The bore size of a component to be used in an assembly is a critical dimension.

Question

The bore size of a component to be used in an assembly is a critical dimension. Random samples of 4 are chosen. The summary information below is for 25 Samples. sigma^25_i=1 xi bar = 107.5 sigma^25_i=1 Ri bar = 12.5 The specification limits are 4.40.2 mm lumes a. Assuming the process is in control, estimate the process standard deviation b. Find the Xbar and R chart control limits c. What are your conclusion regarding the ability of the process to produce items conforming to specifications? Find Cp and Cpk. d. Assuming that if an item exceeds the upper specification limit it can be reworked and if it is below the lower limit it must be scrapped, what percent scrap and rework is the process now producing? e. It the process were centered on at u = 4.4, what would be the effect on the percent scrap or rework? f. The unit cost of scrap and rework are $2.40 and $0.75 respectively. The Daily production rate is 1200. Find the total cost of scrap and rework

Explanation / Answer

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For the given question the subgroup size is n=4

Number of subgroups (k) = 25

Also given, i=125 Xbari =107.5 and i=125Ri = 12.5

Thus, we have the average range = Rbar=i=125Ri/k = 12.5/25 = 0.5

And the average of the averages i=125 Xbari/k = 107.5/25 = 4.3

(a)It is given that the process is in statistical control, so the process standard deviation in this case is given by =Rbar/d2 =0.5/2.059=0.242836.

*The value of d2 depends on the subgroup size n. Here, n=4 so d2=2.059

Thus, the process standard deviation when the process is in control is 0.242836.

(b) The control limits for R-chart is given as

LCLR = D3Rbar and UCLR = D4Rbar

The value of D3 andD4 for a subgroup of size 4 are 0 and 2.282 respectively.

So, we have LCLR = 0 and UCLR =1.141.

And the control limits for Xbar-chart is given as

LCLx = Xbarbar -A2Rbar and UCLX = Xbarbar +A2Rbar

             The value of A2 for a subgroup of size 4 is 0.729.

            So, we have LCLx = 3.9355 and UCLX =4.6645.

(c)The ability of the process to produce items which are within the specification limits(process capabilities) are given by 6 where is the process standard deviation. Since the value of is not known we used the estimate derived in part (a) which is 0.242836.So, the process capability is 6x0.242836=1.457016.

        CP is given by USL-LSL/6..

        The specification limit for the given question are USL=4.4 and LSL=0.2.

        So, we have CP =(4.4-0.2)/1.457016 =2.88260.

        And CPK is given by min[(USL-µ)/3, (µ-LSL)/3].

        So, we have Cpk =min[(4.4-4.3)/3x0.242836, (4.3-0.2)/3x0.242836]

                             CPK = min(0.137266, 5.62794) = 0.137266

(d) The percentage of items produced above the upper specification is given by

Zupper = (USL-Xbarbar)/ = (4.4-4.3)/0.242836 = 0.0242836

ZLower = (Xbarbar-LSL)/ = (4.3-0.2)/0.242836 = 16.8838

*Note that the above are percentages.

Thus, the percent of rework being produced is 0.0242836 and the per-cent of rework being produced is 16.8838.

(e) If the process is centred at 4.4 then the change in scrap and rework is

Zupper = (USL-Xbarbar)/ = (4.4-4.4)/0.242836 = 0

ZLower = (Xbarbar-LSL)/ = (4.4-0.2)/0.242836 = 17.2956

(f) Given the daily production rate is 1200 units.

       And the cost of scrap and rework is $2.40 and $0.75 respectively.

       Using the percentages in (d) we derive at an estimate of the number of scrap and rework        produced in a day.

Number of scrap units =1200x0.168838 = 202.6056~203

Number of rework units =1200x0.000242836 = 0.2914032~0

So, the total cost of scrap and rework is 203x$2.40 + 0x$0.75 =$487.2

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