Question

Question 1

Notex Manufacturing makes various batteries used in mobile devices. The company has a major customer so batteries are shipped in bulk to this customer. The company also distributes these batteries to retail stores as replacement parts. The batteries are packaged individually to retail stores. In all, the company makes about 15 different batteries. Currently, the company does not use any forecasting to predict the demand for the batteries. Instead, it has employed rule of thumb to decide about the volume of the production. This has caused some issues for the company including stock-out for some types of batteries or overstocking for some others. The other problem is an increase in the price of raw materials though the company believes it is a temporary condition. Due to complaints from suppliers and also customers the company has decided to introduce a systematic approach toward forecasting.

Therefore, the company has decided to forecast two most important products. The following table (see next page) shows the data on product demand for the two products from order records for the previous 19 months.

Question

Which forecasting method/s do you suggest for the two products? Briefly explain why? Forecast for the next month for each product. (Use MAD for measuring error).

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec	Jan	Feb	Mar	Apr	May	Jun	Jul
Product 1	33	37	38	40	42	47	43	49	51	55	62	68	69	70	65	55	50	51	49
Product 2	33	34	35	35	36	36	37	39	81*	38	40	38	39	40	39	40	41	42	41

* Unusual order due to flooding of customer's warehouse.

Question 2

UFE Clubs produces ball bearings. The diameter of the ball bearings is very important for the quality purposes. The following tables shows the diameters of 6 randomly selected samples. Each sample contains four observations.

	Observations
Sample	1	2	3	4
1	.604	.612	.588	.600
2	.597	.601	.607	.603
3	.581	.570	.585	.592
4	.620	.605	.595	.588
5	.590	.614	.608	.604
6	.585	.583	.617	.579

a. Is the production process of the ball bearings are under control both in terms of central tendency and dispersion. Show all calculations.

b. A customer for ball bearings places an order with diameter of 0.600 +/- 0.050. What is the Process Capability Index? The Process Capability Ratio?

c. If the firm is seeking four-sigma performance, is the process capable of producing the ball bearings?

Answer 1

QUESTION 1:

he plotting of the data is shown below. Based on the trends we can choose the forecasting methods.

From the graph we can see that product 1 shows some variety in trend as the month progress. Due to this nature of variation in trend,

The best method will be to use exponential smoothing or moving average method to forecast the demand for Product 1.

In case of product 2, we see that except for that anomaly that took place in September, the trend is pretty much a straight line. This is why the best approach will be linear regression method of forecasting.

QUESTION 2:

(a)

Repeat the following calculations in Excel to find the control limits of X-bar and R-charts.

SAMPLE	X1	X2	X3	X4	X-BAR	RANGE
1	0.604	0.612	0.588	0.600	0.601	0.024
2	0.597	0.601	0.607	0.603	0.602	0.010
3	0.581	0.570	0.585	0.592	0.582	0.022
4	0.620	0.605	0.595	0.588	0.602	0.032
5	0.590	0.614	0.608	0.604	0.604	0.024
6	0.585	0.583	0.617	0.579	0.591	0.038
				AVG--->	0.597	0.025
					X-double-bar	R-bar
n=	4
A2=	0.729
D3=	0
D4=	2.282
UCL_X	0.61523
LCL_X	0.57878
UCL_R	0.05705
LCL_R	0.00000

Both the control charts show that the sample means and sample ranges are in control. So, the production process of the ball bearings is under control both in terms of central tendency and dispersion.

(b)

USL = 0.60 + 0.05 = 0.65
LSL = 0.60 - 0.05 = 0.55

An estimate of the process standard deviation can be found by range method which is as follows:

σ = R-bar / d₂ = 0.025 / 2.059 = 0.012

The estimate of process mean in the mean for the sampling distribution i.e. μ = X_double-bar = 0.597

So, since the process mean is shifted towards the LSL

C_pk = (μ - LSL) / 3σ = (0.597 - 0.55) / (3*0.012) = 1.3

C_p = (USL - LSL) / 6σ = (0.65 - 0.55) / (6*0.012) = 1.4

(c)

For a 4-sigma process the process capability index should at least be 1.333. But the C_pk here is 1.3 which is just marginally less. So, the process is marginally capable of meeting this standard.

Note: The coefficients A2, D3, D4, and d2 are to be noted from a standard 3-sigma table as follows: