Determine the loss-by-defect and loss-by-dispersion, Mechanical Engineering

Determine the loss-by-defect and loss-by-dispersion

Given, Annual production = 1,00,000 units

Specification = 20 ± 4  i.e. m = 20, Δ = 4

Cost of repairing or resetting a product out-of-specification is Rs. 100.

a. Process I,

17_Determine the loss-by-defect and loss-by-dispersion 1.png = 20, σ = 1.33

b. Process II,

17_Determine the loss-by-defect and loss-by-dispersion 1.png = 18, σ = 0.66

c. Process III,

17_Determine the loss-by-defect and loss-by-dispersion 1.png = 17, σ = 0.40

Determine the loss-by-defect and loss-by-dispersion.

Solution

Process I

Given specifications 20±4

∴   USL = 24

    LSL = 16

Given process average (17_Determine the loss-by-defect and loss-by-dispersion 1.png) is mean

 centred at target m = 20 and σ = 1.33.

2215_Determine the loss-by-defect and loss-by-dispersion 2.png

= Min {24 - 20 /3 × 1.33, 20 - 16 /3 × 1.33}           

As both values are equal, we might use either of them as minimum value.

∴          C pk  =  4/ (3 × 1.33) = 1

Loss-by-defect

Loss = proportion out of specification × total number × cost of product

= 0.0027 × 1,00,000 × 100

= 0.27 × 1,00,000

Loss-by-dispersion

Loss = Loss per piece × number of products

1642_Determine the loss-by-defect and loss-by-dispersion 3.png

k =    A/ Δ2 = 100/42  = 6.2

Process II

∴          Loss = 6.2 [(20 - 20)2 + 1.332] × 1,00,000

= 10.97 × 1,00,000

 The process average is observed to be centred at 18 with σ = 0.66

2319_Determine the loss-by-defect and loss-by-dispersion 4.png

= Min { 24 - 18  /3 × 0.66      , 18 - 16/3 × 0.66}     

C pk  =  18 - 16 / 3 × 0.66 = 1.01 ≈ 1

Loss-by-defect

Loss = proportion out of specification × total number × cost of product

Standard normal variable at LSL

 At USL

Z 1 = 16 - 18/  0.66

 = - 3.03

Z2   =  24 - 18/0.66 = 9.09

∴          Proportion out of specification from tables,

= F (- 3.03) + F (9.09)

= 0.00122 + 0

= 0.00122

∴          Loss = 0.00122 × 100000 × 100

= 0.122 × 105 Rs.

Loss-by-dispersion

Loss = Loss per piece × Number of products

1723_Determine the loss-by-defect and loss-by-dispersion 5.png

k =    A/ Δ2

= 100 = 6.25

∴          Loss = 6.25 [(18 - 20)2 + 0.662] × 1,00,000

                     = 27.7 × 105

Process III

x = 17, σ = 0.40

562_Determine the loss-by-defect and loss-by-dispersion 6.png

= Min {24 - 17/3 ´0.4  , 17 - 16 /3´0.4}

= min {5.83, 0.83}

∴          PCI = 0.83

At LSL Z = 16 - 17 /0.4 = - 2.5

At USL Z = 24 - 17 /0.4 = 17.5

∴          Proportion out of specification, from tables

= F (- 2.5) + F (17.5)

= - F (2.5) + F (17.5)

= 0.00621 + 0

= 0.00621

∴ Loss by defect = Proportion out of specification × Total product

× Cost of product

= 0.00621 × 100000 × 100

= 0.621 × 105

Loss-by-dispersion

1539_Determine the loss-by-defect and loss-by-dispersion 7.png

= 6.25 [(17 - 20)2 + 0.42] × 100000

= 57.25 × 105

Posted Date: 12/26/2012 8:05:27 AM | Location : United States







Related Discussions:- Determine the loss-by-defect and loss-by-dispersion, Assignment Help, Ask Question on Determine the loss-by-defect and loss-by-dispersion, Get Answer, Expert's Help, Determine the loss-by-defect and loss-by-dispersion Discussions

Write discussion on Determine the loss-by-defect and loss-by-dispersion
Your posts are moderated
Related Questions
Clausius inequality - thermodynamics: Whenever closed system undergoes cyclic process, cyclic integral is less than zero (that is negative) for an irreversible cyclic process

WHATIS CG OF A WIRE BENT IN THE FORM OF A QUARTER ELLIPSE


Selecting Dimension Commands                                            The Linear Dimension Commands As the name suggests the Linear dimension commands are used to

Brinel l Hardness Test: Insert the Ball of diameter D in ball holder of machine. Make specimen surface free from dust and dirt. Make contact between the surface and ball b

I am attending Thermodynamics class at the university. The professor asked me to write a research on Renewable Energy: Solar Power. The report should be of 10 pages and at least 10

Explain the Adiabatic System - Thermodynamic Adiabatic System: A system along with adiabatic walls can just only exchange work and not heat with the surrounding is known as Adi

Expected quarterly sales in Year 1 (vehicles) The work content of a single unit of each product in each process has been estimated at: Using MS-Excel, calculate the workloads

Design for a water treatment plant: Produce as individuals a single tab Excel spreadsheet which shows a rough initial design for a water treatment plant taking 15

Q. What is IFD in project management? The P&IDs now reflect the next intermediate stage of development and incorporate any changes requested by the Customer during the IFA revi