 17th Feb 2024
 06:03 am
On November 1, 2021, John Adams, a customer service representative of Americo Drilling Supplies (ADS), was summoned to the Houston office of Drilling Contractors, Inc. (DCI), to inspect three boxcars of mudtreating agents that ADS had shipped to the Houston firm. DCI had complained that the 25pound bags of treating agents received from ADS were lightweight.
The lightweight bags were initially detected by one of DCI’s receiving clerks, who noticed that the railroad side scale tickets indicated that the net weights were significantly less on all three of the boxcars than those of identical shipments received on October 25, 2021. ADS’s traffic department was called to determine if lighterweight dunnage or pallets were used on the shipments. (This might explain the lighter weights.) ADS indicated, however, that no changes had been made in the loading or palletizing procedures. Hence, DCI randomly checked 25 of the bags and discovered that the average net weight was almost 24.5 pounds. Consequently, they concluded that the sample indicated a significance shortweight. ADS was then contacted, and Adams was sent to investigate the complaint and to issue credit to DCI.
DCI, however, was not completely satisfied with the issuance of credit for the short shipment. The charts followed by their mud engineers on the drilling platforms were based on 25pound bags of treating agents. Lighterweight bags might result in poor chemical control during the drilling operation and might adversely affect drilling efficiency. (Mudtreating agents are used to control the pH and other chemical properties of the open during drilling operations.) This could cause severe economic consequences because of the extremely high cost of oil and natural gas drilling operations. Consequently, special use instructions had to accompany the delivery of these shipments to the drilling platforms. Moreover, the lightweight shipments had to be isolated in DCI warehouse, causing extra handling and poor space utilization. Hence, Adams was informed that CDI might seek a new supplier of mudtreating agents if in the future it received bags that deviated significantly below 25 pounds.
The quality control department at ADS suspected that the lightweight bags may have resulted from “growing pains” at the Orange, Texas plant. Because of the earlier energy crises, oil and natural gas exploration activity had greatly increased. This increased activity, in turn, created increased demand for products produced by related industries, including drilling muds. Consequently, ADS had to expand from one shift (6 A.M. to 2 P.M.) to a twoshift (2 P.M. to 10 P.M.) operation in Mid2014, and finally to a threeshift operation (24 hours per day) in January of 2018.
The additional night shift bagging crew was staffed entirely by new employees. The most experienced foremen were temporarily assigned to supervise the night shift. Most emphasis was placed on increasing the output of bags to meet the everincreasing demand. It was suspected that only occasional reminders were made to doublecheck the bag weight feeder. (A double check is performed by systematically weighting a bag on a scale to determine if the proper weight is being loaded by the weightfeeder. If there is significant deviation from 25 pounds, corrective adjustments are made to the weightrelease mechanism.)
To verify this expectation, the quantity control staffs at ADS randomly sampled bags and prepared the following table (see Excel dataset). Note that four bags were sampled and weighted each hour (sample of size 4).
ANALYSIS
Assume you are John Adams of ADS. Based on the following analysis, prepare a report to be submitted to both ADS and DCI executives regarding the status of the filling process at ADS and recommend method to improve quality control at the filling station and estimate the amount of credit to DCI. Use the data provided in the Excel data file.
 Calculate the Range Column in Excel (LargestSmallest) and find mean and standard deviations for Average Weight, Smallest, Largest, and Range columns.
Average (Xbar) 
Smallest 
Largest 
Range 

Mean 
24.6 
23.7 
25.4 
1.7 
Standard Deviations 
0.5450 
0.7678 
0.5965 
0.6341 
 What is the standard deviation of individual bags if sample averages are based on 4 bags? Explain your finding. Hint: ?_{xbar} = ?/sqrt(n). Having ?_{xbar}_{ }based on part (a) calculation, solve for ? given n= 4. So ? = (?_{xbar})*(sqrt(n))
Average (Xbar) 
Smallest 
Largest 
Range 

Mean 
24.6 
23.7 
25.4 
1.7 
Standard Deviations (?_{xbar})*(sqrt(n)) 
0.5450 
0.7678 
0.5965 
0.6341 
Standard Deviations (σ) 
1.09 
1.54 
1.19 
1.27 
 Construct a time series plot of all four variables (two graphs: one includes avg., smallest, and largest and the other has range data) and discuss your findings based on the graphs.
 Construct Xbar and R Charts, graph, and discuss your findings based on the control charts. Make sure to include the control limit calculations below. Is the process out of control? Why? Explain. .6
n = 4
A2 = 0.729
D3 = 0
D4 = 2.282
Xbar = 24.6
Rbar = 1.7
X – chart
UAL = Xbar +A2 * Rbar
= 24.6 + 0.729 * 1.7
= 24.6 + 1.2393
= 25.8393
LAL = Xbar – A2 * Rbar
= 24.6 – 0.729 * 1.7
= 24.6 – 1.2393
= 23.3607
R – chart
URL = D4 * Rbar
= 2.282 * 1.7
= 3.8794
LRL = D3 * Rbar
= 0 * 1.7
= 0
The process is not out of control, according to the R chart, which depicts process variation. This was true for all three shifts. The XChart, commonly known as the process mean chart, is outside of the predetermined range, indicating that the process is out of control. We can see that during the night shift, the process mean chart or x chart is breaching the lower control limit.
 Is there any differences between performances of three shifts? Hint: (Carve out the morning, afternoon, and night shift data into three columns and graph their averages. Explain your findings based on graph.
We can observe that “Afternoon” curve is very close to the mean whereas the “Morning” and “Night” shift data displays significant deviation from the mean.
 Use One Way Analysis of Variance (ANOVA) to compare the three groups (shifts) to test if there is significant differences in the performance of three shift averages. Use Alpha =0.01. Indicate the Null and Alt. hypotheses and explain your findings based on ANOVA table.
Anova: Single Factor 









SUMMARY 


Groups 
Count 
Sum 
Average 
Variance 


Average (Xbar) 
72 
1771.8 
24.60833 
0.297042 


Smallest 
72 
1707.8 
23.71944 
0.589476 


Largest 
72 
1830.7 
25.42639 
0.355773 








ANOVA 


Source of Variation 
SS 
df 
MS 
F 
Pvalue 
F crit 
Between Groups 
104.9519 
2 
52.47597 
126.7239 
5.55781E37 
4.706187 
Within Groups 
88.20264 
213 
0.414097 





Total 
193.1546 
215 




 Assume Tolerance limits of 25±1.75 lbs. is specified in the sales contract and find the Process Capability index C_{pk}. Discuss if this process is capable to meet contractual agreement.
USL (Upper specification Limit) = 25 + 1.75 = 26.75
LSL (Lower specification Limit) = 25 – 1.75 = 23.25
C_{pk} = min ((USL – Xbar)/(3 * σ), (Xbar – LSL)/(3 * σ))
C_{pk} = min ((26.75 – 24.6) / (3 * 1.09), (24.6 – 23.25) / (3 * 1.09))
C_{pk} = min (0.657, 0.413)
C_{pk} = 0.413
Since the C_{pk} is less than 1. So, this is not a capable process.
 Redo part g for all three shifts separately and determine their C_{pk}. Note that you must carve out the three shifts and find the mean and standard deviations for each shift separately and go through part b calculations for each shift to find ? for each shift.
Morning 
AfterNoon 
Night 

Mean 
25.2 
24.6 
24.0 
Standard Deviations (σx_bar) 
0.23 
0.18 
0.35 
Standard Deviations (σ) 
0.47 
0.35 
0.71 
USL = 26.75
LSL = 23.25
C_{pk} = min ((USL – Xbar)/(3 * σ), (Xbar – LSL)/(3 * σ))
For Morning Shift
C_{pk} = min ((26.75 – 25.2)/(3 * 0.47), (25.2 – 23.25)/(3 * 0.47))
C_{pk} = min (1.10, 1.38)
C_{pk} = 1.10
For AfterNoon Shift
C_{pk} = min ((26.75 – 24.6)/(3 * 0.35), (24.6 – 23.25)/(3 * 0.35))
C_{pk} = min (2.05, 1.29)
C_{pk} = 1.29
For Night Shift
C_{pk} = min ((26.75 – 24.0)/(3 * 0.71), (24.0 – 23.25)/(3 * 0.71))
C_{pk} = min (1.29, 0.35)
C_{pk} = 0.35
For morning shift and afternoon shift the Cpk does not lies between 0 and 1 that implies that the process is not capable. But for night shift it lies between 0 and 1 that implies that the process is capable in night shift.
 If the process average is adjusted to 25.0 lbs. and process standard deviation (answer in part b) is reduced by 60%, what is the new C_{pk}? Are you comfortable for making such recommendation to management? Explain.
USL = 25 + 1.75 = 26.75
LSL = 25 – 1.75 = 23.25
As per given new condition, the new standard deviation will be
= 1.09 – 0.60 * 1.09 = 0.44
C_{pk} = min ((USL – Xbar)/(3 * σ), (Xbar – LSL)/(3 * σ))
C_{pk} = min ((26.75 – 25.0)/(3*0.44), (25.0 – 23.25)/(3*0.44))
C_{pk} = min (1.33, 1.33)
C_{pk} = 1.33
Since, the C_{pk }value is larger than 1, so the process is not capable.
 Find the control limits for an Xbar and R chart if a new improved process has average of 25.0 lbs. and Rbar of 1.0 lbs. Assume n=5 for new process control. No graph needed.
Given n = 5,
So,
A2 = 0.577
D3 = 0
D4 = 2.114
Xbar = 25.0
Rbar = 1.0
X – chart
UAL = Xbar + A2 * Rbar
= 25.0 + 0.577 * 1.0
= 25.0 + 0.577
= 25.577
LAL = Xbar – A2 * Rbar
= 25.0 – 0.577 * 1.0
= 25.0 – 0.577
= 24.423
R chart
URL = D4 * Rbar
= 2.114 * 1.0
= 2.114
LRL = D3 * Rbar
= 0.0 * 1.0
= 0.0
 Estimate the amount of credit ($) must be given to DCI (for the past 36 months) if it purchased 30,000 bags per month at cost of $8 per lb.
Estimate for Credit 

Xbar 
24.6 
Deficit per Bag 
0.46 
No of Bags in a Month 
30,000 
Total Deficit 
13,800 
No of Months 
36 
Total Deficit in 10 months 
496,800 
Cost per Lb 
8 


Total Credit 
3,974,400 
 Provide complete conclusion regarding your findings and make recommendations regarding the filling process in your conclusion paragraph.
The process variation chart (RChart) and process mean chart (XChart) both indicate that the process is out of control. When the Cpk (process capability index) is low (between 0 and 1), during the night shift, the process is capable.