A large manufacturing company manufactures fuel injection pumps for diesel engin
ID: 3200941 • Letter: A
Question
A large manufacturing company manufactures fuel injection pumps for diesel engines. These pumps were used in a variety of vehicles starting from small cars to locomotives. These are high-precision items and the company is equipped with high accuracy machines and a highly competent workforce of around 1200 people. The problem that arose was the assembly of monobock elements for the fuel injection pumps. Each element consists of a barrel in which a plunger operates. The element pressurizes fuel to around 1300 bar for injection into the engine. The clearance between the outer diameter of the plunger and the inner diameter of the barrel is critical. If the clearance is more than specified, there can be fuel leakages and if the clearance is less than the specified value, the fuel injection pump may jam during operation and the diesel engine will stop working. Hence, the clearance is a ‘critical to quality’ characteristic (CTQ). A grinding operation known as ‘finish match grinding’ is used to grind a minute amount of material from each plunger to match its barrel before assembly. Before the project, the first pass yield of the process was as low as 85%. Second, there were on average 34 customer complaints per month from the field regarding fuel leakages and pump jamming in fuel injection pumps supplied by the company. Once this field complaint occurs, the vehicle cannot be used until the pump is replaced, requiring a tow to a garage. This was leading to total customer dissatisfaction and a negative impact on business [Antony et al., 2011]. As part of a six sigma approach to solve the problem, a DOE with four factors at three levels with replications was conducted. The ANOVA table and main/interactions plots are shown below. Answer the following questions:
1.What are the assumptions needed for conducting a DOE?
2.State the hypotheses that should be tested in this experiment.
3.What can be concluded from the ANOVA table about the process and the factors?
4.What is the optimal value for each of the factors considered in this study?
Source DF Sequential Ss Adjusted SS Adjusted MS F p-Value Dressing frequency 154.192 8.83 0.006 77.096 154.192 Grinding stock 2.31 0.150 20.133 40.266 Grinding feed 1.76 0.221 30.716 15.358 30.716 Dressing feed rate 3.49 0.071 30.423 60.846 Dressing frequency x grinding feed 0.022 4.65 40.571 162.069 162.069 Grinding stock x grinding feed 0.032 35.706 142.826 42.826 8.728 10 87.279 87.279 Total 678.195 26 Grinding stock Dressing frequency 81.0 79.5 78.0 76.5 75.0 45 35 Dressing feed rate Grind feed 81.0 79.5 780 75.0 120 100 frequency Dnessing frequency Grinding stock Grinding Grinding feedExplanation / Answer
a)The assumption of ANOVA with replication can be summed as follows
b)The null and alternative hypothesis can be written as
Null Hypothesis:
The population mean of the dressing frequency are equal.
The population mean of the grinding stock are equal
The population mean of the grinding feed are equal.
The population mean of the dressing feed rate are equal.
There is no interaction between dressing frequency and grinding feed.
There is no interaction between grinding stock and grinding feed.
Alternative Hypothesis:
The population mean of the dressing frequency are not equal.
The population mean of the grinding stock are not equal
The population mean of the grinding feed are not equal.
The population mean of the dressing feed rate are not equal.
There is interaction between dressing frequency and grinding feed.
There is interaction between grinding stock and grinding feed.
3)From the above analysis of two way ANOVA with replication the significant factors are those which has a p-value of less than 0.05 which can be summarised as
4)The optimal value of factor is the value having largest Signal to Noise ratio as it corresponds to better performance.
Optimal value for each factor is shown below
a)Decreasing frequency =level 30 (value=81)
b)Grinding stock =level 45 (value = 79.5)
c)Grinding feed =Level 2 (value =79.5)
d)Decreasing feed rate =100(value=81)
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