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What is problem reliability design

varsha222001 profile image varsha222001 ・2 min read

We are to design 3 stage systems with devices D1, D2, D3. The costs are $30, $15, $20 respectively. The cost of the system to be no more than $105. The reliability of each device type is 0.9, 0.8, and 0.5 respectively.

Cost Reliability Max number of devices
D1 30 0.9 2
D2 15 0.8 3
D3 20 0.5 3

Total cost used= 30+15+20=6530+15+20=65
Avalialbe amount = 105-65=40105−65=40
(c-(c1+c2+c3))/c2=1+1(c−(c1+c2+c3))/c2=1+1

Maximum number of additional D1 devices we can purchase= 40/30=140/30=1
Maximum number of additional D2 devices we can purchase= 40/15=240/15=2
Maximum number of additional D3 devices we can purchase= 40/20=240/20=2
Let S0={(1,0)}
S11 ={ (0.9,30)}S11=(0.9,30)

Malfunction of D1 device = 1-0.9=0.1

Relialbility of two D1 devices = 1- 0.1 * 0.1=0.99

S21= { (0.99,60)}

S1={ (0.9,30), (0.99,60)}
S12={(0.72,45), (0.792,75)}S12=(0.72,45),(0.792,75)

Relialbility of two D2 devices=1-(1-0.8)2=0.96

S22={(0.864,60)}

The tuple (0.9504,90) is eliminated as with the remaining amount of $15 we cannot purchase a D3 device. Reliability of three D2 devices=1-(1-0.8)3=0.992

S32= {(0.8928,75)}

The tuple (0.792,75) is eliminated as the tuple (0.864,60) dominates (0.792,75).

S2={(0.72,45), ((0.864,60), (0.8928,75)}
S13={(0.36,65), (0.432,80), (0.4464,95)}S13=(0.36,65),(0.432,80),(0.4464,95)

Reliability of two D3 devices = 1-(1-0.5)2 =0.75 S32={(0.54,85),(0.648,100)}

Rliability of three D3 devices = 1-(1-0.5)3 =0.875 S33={(0.63,105)}

The tuple (0.63,105) is dominated by (0.648,100) and the tuple (0.4464,95) is dominated by (0.54,85)

S3={(0.36,65), (0.432,80),( 0.54,85),(0.648,100)}}
Tracing back through Si’s we can determine that 1 D1 device, 2 D2 devices, and 2 D3 devices give the highest reliability.

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