]
(d) For the case of 19 storage buffers (each stage is one station),
[latex]F_{1} = F_{2} = . . . = F_{20} = 1(0.005) = 0.005[/latex] and

[latex]T_{p} = 1.2 + 0.005(8) = 1.24 min/pc[/latex]

[latex]E_{∞} = 1.2/1.24 = 0.9677[/latex] and [latex]R_{p} = 0.9677(60) /1.2 = 48.39 pc/hr[/latex]
Comment: This last value is very close to the ideal production rate of [latex]R_{c} = 60/1.2 = 50 pc/hr.[/latex]

Question 16A.4: Transfer Lines with More Than One Storage Buffer For the sam...

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