4 Engineering Example 1

4.1 Reliability in a communication network

Introduction

The reliability of a communication network depends on the reliability of its component parts. The reliability of a component can be represented by a number between 0 and 1 which represents the probability that it will function over a given period of time.

A very simple system with only two components C 1 and C 2 can be configured in series or in parallel. If the components are in series then the system will fail if one component fails (see Figure 4)

Figure 4 :

{Both components 1 and 2 must function for the system to function}

If the components are in parallel then only one component need function properly (see Figure 5) and we have built-in redundancy.

Figure 5 :

{Either component 1 or 2 must function for the system to function}

The reliability of a system with two units in parallel is given by 1 − ( 1 − R 1 ) ( 1 − R 2 ) which is the same as R 1 + R 2 − R 1 R 2 , where R i is the reliability of component C i . The reliability of a system with 3 units in parallel, as in Figure 6, is given by

1 − ( 1 − R 1 ) ( 1 − R 2 ) ( 1 − R 3 )

Figure 6 :

{At least one of the three components must function for the system to function}

Problem in words

  1. Show that the expression for the system reliability for three components in parallel is equal to    R 1 + R 2 + R 3 − R 1 R 2 − R 1 R 3 − R 2 R 3 + R 1 R 2 R 3
  2. Find an expression for the reliability of the system when the reliability of each of the components is the same i.e. R 1 = R 2 = R 3 = R
  3. Find the system reliability when R = 0.75
  4. Find the system reliability when there are two parallel components each with reliability R = 0.75 .

Mathematical statement of the problem

  1. Show that 1 − ( 1 − R 1 ) ( 1 − R 2 ) ( 1 − R 3 ) ≡ R 1 + R 2 + R 3 − R 1 R 2 − R 1 R 3 − R 2 R 3 + R 1 R 2 R 3
  2. Find 1 − ( 1 − R 1 ) ( 1 − R 2 ) ( 1 − R 3 ) in terms of R when R 1 = R 2 = R 3 = R
  3. Find the value of (b) when R = 0.75
  4. Find 1 − ( 1 − R 1 ) ( 1 − R 2 ) when R 1 = R 2 = 0.75 .

Mathematical analysis

  1. 1 − ( 1 − R 1 ) ( 1 − R 2 ) ( 1 − R 3 ) ≡ 1 − ( 1 − R 1 − R 2 + R 1 R 2 ) ( 1 − R 3 )

    = 1 − ( ( 1 − R 1 − R 2 + R 1 R 2 ) × 1 − ( 1 − R 1 − R 2 + R 1 R 2 ) × R 3 )

    = 1 − ( 1 − R 1 − R 2 + R 1 R 2 − ( R 3 − R 1 R 3 − R 2 R 3 + R 1 R 2 R 3 ) )

    = 1 − ( 1 − R 1 − R 2 + R 1 R 2 − R 3 + R 1 R 3 + R 2 R 3 − R 1 R 2 R 3 )

    = 1 − 1 + R 1 + R 2 − R 1 R 2 + R 3 − R 1 R 3 − R 2 R 3 + R 1 R 2 R 3

    = R 1 + R 2 + R 3 − R 1 R 2 − R 1 R 3 − R 2 R 3 + R 1 R 2 R 3

  2. When R 1 = R 2 = R 3 = R the reliability is

    1 − ( 1 − R ) 3 which is equivalent to 3 R − 3 R 2 + R 3

  3. When R 1 = R 2 = R 3 = 0.75 we get

    1 − ( 1 − 0.75 ) 3 = 1 − 0.2 5 3 = 1 − 0.015625 = 0.984375

  4. 1 − ( 0.25 ) 2 = 0.9375

Interpretation

The mathematical analysis confirms the expectation that the more components there are in parallel then the more reliable the system becomes (1 component: 0.75;  2 components: 0.9375;  3 components: 0.984375). With three components in parallel, as in part (c), although each individual component is relatively unreliable ( R = 0.75 implies a one in four chance of failure of an individual component) the system as a whole has an over 98 % probability of functioning (under 1 in 50 chance of failure).

Exercises
  1. Remove the brackets from each of the following expressions:
    (a) 2 ( m n ) , (b) 2 ( m + n ) , (c) a ( m n ) , (d) a ( m + n ) , (e) a ( m − n ) ,
    (f) ( a m ) n , (g) ( a + m ) n , (h) ( a − m ) n , (i) 5 ( p q ) , (j) 5 ( p + q ) ,
    (k) 5 ( p − q ) , (l) 7 ( x y ) , (m) 7 ( x + y ) , (n) 7 ( x − y ) , (o) 8 ( 2 p + q ) ,
    (p) 8 ( 2 p q ) , (q) 8 ( 2 p − q ) , (r) 5 ( p − 3 q ) , (s) 5 ( p + 3 q ) (t) 5 ( 3 p q ) .
  2. Remove the brackets from each of the following expressions:
    1. 4 ( a + b ) ,
    2. 2 ( m − n ) ,
    3. 9 ( x − y ) ,
  3. Remove the brackets from each of the following expressions and simplify where possible:
    1. ( 2 + a ) ( 3 + b ) ,
    2. ( x + 1 ) ( x + 2 ) ,
    3. ( x + 3 ) ( x + 3 ) ,
    4. ( x + 5 ) ( x − 3 )
  4. Remove the brackets from each of the following expressions:
    (a) ( 7 + x ) ( 2 + x ) , (b) ( 9 + x ) ( 2 + x ) , (c) ( x + 9 ) ( x − 2 ) , (d) ( x + 11 ) ( x − 7 ) ,
    (e) ( x + 2 ) x , (f) ( 3 x + 1 ) x , (g) ( 3 x + 1 ) ( x + 1 ) , (h) ( 3 x + 1 ) ( 2 x + 1 ) ,
    (i) ( 3 x + 5 ) ( 2 x + 7 ) , (j) ( 3 x + 5 ) ( 2 x − 1 ) , (k) ( 5 − 3 x ) ( x + 1 ) (l) ( 2 − x ) ( 1 − x ) .
  5. Remove the brackets from ( s + 1 ) ( s + 5 ) ( s − 3 ) .
    1. 2 m n ,
    2. 2 m + 2 n ,
    3. a m n ,
    4. a m + a n ,
    5. a m − a n ,
    6. a m n ,
    7. a n + m n ,
    8. a n − m n ,
    9. 5 p q ,
    10. 5 p + 5 q ,
    11. 5 p − 5 q ,
    12. 7 x y ,
    13. 7 x + 7 y ,
    14. 7 x − 7 y ,
    15. 16 p + 8 q ,
    16. 16 p q ,
    17. 16 p − 8 q ,
    18. 5 p − 15 q ,
    19. 5 p + 15 q ,
    20. 15 p q
    1. 4 a + 4 b ,
    2. 2 m − 2 n ,
    3. 9 x − 9 y
    1. 6 + 3 a + 2 b + a b ,
    2. x 2 + 3 x + 2 ,
    3. x 2 + 6 x + 9 ,
    4. x 2 + 2 x − 15
  4. On removing brackets we obtain:
    (a) 14 + 9 x + x 2 , (b) 18 + 11 x + x 2 , (c) x 2 + 7 x − 18 , (d) x 2 + 4 x − 77
    (e) x 2 + 2 x , (f) 3 x 2 + x , (g) 3 x 2 + 4 x + 1 (h) 6 x 2 + 5 x + 1
    (i) 6 x 2 + 31 x + 35 , (j) 6 x 2 + 7 x − 5 , (k) − 3 x 2 + 2 x + 5 , (l) x 2 − 3 x + 2
  5. s 3 + 3 s 2 − 13 s − 15