Case Study:Acceptable Pins

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Case on Data Interpretaion in Uncertainity

Transcript of Case Study:Acceptable Pins

DATA INTERPRETATION IN UNCERTAINITY

ACCEPTABLE PINS

ACCEPTABLE PINS PRODUCT OF COMPANY DATA ON HAND MEAN, ( )= 1.012 MANUFACTURE & SUPPLY OF PINS

STANDARD DEVIATION, () = 0.018 CUSTOMER DESIRED LENGTH = 1.00 ACCEPTABLE TOLERANCES = 0.02 NORMALLY DISTRIBUTED

# All the dimensions are in inches

ACCEPTABLE PINS1. WHAT PERCENTAGE OF THE PINS WILL BE ACCEPTABLE TO THE CONSUMER? Now, X U N(1.012, 0.018)

Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/0.018)) < (X- 1.012 )/ 0.018 < (1.02- 1.012)/0.018)] = P[ -1.777 < Z < 0.444 ] = P[ -1.777 < Z < 0 ] + P[ 0 < Z < 0.444 ] = 0.4620 + 0.1720 = 0.6340 ---------(From Tables) [Say, Z = (X- )/ ]

# All the dimensions are in inches

ACCEPTABLE PINS

-1.777 = 1.012

0.444

63.40% OF PINS ARE ACCEPTABLE TO THE CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINS2. IF THE LATHE CAN BE ADJUSTED TO HAVE THE MEAN OF THE LENGTHSTO ANY DESIRED VALUE, WHAT SHOULD IT BE ADJUSTED TO? WHY?

SUPPOSE, =1.00 (ADJUSTED TO DESIRED VALUE) & = 0.018 Now, X U N(1.00, 0.018)

Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/0.018)) < (X- 1.00)/ 0.018 < (1.02- 1.00)/0.018)] [Say, Z = (X- )/ ] = P[ -1.111 < Z < 1.111] = P[ -1.111 < Z < 0 ] + P[ 0 < Z < 0.111 ] = 2(0.3665) = 0.7330 ---------(From Tables)

# All the dimensions are in inches

ACCEPTABLE PINS

-1.111 = 1.00

1.111

73.30% OF PINS ARE ACCEPTABLE TO THE CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINS3. SUPPOSE THE MEAN CANNOT BE ADJUSTED BUT THE STANDARDDEVIATION CAN BE REDUCED. WHAT MAXIMUM VALUE OF THE STANDARD DEVIATION WOULD MAKE 90% OF THE PARTS ACCEPTABLE TO THE CONSUMER (ASSUME THE MEAN TO BE 1.012.)

SUPPOSE, =1.012 & (VALUE NOT KNOWN) Now, X U N(1.00, ) given, P(0.98 < X < 1.02) = 0.90

Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/ ) < (X- 1.00)/ < (1.02- 1.00)/ )] [Say, Z = (X- )/ ] 0.90 = P[ -0.032/ < Z < 0.008/ ] 0.5 + 0.40 = P[[ -0.032/ < Z < 0 ] + P[ 0 < Z < 0.008/ ] [ASSUME, P[[ -0.032/ < Z < 0 ] ] = 0.5 0.5 + 0.40 = P[ a< Z < 0 ] + P[ 0 < Z < b ]

# All the dimensions are in inches

ACCEPTABLE PINS0.40 = P[ 0 < Z < b ] Say, b = 0.008/ Value of b = 1.28 @ 0.40 1.28 = 0.008 / ---------(From Tables)

Therefore, = 0.00625

# All the dimensions are in inches

ACCEPTABLE PINSa b a IS 4 TIMES THAT OF b

-0.032/ = 1.012

0.008/

MAXIMUM VALUE OF = 0.00625 WILL MAKE 90% OF THE PARTS ACCEPTABLE TO THE CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINS4. REPEAT QUESTION 3, WITH 95% AND 99% OF THE PINS ACCEPTABLE? SUPPOSE WITH 95% OF PINS ACCEPTANCE =1.012 & (VALUE NOT KNOWN) Now, X U N(1.012, ) given, P(0.98 < X < 1.02) = 0.95

Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/ ) < (X- 1.00)/ < (1.02- 1.00)/ )] 0.95 = P[ -0.032/ < Z < 0.008/ ] [Say, Z = (X- )/ ]

0.5 + 0.45 = P[[ -0.032/ < Z < 0 ] + P[ 0 < Z < 0.008/ ] 0.5 + 0.45 = P[ a< Z < 0 ] + P[ 0 < Z < b ] 0.40 = P[ 0 < Z < b ]

# All the dimensions are in inches

ACCEPTABLE PINSSay, b = 0.008/ Value of b = 1.645 @ 0.45 1.645 = 0.008 / ---------(From Tables)

Therefore, = 0.00486

# All the dimensions are in inches

ACCEPTABLE PINSa b a IS 4 TIMES THAT OF b

-0.032/ = 1.012

0.008/

MAXIMUM VALUE OF = 0.00486 WILL MAKE 95% OF THE PARTS ACCEPTABLE TO THE CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINS SUPPOSE WITH 99% OF PINS ACCEPTANCE =1.012 & (VALUE NOT KNOWN) Now, X U Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/ ) < (X- 1.00)/ < (1.02- 1.00)/ )] 0.99 = P[ -0.032/ < Z < 0.008/ ] [Say, Z = (X- )/ ] 0.5 + 0.49 = P[[ -0.032/ < Z < 0 ] + P[ 0 < Z < 0.008/ ] 0.5 + 0.49 = P[ a< Z < 0 ] + P[ 0 < Z < b ] 0.40 = P[ 0 < Z < b ] Say, b = 0.008/ N(1.012, ) given, P(0.98 < X < 1.02) = 0.99

# All the dimensions are in inches

ACCEPTABLE PINSValue of b = 2.33 @ 0.49 2.33 = 0.008 / ---------(From Tables)

Therefore, = 0.00343

# All the dimensions are in inches

ACCEPTABLE PINSa b a IS 4 TIMES THAT OF b

-0.032/ = 1.012

0.008/

MAXIMUM VALUE OF = 0.00343 WILL MAKE 99% OF THE PARTS ACCEPTABLE TO THE CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINS5. IN PRACTICE, WHICH ONE DO YOU THINK IS EASIER TO ADJUST, THEMEAN OR THE STANDARD DEVIATION? WHY?

THE SUMMARIZATION OF ABOVE RESULTS THINK US AS FOLLOWS:Q 1 2 3 4a 4b MEAN STANDARD DEVIATION () () 1.012 1.00 1.012 1.012 1.012 0.018 0.018 0.00625 0.00486 0.00343 LEVEL OF PINS ACCEPTABLE (% ) 63.40 73.30 90 95 99 REMARKSA minor change in results in comparatively lesser impact on % acceptance A minor change in results in comparatively lesser impact on % acceptance

AS PER THE PROBLEM, THE COST OF RESETTING THE MACHINE TO ADJUST THE MEAN ()INVOLVES THE ENGINEERS TIME AND THE COST OF PRODUCTION TIME. THE COST OF REDUCING THE STANDARD DEVIATION () INVOLVES THE ENGINEERS TIME, THE COST OF PRODUCTION TIME, THE COST OF OVERHAULING THE MACHINE AND REENGINEERING THE PROCESS.

SO IN PRACTICE, THE MEAN IS EASIER TO ADJUST

# All the dimensions are in inches

ACCEPTABLE PINS6. ASSUME IT COSTS $150 x2 TO DECREASE THE STANDARD DEVIATION () BY (x/1000) INCH. FIND THE COST OF REDUCING THE STANDARD DEVIATION TO THE VALUES FOUND IN QUESTION 3 AND 4?

From Q3, we know,

[Say, a - ACTUAL STANDARD DEVIATION, d - DERIVED STANDARD DEVIATION & c CHANGE IN STANDARD DEVIATION

=1.012 & d = 0.00625 & P(0.98 < X < 1.02) = 0.90CHANGE IN , c = (a d) = (0.018 0.00625) = 0.01175 c = 11.75/1000 i.e., x = 11.75

# All the dimensions are in inches

ACCEPTABLE PINSFrom given relation, Cost = $150 x2 = $150 X (11.75) 2 Therefore, Cost = $20709.375 THE COST OF REDUCING THE STANDARD DEVIATION IS $20709.375, IF 90% OF THE PARTS ACCEPTABLE TO CONSUMER From Q4a, we know, =1.012 & d = 0.00486 & P(0.98 < X < 1.02) = 0.95 CHANGE IN , c = (a d) = (0.018 0.00486)

# All the dimensions are in inches

ACCEPTABLE PINSc = 0.01314c = 13.14/1000 From given relation, i.e., x = 13.14

Cost = $150 x2= $150 X (13.14) 2 Therefore, Cost = $25898.94

THE COST OF REDUCING THE STANDARD DEVIATION IS $25898.94, IF 95% OF THE PARTS ACCEPTABLE TO CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINSFrom Q4b, we know, =1.012 & d = 0.00343 & P(0.98 < X < 1.02) = 0.99 CHANGE IN , c = (a d) = (0.018 0.00343) = 0.01457

c = 14.57/1000From given relation, Cost = $150 x2 = $150 X (14.57) 2 Therefore, Cost = $31842.735

i.e.,

x = 14.57

THE COST OF REDUCING THE STANDARD DEVIATION IS $31842.735, IF 99% OF THE PARTS ACCEPTABLE TO CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINS7. NOW ASSUME THAT THE MEAN HAS BEEN ADJUSTED TO THE BEST VALUE FOUND IN QUESTION 2 AT A COST OF $80. CALCULATE THE REDUCTION IN STANDARD DEVIATION NECESSARY TO HAVE 90%, 95% AND 99% OF THE PARTS ACCEPTABLE. CALCULATE THE RESPECTIVE COSTS, AS IN QUESTION 6?

From Q2, we know,

=1.00 & (VALUE NOT KNOWN),WITH 90% OF PINS ACCEPTANCEP(0.98 < X < 1.02) = 0.90 Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/)) < (X- 1.00)/ ) < (1.02- 1.00)/)] 0.9 = P[-0.02/ < Z < 0.02/ ] 0.9 = P[-0.02/ < Z < 0 ] + P[ 0 < Z < 0.02/] # All the dimensions are in inches

ACCEPTABLE PINS0.90 = 2 P[ 0 < Z < 0.02/]0.45 = P[ 0 < Z < 0.02/] 0.45 = P[ 0 < Z < b] Say, b = 0.02/

Value of b = 1.645 @ 0.451.645 = 0.02 /

----------

(From Tables)

Therefore, d = 0.01216# All the dimensions are in inches

ACCEPTABLE PINS

-0.02/ = 1.00

0.02/

# All the dimensions are in inches

ACCEPTABLE PINSCHANGE IN , c = (a d)= (0.018 0.01216) = 0.00584 c = 5.84/1000 From given relation, Cost = $150 x2 + $80 ($80 IS THE COST FOR ADJUSTING TO DESIRED VALUE) = $150 X (5.84) 2 + $80 Therefore, Cost = $5195.84 THE COST OF REDUCING THE STANDARD DEVIATION IS $5195.84, IF 90% OF THE PARTS ACCEPTABLE TO CONSUMER i.e., x = 5.84

# All the dimensions are in inches

ACCEPTABLE PINSFrom Q2, we know, =1.00 & (VALUE NOT KNOWN),WITH 95% OF PINS ACCEPTANCE P(0.98 < X < 1.02) = 0.95 Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/)) < (X- 1.00)/ ) < (1.02- 1.00)/)] 0.95 = P[-0.02/ < Z < 0.02/ ] 0.95 = P[-0.02/ < Z < 0 ] + P[ 0 < Z < 0.02/] 0.95 = 2 P[ 0 < Z < 0.02/] 0.475 = P[ 0 < Z < 0.02/]

# All the dimensions are in inches

ACCEPTABLE PINS0.475 = P[ 0 < Z < 0.02/]Say, b = 0.02/ Value of b = 1.96 @ 0.475 1.96 = 0.02 / ---------(From Tables)

Therefore, d = 0.01020CHANGE IN , c = (a d) = (0.018 0.01020)

= 0.00780c = 7.80/1000 i.e., x = 7.80

# All the dimensions are in inches

ACCEPTABLE PINSFrom given relation,Cost = $150 x2 + $80 ($80 IS THE COST FOR ADJUSTING TO DESIRED VALUE) = $150 X (7.80) 2 + $80

Therefore, Cost = $9206 THE COST OF REDUCING THE STANDARD DEVIATION IS $9206, IF 95% OF THE PARTS ACCEPTABLE TO CONSUMER

# All the dimensions are in inches

ACCEPTABLE PINSFrom Q2, we know, =1.00 & (VALUE NOT KNOWN),WITH 99% OF PINS ACCEPTANCE P(0.98 < X < 1.02) = 0.99 Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/)) < (X- 1.00)/ ) < (1.02- 1.00)/)] 0.99 = P[-0.02/ < Z < 0.02/ ] 0.99 = P[-0.02/ < Z < 0 ] + P[ 0 < Z < 0.02/] 0.99 = 2 P[ 0 < Z &