Unit 3 Emotional Intelligence and Spiritual Intelligence.pdf
Realibity design
1. Reliability Design Problem
In reliability design, the problem is to design a system that is composed
of several devices connected in series. If we imagine that r1 is
the reliability of the device. ... Then they make use of such devices at
each stage, that result is increase in reliability at each stage.
2. RELIABILITY DESIGN
• Solve a problem with a multiplicative optimization function
• Several devices are connected in series
• ri be the reliability of device Di
• Reliability of the entire system
• Duplicate : multiple copies of the same device type are connected in parallel
use switching circuits
3. RELIABILITY DESIGN
n devices Di, 1<=i<=n, connected in series
Multiple devices connected in parallel in each stage
4. Reliability Design
Devices D1 D2 D3 D4
Cost C1 C2 C3 C4
Reliability R1 R2 R3 R4
• Total Reliability is
• R= (0.6)*(0.7)*(0.8)*(0.9)
• R=0.3024
• So we need to increase the
reliability.
• But how ??
• By Increase the no of Devices
instances.
Devices D1 D2 D3 D4
Cost C1 C2 C3 C4
Reliability 0.6 0.7 0.8 0.9
5. Reliability Design
Devices D1 D2 D3 D4
Cost C1 C2 C3 C4
Reliability R1 R2 R3 R4
Suppose r1 is the reliability =0.9
Of success.
Then 1-r1 is the failure probability.
1-0.9= 0.1
Now suppose three instances of device D1
Probability of all three devices failed is = ( 1-
r1)3
= ( 0.1)3
= 0.001
So reliability of Stage is = 1-(1-r)3
=.999
So reliability is increased.
Devices D1 D2 D3 D4
Cost C1 C2 C3 C4
Reliability 0.6 0.7 0.8 0.9
6. Reliability Design Problem
C =105 given Amount
Solution:
𝐶𝑖 = 𝐶1 + 𝐶2 + 𝐶3
= 30+15+20=65
Remaining Amount = C- 𝑐𝑖
=105-65
=40
Di Ci Ri Ui
D1 30 0.9
D2 15 0.8
D3 20 0.5
7. Reliability Design Problem
How Can Calculate Upper Bound
Ui =
𝐶− 𝑐𝑖
𝑐𝑖
+1
U1=((105-65)/30)+1=2
U2=((105-65)/15)+1=3
U3=((105-65)/20)+1=3
Di Ci Ri Ui
D1 30 0.9
D2 15 0.8
D3 20 0.5
8. Reliability Design Solution
Need to prepare ordered Pair of
Reliability & Cost.
(R , C)
S0
0={1,0}
Now Consider D1 one instance
S1
1={0.9,30}
Consider D1 second instance
S1
2={1-(1-(0.9))2,60}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
Di Ci Ri Ui
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3
10. Reliability Design Solution
S1
1={0.9,30}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
S2
1={0.72,45), (0.792,75)}
Now consider second Copy of D2
Reliability when we take two instance of D2 is = 1-(1-r)2
= 1-(1-0.8)2 = 1-(0.2)2=1-0.04= 0.96
S2
2={0.9*0.96,30+30), (0.99*0.96,60+30)}
S2
2={0.864,60), (0.9504,90)}
Suppose we take (0.9504,90) then we can not buy 3 device
because it restrict to buy D3
S2
2={0.864,60)}
Di Ci Ri Ui
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3
11. Reliability Design Solution
S1
1={0.9,30}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
S2
1={(0.72,45), (0.792,75)}
S2
2={(0.864,60)}
Now consider second Copy of D2
Reliability when we take three instance of D2 is = 1-(1-r)3
= 1-(1-0.8)3 = 1-(0.2)3=1-0.008= 0.992
S2
3={0.72,45), (0.792,75)}
S2
3={0.9*0.992,30+45), (0.99*0.992,60+45)}
S2
3={(0.8982,75), (0.9820,105)}
Suppose we take (0.9820,105) then we can not buy 3 device because it restrict to buy D3
S2={(0.72,45), (0.792,75), (0.864,60), (0.8982,75)}
By Apply Dominance rule
Can not increase reliability with decrease the cost.
S2={(0.72,45),(0.864,60), (0.8982,75)}
Di Ci Ri Ui
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3
13. Reliability Design Solution
S1
1={0.9,30}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
S2
1={(0.72,45), (0.792,75)}
S2
2={(0.864,60)}
S2
3={(0.8982,75)}
S2={(0.72,45),(0.864,60), (0.8982,75)}
S3
1={(0.36,65),(0.432,80), (0.4464,95)}
Now consider second Copy of D3
Reliability when we take two instance of D3 is = 1-(1-r)2
= 1-(1-0.5)2 = 1-(0.5)2=1-0.25= 0.75
S3
2={(0.72*0.75,45+ 40),(0.864*0.75,60+40), (0.8982*0.75,75+40)}
S3
2={(0.54,85),(0.648,100), (0.6714,115)}
(0.6714,115) it exceed the given cost.
Di Ci Ri Ui
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3
14. Reliability Design Solution
S1
1={0.9,30}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
S2
1={(0.72,45), (0.792,75)}
S2
2={(0.864,60)}
S2
3={(0.8982,75)}
S2={(0.72,45),(0.864,60), (0.8982,75)}
S3
1={(0.36,60),(0.432,80), (0.4464,95)}
S3
2={(0.54,85),(0.648,100), (0.6714,115)}
Now consider second Copy of D3
Reliability when we take third instance of D3 is = 1-(1-r)3
= 1-(1-0.5)3 = 1-(0.5)3=1-0.125= 0.875
S3
3={(0.72* 0.875 ,45+60),(0.864* 0.875,60+60), (0.8982* 0.875,75+60)}
S3
3={(0.63,105),(?,120), (?,135)}
((?,120), (?,135) it exceed the given cost.
Di Ci Ri Ui
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3
15. Reliability Design Solution
S1
1={0.9,30}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
S2
1={(0.72,45), (0.792,75)}
S2
2={(0.864,60)}
S2
3={(0.8982,75)}
S2={(0.72,45),(0.864,60), (0.8982,75)}
S3
1={(0.36,60),(0.432,80), (0.4464,95)}
S3
2={(0.54,85),(0.648,100), (0.6714,115)}
S3
3={(0.63,105),(?,120), (?,135)}
S3={(0.36,60),(0.432,80), (0.4464,95), (0.54,85), (0.63,105),(0.648,100)}
Reliability increase so cost should be increase that why eliminate the(0.4464,95), (0.63,105)
S3={(0.36,60),(0.432,80), (0.54,85), (0.648,100)}
Di Ci Ri Ui
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3
16. Reliability Design Solution
S1
1={0.9,30}
S1
2={0.99,60}
S1={(0.9,30),(0.99,60)}
S2
1={(0.72,45), (0.792,75)}
S2
2={(0.864,60)}
S2
3={(0.8982,75)}
S2={(0.72,45),(0.864,60), (0.8982,75)}
S3
1={(0.36,60),(0.432,80), (0.4464,95)}
S3
2={(0.54,85),(0.648,100), (0.6714,115)}
S3
3={(0.63,105),(?,120), (?,135)}
S3={(0.36,60),(0.432,80), (0.54,85), (0.648,100)}
After getting S3 The Maximum Reliability is = 0.648
Maximum Cost is =100
How many device instances can be used so the answer is S3
2, S2
2, S1
1
Di Ci Ri Ui
D1 30(1) 0.9 2
D2 15(2) 0.8 3
D3 20(2) 0.5 3
