DPMO Example 1
Calculate the DPMO for the Valve manufacturing process. Values are given in Table 1.1 below.
Given,
Lower specification limit (LSL) = 5.95
Upper specification limit (USL) = 6.05
Observe image 1.4, the area on the left side of 6.05 is 0.995599. Thus, the area on the right side of 6.05 is 0.004401 (i.e., 1 – 0.995599).
Hence, the proportion of defectives
= Area on the left side of 5.95 + Area on the right side of 6.05
= 0.0078001 + 0.004401
= 0.012201
Hence, DPMO = 0.012201 * 1,000,000 = 12,201. or 3.4 Sigma
Valve Diameters - Table 1.1
5.99 | 6.04 | 6.01 | 6 | 6.03 |
5.99 | 6 | 6.03 | 5.97 | 5.98 |
5.99 | 6 | 6.02 | 6.01 | 6.02 |
6.04 | 5.99 | 5.99 | 5.97 | 6.01 |
6.02 | 6.01 | 5.99 | 5.98 | 6 |
6.01 | 5.99 | 6.01 | 6.05 | 6.03 |
6 | 5.98 | 5.97 | 6 | 6 |
5.99 | 5.99 | 5.98 | 5.99 | 6.03 |
6.01 | 5.99 | 6.03 | 6.01 | 5.99 |
5.99 | 5.98 | 6.01 | 5.99 | 6.01 |
6 | 6.01 | 6 | 6.03 | 6.03 |
6 | 5.99 | 6 | 6.02 | 6.01 |
5.98 | 6.01 | 6 | 5.99 | 5.99 |
6.01 | 6.02 | 5.98 | 5.98 | 5.99 |
5.98 | 5.96 | 5.95 | 6.01 | 6.03 |
6.01 | 6 | 5.97 | 6.02 | 6.05 |
5.97 | 5.95 | 5.98 | 6.01 | 5.99 |
6.02 | 5.99 | 6 | 6.01 | 6 |
6 | 5.99 | 6.03 | 6.03 | 5.98 |
5.97 | 6 | 5.97 | 5.99 | 5.97 |
6.01 | 6 | 5.99 | 6.02 | 6 |
6.01 | 6.01 | 5.99 | 5.97 | 6 |
5.99 | 5.96 | 5.99 | 6.05 | 6.01 |
6 | 5.98 | 6 | 5.97 | 6.01 |
5.99 | 5.98 | 5.98 | 5.98 | 6 |
6.01 | 6.03 | 6 | 6 | 6.01 |
5.95 | 6.01 | 6 | 5.97 | 6.02 |
6 | 5.97 | 5.99 | 5.99 | 6 |
5.98 | 5.98 | 6 | 6 | 6 |
5.95 | 6 | 6 | 6.01 | 6 |
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DPMO Example 2
Roller engineering. manufactures SS ball bearings 25 mm in diameter. Allowable tolerance as per customer is plus-minus 6mm from 25 mm. Find (DPMO).
Assume data is normally distributed and Process mean = 28 mm, Process standard deviation = 2 mm
Solution
Process target = 25 mm (given)
Process tolerance = ± 6 mm (given)
Hence, LSL = 25 – 6 = 19 mm, and USL = 25 + 6 = 31 mm
Hence, the proportion of defectives
= Area on the left side of 19 + Area on the right side of 31
= 0.0000034 + 0.066807
= 0.0668104
Hence, DPMO = 0.0668104* 1,000,000 = 66810
DPMO Example 3
Suppose that the Pizza King restaurant delivers pizza in 1,750 seconds on average, with a standard deviation of 90 seconds. The promised delivery time is 30 minutes or less. else you get free pizza.
Calculate the DPMO for the pizza delivery process. Also, find the sigma level of the process? Assume delivery time is normally distributed.
Solution:
LSL = None
USL = Target = 30 min. = 1,800 seconds
Process mean = 1,750 seconds (given)
Process standard deviation = 90 seconds
Notice in image 1.9 that the area on the left side of 1,800 is 0.710743. Thus, the area on the right side of 1,800 is 1 - 0.710743 = 0.289257
Hence, the proportion of defectives
= Area on the right side of 1,800
= 0.289257
Hence, DPMO = 0.289257 * 1,000,000 = 289,257.
DPMO value of 289,257 equates to the sigma level between 2 and 3.
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