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In: Operations Management

Process Capability (Cpk and Cp) The operations manager of an insurance claims- processing department wants to...

Process Capability (Cpk and Cp) The operations manager of an insurance claims- processing department wants to determine the claims-processing capability of the department. Claims usually take a minimum of four days to handle. The company has a commitment to handle all claims within 10 days. On the average, claims are processed in eight days and processing has a standard deviation of one day. a. Compute Cp and Cpk for the claims-processing department. Based on these computations, should the claims department improve its process? b. Using the same data, recompute Cpk, but use an average claims-processing time of seven days instead of eight days. c. Using the original data, recompute Cpk, but use a standard deviation of 2/3 of a day. Which change made the most improvement—the change in mean in part b or the change in standard deviation? Can you explain the results?

Solutions

Expert Solution

Given are the following data :

Upper specification limit for claim handling = USL = 10 DAYS

Lower specification limit for claim handling = LSL = 4 days

Answer to question a :

Given are,

Process average = m = 8 days

Process standard deviation = Sd = 1 day

Therefore ,

Cp = ( USL – LSL) / 6 X Sd = ( 10 – 4 )/ 6 x 1 = 6 /6 = 1

Cpk = Minimum ( ( USL – m) /3xSd , ( m – LSL) /3x Sd)

= Minimum ( ( 10 – 8)/3 x 1 , ( 8 – 4)/3 x 1)

= Minimum ( 2/3 , 4/3)

= 2/3

For a process to be capable , following conditions are to be met :

Cp > = 1
Cpk >= 1 ( preferably Cpk >= 1.33)

Because Cpk is only 2/3 , the process is not upto the mark and the claim department must improve the process.

Answer to question b :

When average claim processing time get revised to 7 days instead of 8 days, revised value of m = 7 days

Revised Cpk = Minimum ( ( 10 – 7)/3 x 1 , ( 7 -4 ) / 3 x 1 )

= Minimum ( 1, 1 )

= 1

Answer to question c :

Revised standard deviation = Sd = 2/3

Revised Cpk = Minimum ( ( 10- 8 ) / (3 x 2/3) , ( 8 – 4) / ( 3 x 2/3)

= Minimum ( 1, 2)

= 1

In both cases ( i.e. question #b , question #c ) Cpk yields the same result which is 1

However it gives an improved Cpk of 1 vis a vis earliest case of 2/3. Which improves the process capability . The process become “just” capable by virtue of Cpk = 1 . However if threshold Cpk is set at 1.33, there is scope of improvement

keosha answered 1 year ago

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