Process Capability Cp Formula

Last Updated : 06 Jan, 2024

A process is a series of steps/tasks to convert some input to an output. Capability on the other hand means to produce something with some set of specifications consistently. Thus we say that Process Capability is a statistical measurement of a process’s ability to produce parts of the desired specification consistently.

With the help of historical data we get the answer to the question – “Can this process produce parts of desired specifications on a daily basis”. Process capability studies control graph, which helps determine if a process is in a controlled or uncontrolled state.

Before we dive deep into various concepts, let us know some basic definitions

Upper Specification Limit – This is the highest limit or the upper limit of a measurement or a reading
Lower Specification Limit – This is the lowest limit or the lower limit of a measurement or a reading
Upper Control Limit – The upper limit of the control line, any reading above this should be investigated for unusual variation
Lower Control Limit – The lower limit of the control line, any reading below this should be investigated for unusual variation

C_pand the C_pkIndex

For any process, the final aim is to have a control chart that is

Centered around the mean or nominal value
Spread over a narrow specification width

To achieve this we take the help of C_pand the C_pk index or also called the Process Capability indexes.

The formula for both is as follows:

$C_p = \frac{\text{Specification Width}}{\text{Process Width}} = \frac{\text{USL - LSL}}{\text{UCL-LCL}} = \frac{\text{USL - LSL}}{6\sigma}\\$

$C_{pk} = \frac{\text{Distance from mean to nearest spec limit}}{\text{Distance of mean to control limit}} = \min({\frac{|\bar{\bar{x}}-UCL|}{3\sigma},{\frac{|\bar{\bar{x}}-LCL|}{3\sigma}}})$

C_pdetermines the spread of the control chart, how narrow or wide the specification limits are.

C_pk determines the spread of the control chart relative to the specification limits.

Now let us take an analogy that can help us understand the two indexes.

Let us consider a scenario where a bunch of drivers, all having a different experience in driving, are told to park cars in a garage. All cars are identical in width, i.e., their specification(at least the width) is the same and this width is less than that of the garage(but not by much).

In this C_pdetermines the width of the queue of cars and C_pkdetermines how central they are to the garage door

Case no.	Cp	Cpk	Result
Case 1	Cp < 0.7	Cpk < 0.7	Here the width of the queue is narrow (Cp<0.5) but the cars aren’t central (Cpk<0.5) thus the cars won’t fit inside making this case not capable.
Case 2	0.7<Cp<1.33	0.7<Cpk<1.33	Here the width of the queue is less than in the previous case, and the cars are relatively more central, thus the cars just fit inside making this case just capable
Case 3	Cp>1.33	Cpk < 0.7	Here the width of the queue is the least but the cars aren’t centra, thus the cars wouldn’t fit inside making this case not capable
Case 4	Cp>1.33	Cpk>1.33	Here the width of the queue is the least and the cars are as central as they can be, thus the cars would easily fit inside making this case capable

Note: that these are just arbitrary numbers just give an analogy.

From the above table, we can conclude that having a higher Cp and Cpk is always a better option, as it narrows down the control chart bringing all the readings closer to the mean and also centralizing it between the two control limits. This ensures that the number of defective products/parts produces is as low as possible.

Sample Problems

Problem 1: Food served at a restaurant should be between 40°C and 50°C when it is delivered to the customer. The process used to keep the food at the correct temperature has a process standard deviation of 2°C and the mean value for these temperatures is 43. What is the process capability index of the process?

Solution:

$\bar{x}$ = 43

USL = 50

LSL = 40

σ = 2

$C_{pk} = \min(\frac{\bar{x}-LSL}{3\sigma},\frac{USL-\bar x}{3\sigma})\\ C_{pk} = \min(\frac{43-40}{3\times 2},\frac{50-43}{3\times 2})$

C_pk = min (0.5, 1.16)

= 0.5

Problem 2: Calculate the C_pof a control chart whose USL, LSL, and SD are 15,5,2 respectively.

Solution:

USL = 15

LSL = 5

σ = 2

$C_p = \frac{\text{USL} -\text{LSL}}{6\times \sigma}\\ C_p = \frac{15-5}{6\times 2}$

C_p = 10/12

= 0.833

Problem 3: A control chart of a company gives the following readings;

Parameter	Reading
USL	117
UCL	71
LSL	37
LCL	43
Standard Deviation	4.68
Mean	53

Calculate the process capability indexes(C_pand C_pk)

Solution:

From the table we see that

USL = 117

UCL = 71

LCl = 43

LSL = 37

σ = 4.68

$\bar{x}$ = 53

$C_p = \frac{\text{USL}-\text{LSL}}{\text{UCL}-\text{LCL}}\\ C_p = \frac{117-37}{71-43} = 2.85\\ C_{pk} = \min(\frac{\bar{x}-LSL}{3\sigma},\frac{USL-\bar x}{3\sigma})\\ C_{pk} = \min(\frac{53-37}{3\times 4.68},\frac{117-53}{3\times4.68})$

C_pk = min (1.14 , 4.55)

= 1.14