Operations and Supply Chain Management The Core

​
336 OPERATIONS AND SUPPLY CHAIN MANAGEMENT
To find the probability of a defective unit, we need to find the Z-scores of the LCL and
USL with respect to the current process:
LSL − X ​
​Z​ LSL ​ = ​ ________ ​= ​ _________ 5.97 − 5.99
​ = − 2.00 NORM.S.DIST(− 2.00) = 0.02275​
σ 0.01
2.275 percent of the output will be too small.
USL − X ​
​Z​ LSL ​ = ​ ________ ​= ​ _________ 6.03 − 5.99
​ = 4.00 NORM.S.DIST(4.00) = 0.999968​
σ 0.01
The probability of too large a unit is 1 − 0.999968 = 0.000032, so 0.0032 percent of output
will be too small.
The probability of producing a defective unit is 0.02275 + 0.000032 = 0.022782, so
2.2782 percent of output will be defective. As a numerical example, 22,782 out of every
million units will be defective.
b. ​ X ​ = 6.00 LSL = 6.00 − 0.03 = 5.97 USL = 6.00 + 0.03 = 6.03 σ = 0.01​
​C​ pk ​ = min ​ ​ _ 6.00 − 5.97
​ or ​ _ 6.03 − 6.00
​ = min [1.00 or 1.00] = 1.00​
[ 0.03
0.03
​]
​Z​ LSL ​ = ​ ________ LSL − X ¯ ​ ​
​= ​ _________ 5.97 − 6.00
​ = − 3.00 NORM.S.DIST(− 3.00) = 0.00135​
σ 0.01
Only 0.135 percent of the output will be too small.
​Z​ USL ​ = ​ ________ USL − X ¯ ​ ​
​= ​ _________ 6.03 − 6.00
​ = 3.00 NORM.S.DIST(3.00) = 0.99865​
σ 0.01
The probability of too large a unit is 1 − 0.99865 = 0.00135, so 0.135 percent of the output
will be too large.
The probability of producing a defective unit is 0.00135 + 0.00135 = 0.0027, so
0.27 percent of the output will be defective. As a numerical example, 2,700 out of every
million units will be defective. That’s about a 90 percent reduction in defective output just
from adjusting the process mean!
Because the process is exactly centered on the target and the specification limits are three
standard deviations away from the process mean, this adjusted process has a C pk = 1.00. In
order to do any better than that, we would need to reduce the variation in the process, as
shown in part (c).
c. ​ X ​ = 6.00 LSL = 6.00 − 0.03 = 5.97 USL = 6.00 + 0.03 = 6.03 σ = 0.005​
​C​ pk ​ = min ​ ​ _ 6.00 − 5.97
​ or ​ _ 6.03 − 6.00
​ = min [2.00 or 2.00] = 2.00​
[ 0.015
0.015
​]
We have doubled the process capability index by cutting the standard deviation of the process
in half. What will be the effect on the probability of defective output?
LSL − X ​
​Z​ LSL ​ = ​ ________ ​= ​ _________ 5.97 − 6.00
​ = − 6.00 NORM.S.DIST(− 6.00) = 0.0000000009866
σ 0.005
​
USL − X ​
​Z​ USL ​ = ​ ________ ​= ​ _________ 6.03 − 6.00
​ = 6.00 NORM.S.DIST(6.00) = 0.9999999990134
σ 0.005
Following earlier logic, the probability of producing a defective unit in this case is just
0.000000001973, a very small probability indeed! Using the earlier numerical example, this
would result in only 0.001973 defective units out of every million. By cutting the process
standard deviation in half, we could gain far more than a 50 percent reduction in defective