# Paper Solved

Topics: Arithmetic mean, Average, Sample size Pages: 25 (5615 words) Published: March 1, 2013
Fourth term End term exam 2011 (1)
Section B
2. b
From a factory producing metal sheet, following defective were analyses in every lot size. Prepare relevant chart only UCL and LCL. Lot Size| 100| 125| 90| 120| 110|
Defective| 6| 7| 4| 8| 10|

Solution
Since the Lot size are different we have to use ‘u’ chart. Sample NO| LOT Size| Defective| Proportion of defective| UCL| LCL| 1| 100| 6| 0.06| 0.140245| -0.0118|

2| 125| 7| 0.056| 0.06422| -0.00378|
3| 90| 4| 0.04 | 0.144358| -0.01592|
4| 120| 8| 0.066666667| 0.133621| -0.00518|
5| 110| 10| 0.090909091| 0.136707| -0.00827|
Total| 545| 35| | | |

u =sum of defects in all lots / total number of items in all lots = 35/545 = 0.064
UCL = u + 3√(u/ni)
LCL = u - 3√(u/ni)
UCL always <1.0
And LCL always >0
Since here all the LCL value <0 we have to ignore LCL values.

Proportion of Defective
Proportion of Defective
Lot
Lot
Center Line (u)
Center Line (u)

4. a) Topaz Industries manufactures Z brand radar scanner used to detect speed traps. The printed circuit boards in the scanners are purchased from an outside vendor. The vendor produces the boards to an AQL of 2% defective and is willing to run 5% risk(alpha) of having lots of this level or fewer defectives rejected. Topaz’s consider lot of 8% or more defective lot tolerance unacceptable and time. A large shipment has just been delivered. what values of n and c should be selected to determine the quality of this lot? Solution:

The negotiated consumer’s and producer’s risks, corresponding to LTPD 8% and AQL of 2%, are 10% and 5% each.

Therefore P0.10/P0.95 = 0.08/0.02 = 4
Therefore c = 4 gives the best fit.

Now as per the table, the sample size can be calculated at the producer’s risk point : np0.95 =1.970 n= 1.970/0.02 = 98.5=99
Therefore the sampling plan is (99,4)

6) Elisa watches assemble four different types of watches on their assembly line. The assembly line in done in batches. Given the following data, in what batch sizes should each type of watch be produced? The company has 300 working days in a year. Find economic batch quantity? Watch| Annual dd - Unit| Set up cost(Per set up)| Carrying cost per unit per yr| Assembling rate(no/day)| TAG-H| 9000| 1000| 40| 100|

TITAN| 5000| 1000| 50| 120|
GUCCI| 10000| 1000| 120| 100|
FOSSIL| 3000| 1000| 30| 90|

Solution:
There are 300 working days in a year.
Based on the above given information, following gives the rates of consumption of the products, the economic batch quantitates for each watch, the days for which a manufactured batch will last at the given consumption rate and the individual production times for each batch of product (along with the total time for production for a batch-each of the entire range of products). I| II| III| IV| V| VI| VII| VIII|

Watch| Annual dd| Assembling rate(unit/day)| Daily consumption rate(unit/day)| Economic Batch Quantity| No of days required to produce a batch quantity| No of days required to manufacture annual requirements| No of days taken to consume a batch quantity| TAG-H| 9000| 100| 30| 670.8204| 6.708| 90| 22.36|

TITAN| 5000| 120| 16.6666| 447.2136| 3.726| 41.66| 26.83| GUCCI| 10000| 100| 33.33333| 408.2483| 4.08| 100| 12.24| FOSSIL| 3000| 90| 10| 447.2136| 4.96| 33.33| 44.72| Total| 19.48| 265| |

Calculation of Daily consumption rate by the formula Column II / 300 Calculation of EBQ = √(2 X Annual dd X Setup Cost) / Carrying cost Calculation of No of days required to produce a batch quantity = Column V/ Column III Calculation of No of days required to manufacture annual requirements = Column II/ Column III Calculation of No of days taken to consume a batch quantity = Column V/ Column IV

Above table gives the total production days required to...