# k-out-of-m network

In this case, the network/system is composed of m active units, at least k units out of m active units must function normally for the successful system operation. The block diagram of a k-out-of-m unit network/system is shown in Fig. 3.4, and each block in the diagram represents a unit. It is to be noted that the series and parallel networks are special cases of this network for k = m and k = 1, respectively.

By using the binomial distribution, for identical and independent units, we write the following expression for reliability of k-out-of-m unit network shown in Fig. 3.4:

W1 , _{x }*f m *

(3.33)

*R _{k/m} = 2^[iRfl-Ry"-*

*i=k ^{V 7}*

*Figure 3.4* Block diagram of a k-out-of-m unit network/system.

where

; 1 '

J (»!-/)!/!

(3.34)

*R _{k}/„,* is the k-out-of-m network/system reliability. R is the unit reliability.

For constant failure rates of the identical units, by using Equations (3.11) and (3.33), we obtain

nt

*i=k ' ‘*

(3.35)

where

Ri/_{ra}(t) is the k-out-of-m network/system reliability at time t.

X is the unit constant failure rate.

By substituting Equation (3.35) into Equation (3.13), we get

*~ m*

*MTTF _{k/m} =* J X('T‘

^{(1}“ o L <-*

lyi

(3.36)

where

*MTTF _{k}/,„* is the k-out-of-m network/system mean time to failure.

Example 3.7

Assume that an engineering system has three active, independent, and identical units in parallel. At least two units must operate normally for the successful operation of the engineering system. Calculate the engineering system mean time to failure if the unit constant failure rate is 0.0005 failures per hour.

By substituting the specified data values into Equation (3.36), we obtain

*MTTF2/3*

- 1 P
^{1} - (0.0005) L 2
^{+}3

= 1666.66 hours

Thus, the engineering system mean time to failure is 1666.66 hours.

# Standby system

This is another reliability network/system in which only one unit operates and k units are kept in their standby mode. The system is composed of (k + 1) units, and as soon as the operating unit fails, the switching mechanism detects the failure and turns on one of the standby units. The system fails when all its standby units fail.

The block diagram of a standby system with one operating and k standby units is shown in Fig. 3.5. Each block in the diagram represents a unit. By using Fig. 3.5, for identical and independent units, perfect switching mechanism and standby units, and time-dependent unit failure rate, we get the following expression for the standby system reliability [1, 6]:

M0 =---^{k}-----jy--------^{k} (3.37)

where

R_{ss} (t) is the standby system reliability at time t.

X(t) is the unit time-dependent failure rate or hazard rate, k is the number of standby units.

For constant unit failure rate (i.e., X(t) = X), Equation (3.37) becomes

*'¿W'e-'-'*

*R„* (f) = 4=o--------- (3.38)

*i!*

*Figure 3.5* Block diagram of a standby system with one operating and k standby units.

where

X is the unit constant failure rate.

By inserting Equation (3.38) in Equation (3.13), we obtain

(3.39)

where

*MTTF,,* is the standby system mean time to failure.

Example 3.8

Assume that an engineering system is composed of a standby system having three independent and identical units: one operating and other two on standby. The unit constant failure rate is 0.0008 failures per hour.

Calculate the standby system mean time to failure if the switching mechanism is perfect and the standby units remain as good as new in their standby modes.

By inserting the given data values into Equation (3.39), we get

*MTTF _{SS} =*

(2 + 1) (0.0008)

= 3750 hours

Thus, the standby system mean time to failure is 3750 hours.

# Bridge netzvork

Sometimes units/parts in engineering systems may form a bridge network, as shown in Fig. 3.6. Each block in Fig. 3.6 diagram represents a unit/part, and all units/parts are labeled with numerals.

For independent units/parts, the bridge network shown in Fig. 3.6, reliability is expressed by [7]

*Ri„, = 2.R1R2R3R4R5 + R1R3R5* + R2R3R4 + *R2R5* + R1R4

- - R
_{1}R_{2}R_{3}R_{4}- R1R_{2}R_{3}R_{5}- R2R3R4R5 - R1R2R4R5^{(340)} - —
*R3R4R5R1*

*Figure 3.6* Five nonidentical units bridge network.

where

*Ri„,* is the bridge network reliability.

*R,* is the unit i reliability, for i = 1,2,3,4,5.

For identical units, Equation (3.40) becomes

R(„, = R^{5}-5R^{4}+2R^{3} + 2R^{2}

(3.41)

where

R is the unit reliability.

For constant failure rates of all five units, and using Equations (3.11) and (3.41), we get

R_{(}„, (t) = 2c“^{5W} - Sc“^{4}*^{2} + 23W + 2e-^{2W} (3.42)

where

R_{(>}„ (f) is bridge network reliability at time t.

X is the unit constant failure rate.

By substituting Equation (3.42) into Equation (3.13), we get

*MTTF _{b}„ =* j(25X/ - Se

^{-4}^ + 2c“

^{3}'

^{J}+

*2e-*o 49

^{ni})dt(3.43)

"60X

where

*MTTF _{b}„* is the bridge network mean time to failure.

**Example 3.9**

Assume that an engineering system with five independent and identical units form a bridge network. Calculate the bridge network's reliability for a 200-hour mission and mean time to failure, if each unit's constant failure rate is 0.0004 failures per hour.

By substituting the specified data values into Equation (3.42), we get

R_{(}„, (200) = 2e‘^{5}(^{000M}>^{(200}) - Se^^{00004}«^{200}’ + 2_{i}-^{3}(°^{>(20()> + 2e'2(0X200)}

= 0.9874

Similarly, by substituting the given data values into Equation (3.43), we obtain

49

*MTTFt.,, =* -----r

60(0.0004)

= 2041.66 hours

Thus, the bridge network's reliability and mean time to failure are 0.9874 and 2041.66 hours, respectively.