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Chapter 17

Q. 17.4

Single-Station Automatic Assembly System

A single-station assembly machine performs five work elements to assemble four components to a base part. The elements are listed in the table below, together with the fraction defect rate (q) and probability of a station jam (m) for each of the components added (NA means not applicable).

Element Operation Time (sec) q m p
1 Add gear 4 0.02 1.0
2 Add spacer 3 0.01 0.6
3 Add gear 4 0.015 0.8
4 Add gear and mesh 7 0.02 1.0
5 Fasten 5 0 NA 0.012

Time to load the base part is 3 sec and time to unload the completed assembly is 4 sec, giving a total load/unload time of T_{h} = 7 sec. When a jam occurs, it takes an average of 1.5 min to clear the jam and restart the machine. Determine (a) production rate of all product, (b) yield of good product, (c) production rate of good product, and (d) uptime efficiency of the assembly machine.

Step-by-Step

Verified Solution

(a) The ideal cycle time of the assembly machine is
T_{c} = 7 + (4 + 3 + 4 + 7 + 5) = 30  sec = 0.5  min
Frequency of downtime occurrences is
F = 0.02(1.0) + 0.01(0.6) + 0.015(0.8) + 0.02(1.0) + 0.012 = 0.07

Adding the average downtime due to jams,
T_{p} = 0.5 + 0.07(1.5) = 0.5 + 0.105 = 0.605 min
Production rate is therefore R_{p} = 60/0.605 = 99.2 total assemblies /hr

(b) Yield of good product is the following, from Equation (17.5):

P_{ap} = \prod\limits_{i=1}^{n}{(1-q_{i} + m_{i}q_{i})}

 

P_{ap} = \left\{1 – 0.02 + 1.0(0.02)\right\} \left\{1 – 0.01 + 0.6(0.01)\right\} \left\{1 – 0.015 + 0.8(0.015)\right\} \left\{1- 0.02 + 1.0(0.02)\right\}

= (1.0)(0.996)(0.997)(1.0) = 0.993

(c) Production rate of only good assemblies is
R_{ap} = 99.2(0.993) = 98.5 good assemblies /hr
(d) Uptime efficiency is
E = 0.5/0.605 = 0.8264 = 82.64 %