1. CASE STUDY PROCESS OPTIMISATION STATEMENT OF THE PROBLEM To find the process parameters which influence the quality of PCB wave soldering process relative to customer requirement Soldering quality is defined in terms of various types of defects occurring during production OBJECTIVE To determine the process parameters that will minimize the soldering defects Present level of defects : 6920 ppm

2. CAUSE AND EFFECT DIAGRAM DEFECTIVE SOLDERING MAN METHOD MATERIAL MACHINE Untrained operator S.G. of Flux Pre-Heat temp Dross not Cleaned Masking Tape Quality Quality of Flux Defective Displays Dross in Solder-Bath Conveyor Speed Solder Bath temp Thickness of PCB Density of PCB Date expired PCB Auto Fluxer mot function PCB not cleaned properly PCB not properly baked

4. PROCESS FACTORS SELECTION OF FACTORS AND LEVELS 1.00 0.800 – 1.200 m/min 4. Conveyor Speed (D) 246 O C 240 O C – 260 O C 3. Solder Temperature (C) 180 O C ---- 2. Pre-heat Temperature (B) 0.825 0.820 - 0.840 1. S.G of Flux (A) PRESENT SETTING RANGE FACTOR TOTAL : 10 1 CD 1 BD 1 BC 1 AD 1 AC 1 AB 1 1.60 0.840 D 1 2.58 242 C 1 195 165 B 1 0.838 0.822 A HIGH LOW DEGREE S OF FEREEDOM LEVELS FACTORS

5. L 16 OA 1 2 2 1 2 1 1 2 2 1 1 2 1 2 2 16 2 1 1 2 1 2 2 1 2 1 1 2 1 2 2 15 2 1 1 2 2 1 1 2 1 2 2 1 1 2 2 14 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 13 2 1 2 1 1 2 1 2 1 2 1 2 2 1 2 12 1 2 1 2 2 1 2 1 1 2 1 2 2 1 2 11 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 10 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 9 2 2 1 1 1 1 2 2 1 1 2 2 2 2 1 8 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 7 1 1 2 2 1 1 2 2 2 2 1 1 2 2 1 6 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 5 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 4 2 2 2 2 1 1 1 1 2 2 2 2 1 1 1 3 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 COLUMNS TRIAL NO.

6. LINEAR GRAPHS PROBLEM GRAPH STANDARD GRAPH (L16) C (1) 11 5 6 14 3 13 B (2) D (4) A (15)

7. ASSIGNMENT OF FACTORS AND INTERACTIONS TO COLUMNS OF L16 OA A 15 AC 14 AB 13 ABC 12 AD 11 ACD 10 ABD 9 ABCD 8 BCD 7 BD 6 CD 5 D 4 BC 3 B 2 C 1 FACTOR / INTERACTION ASSIGNED COLUMN NUMBER

8. Y = Response = Sum of two repetitions D C D D C D C B B D C B B C A A A A A A A A B BD CD D BC B C 10.0 1 2 2 1 2 1 1 2 2 1 1 2 1 2 2 16 11.0 2 1 1 2 1 2 2 1 2 1 1 2 1 2 2 15 14.8 2 1 1 2 2 1 1 2 1 2 2 1 1 2 2 14 12.0 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 13 12.3 2 1 2 1 1 2 1 2 1 2 1 2 2 1 2 12 14.6 1 2 1 2 2 1 2 1 1 2 1 2 2 1 2 11 6.9 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 10 6.9 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 9 7.4 2 2 1 1 1 1 2 2 1 1 2 2 2 2 1 8 11.7 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 7 29.7 1 1 2 2 1 1 2 2 2 2 1 1 2 2 1 6 3.2 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 5 9.6 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 4 16.9 2 2 2 2 1 1 1 1 2 2 2 2 1 1 1 3 7.5 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 2 13.5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 COLUMN RUN

9. L16 TOTALS TABLE 5.84 93.5 2 5.91 94.5 1 D 92.7 2 95.3 1 BC 78.5 2 109.5 1 AC 5.53 88.5 2 6.22 99.5 1 C 107 2 81 1 AB 6.24 99.8 2 5.51 88.2 1 B 5.00 80 2 6.75 108 1 A MEAN TOTAL LEVEL FACTOR TOTAL 188.00 98.2 2 89.8 1 ABCD 74.2 2 113.8 1 ACD 98.7 2 89.3 1 ABD 113.1 2 74.9 1 ABC 94.2 2 93.8 1 BCD 86.2 2 101.8 1 CD 113.1 2 74.9 1 BD 78.3 2 109.7 1 AD TOTAL LEVEL FACTOR

10. COMPUTATION OF SS SS Total = sum of (Y i ) 2 – (T 2 /N) T = 188 (Grand Total); N = 32 SS Factor/Int = (T 1 - T 2 ) 2 / N where, T 1 and T 2 = Totals at level 1 and level 2 respectively SS e1 = SS due to primary error = 0 (no vacant column) SS e2 = SS due to secondary error = SS Total - SS Int of all factors EXAMPLE: SS Total = 5.4 2 + 8.1 2 + ……… +5.7 2 - 188 2 /32 = 303.88 SS A = (108-80) 2 /32 = 24.5 SS ABC = (74.9 – 113.1) 2 /32 = 45.601 SS e2 = SS Total - SS A – SS B - ……………- SS ABCD

11. SUMS OF SQUARES SS (Total) = 5.4 2 + 8.1 2 + ……… +5.7 2 - 188 2 /32 = 303.88 SS (TOTAL) – (sum of column 1) = 303.88 – 267.478 = 36.402 SS ( E2 ) No vacant column = 0 SS ( E1 ) ( 89.8 – 98.2 ) 2 / 32 = 2.205 SS ( ABCD ) ( 113.8 – 74.2 ) 2 / 32 = 49.005 SS ( ACD ) ( 89.3 – 98.7 ) 2 / 32 = 2.761 SS ( ABD ) ( 93.8 – 94.2 ) 2 / 32 = 0.005 SS ( BCD ) ( 74.9 – 113.1 ) 2 / 32 = 45.601 SS ( ABC ) ( 101.8 – 86.2 ) 2 / 32 = 7.605 SS ( CD ) ( 74.9 – 113.1 ) 2 / 32 = 45.601 SS ( BD ) ( 109.7 – 78.3 ) 2 / 32 = 30.811 SS ( AD ) ( 94.5 - 93.5 ) 2 / 32 = 0.031 SS ( D ) ( 95.3 – 92.7 ) 2 / 32 = 0.211 SS ( BC ) ( 109.5 – 78.5 ) 2 / 32 = 30.031 SS ( AC ) ( 99.5 – 88.5 ) 2 / 32 = 3.781 SS ( C ) ( 81 – 107 ) 2 / 32 = 21.125 SS ( AB ) ( 88.2 – 99.8 ) 2 / 32 = 4.205 SS ( B ) ( 108 – 80 ) 2 / 32 = 24.5 SS ( A )

12. INITIAL ANOVA TABLE * POOLING 31 303.88 TOTAL 2.275 16 36.402 e2 0 0 0 e1 0.97 * 2.205 1 2.205 ABCD 20.04 45.601 1 45.601 ABC 21.54 49.005 1 49.005 ACD 1.21 * 2.761 1 2.761 ABD 0.002 * 0.005 1 0.005 BCD 3.34 7.605 1 7.605 CD 20.04 45.601 1 45.601 BD 13.54 30.811 1 30.811 AD 0.014 * 0.031 1 0.031 D 0.09 * 0.211 1 0.211 BC 13.20 30.031 1 30.031 AC 1.66 * 3.781 1 3.781 C 9.28 21.125 1 21.125 AB 1.85 * 4.205 1 4.205 B 10.77 24.5 1 24.5 A F V df SS SOURCE

13. REVISED INITIAL ANOVA TABLE TABLE VALUE : F ( 95%, 1, 19 ) = 4.38 F ( 99%, 1, 19 ) = 8.18 1.93 19 36.649 e1 + e2 31 303.88 TOTAL 2.275 16 36.402 e2 0.082 3 0.247 e1 0.97 * 2.205 1 2.205 ABCD 20.04 45.601 1 45.601 ABC 21.54 49.005 1 49.005 ACD 1.21 * 2.761 1 2.761 ABD 3.34 7.605 1 7.605 CD 20.04 45.601 1 45.601 BD 13.54 30.811 1 30.811 AD 13.20 30.031 1 30.031 AC 1.66 * 3.781 1 3.781 C 9.28 21.125 1 21.125 AB 1.85 * 4.205 1 4.205 B 10.77 24.5 1 24.5 A F V df SS SOURCE

14. FINAL ANOVA TABLE F alpha = 5% = 4.28 * SIGNIFICANT SS ‘ = PURE SUM OF SQUARES = SS – df * V ( Pooled ) C = PERCENT CONTRIBUTION = SS ‘ / SS (TOTAL) 31 303.880 TOTAL 22.00 66.88 2.16 23 49.601 POOLED ERROR 14.30 43.44 21.11 * 49.005 1 49.005 ACD 15.42 46.85 22.69 * 2.761 1 2.761 ABD 1.79 5.45 3.52 7.605 1 7.605 CD 14.30 43.44 21.11 * 45.601 1 45.601 BD 9.43 28.65 14.26 * 30.811 1 30.811 AD 9.17 27.87 13.90 * 30.031 1 30.031 AC 6.24 18.97 9.78 * 21.125 1 21.125 AB 7.35 22.34 11.34 * 24.5 1 24.5 A C SS ’ F MS df SS SOURCE

15. ESTIMATION OF EFFECTS MEAN VALUES OF SIGNIFICANT FACTORS AND INTERACTIONS Contribution of A2C1D1 and A2B2C1 is minimum. Hence A2C1D1 and A2B2C1 is selected A1B1C1 = 5.775 A1B1C2 = 5.375 A1B2C1 = 10.35 A1B2C2 = 5.5 A2B1C1 = 6.1 A2B1C2 = 4.8 A2B2C1 = 2.675 A2B2C2 = 6.45 A1C1D1 = 10.8 A1C1D2 = 5.325 A1C2D1 = 4.725 A1C2D2 = 6.15 A2C1D1 = 2.675 A2C1D2 = 6.075 A2C2D1 = 5.425 A2C2D2 = 5.825 B1D1 = 4.35 B1D2 = 6.675 B2D1 = 7.463 B2D2 = 5.012 A1D1 = 7.763 A1D2 = 5.738 A2D1 = 4.05 A2D2 = 5.95 A1C1 = 8.063 A1C2 = 5.438 A2C1 = 4.375 A2C2 = 5.625 A1B1 = 5.575 A1B2 = 7.925 A2B1 = 5.45 A2B2 = 4.55 A2 = 5.0, B2 = 6.24 A1 = 1.75, B1 = 5.51

16. A2C1D1 = 2.675 ( B not present ) A2B2C1 = 2.675 ( D not present ) Therefore the combination ABCD has to be considered Mean of A2C1D1B1 = 3.75 Mean of A2C1D1B2 = 1.60 Mean of A2B2C1D1 = 1.60 Mean of A2B2C1D2 = 3.70 SELECTED COMBINATION = A2B2C1D1 PREDICTED OPTIMUM RESPONSE (U) : U = T ’ + ( A2B2C1D1 – T ’ ) = 5.88 + ( 1.6 – 5.88 ) = 1.6

17. CONFIDENCE INTERVAL FOR CONFIRMATION EXPERIMENT CI = { F(90%, 1. Dfe) * MSe * (1/neff + 1/r) } 1/2 where, neff = N/[ 1 + Total degrees of freedom considered for prediction ] r = number of replicates in the confirmation experiment F (90%, 1, 23) = 2.94 neff = 32 / ( 1 + 1 ) = 16 1/neff + 1/r = 1/16 + 1/5 = 0.2625 CI = { 2.94 * 2.16 * 0.2625 } 1/2 = 1.29 Upper confidence limit = 1.6 +1.29 = 2.09 Lower confidence limit = 1.6 – 1.29 = -0.33 CI FOR CONFIRMATOIN EXPERIMENT WITH TWO REPLICATES CI = { F(90%, 1, 23) * 2.16 * (1/16+1/2) } 1/2 Upper limit = 1.6 + 1.89 = 3.49 Lower limit = 1.6 – 1.89 = -0.29