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
Newton's law of motion. Sol.: The entire system of Dynamics is based on the three laws of motion that are basis assumptions, and were initially formulated by Newton. Fir


Explain the uses of Pier and Well Foundations Well foundations are used to transmit large loads from piers of bridges and other marine structures. These are constructions carr

Ceramics, Refractory And Abrasive Materials Introduction There are some situations in practice where high temperatures are occupied. Besides creating that temperatures en

a) You were assigned to determine the production rate for a polymer extrusion process. The extruder is found to have a barrel diameter = 4.5 in and length = 11 ft. The extruder scr

Evaluate the Support required by body on plane: A body having weight 50KN rests in limiting equilibrium on rough plane, whose slope is 30º. The plane is raised to a slope of

A horizontal line ABCD measuring 9m is acted upon by forces of magnitude 400, 600, 400 and 200 N at points A, B, C, D respectively with downward direction. These point are so locat

Information phase of screw conveyor with auger: Steps: To secure all and complete information and gather facts and from best sources.Our project is selected based on the cu

Machine and Tool Selection The quality of a product largely depends on the type of machine, the type of process, machining conditions and the right selection of tools. The cur