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
Q. Wind Induced Vibration of Vertical Vessels? Vertical vessels having a height-to-diameter ratio (h/D) (does not include insulation thickness but does include skirt height) gr

Carno t Theorem: No heat engine operating in cycle between the two given thermal reservoir, having fixed temperature can be more efficient than reversible engine operating be

Determine the width and depth of the beam: A timber beam of rectangular section is only supported over a span of 5 m. It carries an uniformly distributed load of 15 kN/m over

What is alternating quantity?

Loa d Diagram and BM D fro m the Given SF D: Q: Shear force diagram of simply supported beam is given in the figure below.  Calculate the support reactions of beam and

Explain how chips are prepared in cutting and how are they classified with sketches. What are the factors responsible for creation of different forms of chips?

Safety Benefits - Industrial Safety An industry or a company which follows safety programmes is benefitted in several ways. The benefits are shared among employer and employee

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

Stress-strain carves for ductile materials: A material is said to be ductile in nature, if it elongates before fracture. One this type of material is mild steel. Shape of stre

Inspection of Fasteners: The motorcycle carries number of nuts and bolts. They should be properly tightened for the smooth running of motorcycle. All fasteners are checked