ALL 9 CHAPTERS COVERED
SOLUTIONS MANUAL
, Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.
Waveland Press, Inc., Copyright © 2009
CHAPTER 11
11.1
a)
r λ .1 .02
At2 −0 = + 1 − e − ( λ +r ) t2 → A30 = + 1 − e − (.02 +.1)30 =.8784
b
r + λ r + λ t2
2
gb g .1+.02 .1+.02 2 30 b gb g
r .1
A= → A= =.8333
r+λ .1+.02
b)
A = P1 + P2 = 1 +
LM λ1
+
λ 1λ 2 OP −1
+
λ1
N r r 2
Q r
P1
LM
A = 1+
.02 .022
+ 2
OP −1
+
.02
1+
LM
.02 .022
+ 2
OP −1
=.8065+.2 .8065 =.9678
N .1 .1 Q .1 .1 N .1 Q
c) As = 1 − (1 − Ai )2 → As = 1 − (1−.8333)2 =.9722 (note: assumes two repair crews are
available. For a single repair crew see problem 11.9)
1 − Ai
11.2 Ai = 5 .95 =.98979 and MTTRi ≤ xMTBFi =.0103 MTBFi
Ai
prop: MTBF = 10,000Γ (1 +
. ) = 892 → MTTR ≤ 9.2 hrs
avion: MTBF = 3333
. → MTTR ≤ 3.4 hrs
struc: MTBF = 2000Γ (1 +
. ) = 1771 → MTTR ≤ 18.2 hrs
elec: MTBF = 870Γ (1 +
. ) = 773 → MTTR ≤ 7.9 hrs
env: MTBF = 10,000 → MTTR ≤ 10.3 hrs
11.3 From Eq 11.19:
L λ λ λ OP
A = P + P = M1 + + 1 1 2 L λ λ λ OP = LM1 + + () OP
−1
+ M1 + +
λ1 1 1 2
−1 −1
N r r Q r N r Q N .5 Q
1 2 2 2 2
r .5
L () O
−1
.5 MN PQ =.7813+.2 .7813 =.9376
+ 1+ + 2
.5 .5
11.4 From Eq. 11.19:
A = P1 + P2 = 1 +
LM λ1
+
λ1λ 2 OP −1
+
λ1LM1 + λ + λ λ OP = LM1 + 1 + 1 OP LM1 + 1 OP = 3
1 1 2
−1 −1
N r r2 Q r N r r Q N 2 2Q N 2Q 4
2
11.5
r .5 .4
a) A = = = =.80
r + λ .5+.1 .4 + 1
.4 .1
b) A2 = + 1 − e − (.4 +.1)2 =.80+.1264 =.9264
.4+.1 .4+.1 2 2b gbg
11-1
, Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.
Waveland Press, Inc., Copyright © 2009
r λ − ( λ + r )2 .4 .1 −.5(2 )
c) A( 2) = + e = + e =.8736
r+λ r+λ .4+.1 .4+.1
d) Two components in series:
a) As = A2 =.802 =.64 , b) As = A2 2 =.92642 =.8582 , c) As = A( 2)2 =.87362 =.7632
e) Two components in parallel:
a) As = 1 − (1 − A)2 = 1 − (1−.8 )2 =.96 , b) As = 1 − (1 − A2 )2 = 1 − (1−.9264)2 =.9946
c) As = 1 − (1 − A( 2 ))2 = 1 − (1−.8736 )2 =.9840
f) A = P1 + P2 = 1 +
LM .1 .12
+
OP −1
+
LM
.1 .1 .12
1+ + 2
OP −1
=.7619+.25(.7619) =.9524
N .4 .42 Q .4N .4 .4 Q
11.6
1
λ1=2
r=3/2
Condition State
2
3 primary operating (λ1=2)
1
2
primary failed and secondary operating
λ2=4
(λ2=2)
3 both failed (r=3/2)
3 4 1
2 P1 = P3 → P3 = P1 and 2 P1 = 4 P2 → P2 = P1
2 3 2
P1 + P2 + P3 = P1 1 +
FG 1 4
+
IJ
= 1 → P1 =
6 + 3+ 8 LM OP −1
=
6
H 2 3 K 6 N Q 17
A = P1 + P2 =
6 1 6
+ =
FG IJ
9
=.529
17 2 17 17H K
11.7
State Condition
1 1 operating
1 1
2 degraded
4 3 failed
2 3
2
Steady-state equations where λ 12 = 1, λ 13 = 1, λ 23 = 2,r = 4 :
(1) rP3 = λ 12 P1 + λ 13 P1 = ( λ 12 + λ 13 ) P1 (2) λ 12 P1 = λ 23 P2
(3) λ 23 P2 + λ 13 P1 = rP3 (4) P1 + P2 + P3 = 1
λ 1
Solving (2) for P2 results in: P2 = 12 P1 = P1 =.5 P1
λ 23 2
λ P + λ 13 P1 2(.5P1 ) + P1
P3 from (3) is: P3 = 23 2 = =.5P1
r 4
1
Solving (4) for P1 results in: P1 +.5 P1 +.5 P1 = 1 → P1 = =.5
2
Therefore, P2 = P3 =.25 and Availability=P1+P2=.75 .
11-2
, Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.
Waveland Press, Inc., Copyright © 2009
1 − e − λT 1 − e −.00314 T
11.8 A(T ) = =
λ T + t1 + t2 (1 − e − λt ) .00314 T +.25 + (1 − e −.00314 T )
T in days A(T)
1 .797 8 .955 15 .9579
2 .884 10 .9576 16 .9574
3 .916 11 .9582 18 .9560
4 .932 12 .9585 20 .9544
6 .948 14 .9585
11.9
State Unit 1 Unit 2
λ 1 λ 1 operating operating
2 failed operating
r r 3
2 3 operating failed
r 4 failed failed
4 λ
λ
Steady-state equations:
(1) rP2 + rP3 = 2λP1 (2) λ 1 P1 + rP4 = rP2 + λP2 (3) λ 1 P1 = rP3 + λP3
(4) λP2 + λP3 = rP4 (5) P1 + P2 + P3 + P4 = 1
From (3): P3 =
FG λ IJ P
H λ + rK 1
Solving (1) for P2 results in: P2 =
FG IJ
2λP1 − rP3 2λ
= P1 −
λ
H K r r
P
λ +r 1
λP + λP λ λF λ I
Solving (4) for P results in: P = = P + G JP
r H λ + rK
2 3
4 4 2 1
r r
λ L 2λ F λ IJ P OP + λ FG λ IJ P = 2λ 2
= M P −G
rN r H λ + rK Q r H λ + rK r
1 1 1 2
P1
FG1 + 2λ − λ + λ + 2λ IJ P = 1 → 2
L 2λ + 2λ OP
P = M1 +
2 −1
From (5):
H r λ +r λ+r r K 2 1 1
N r r Q 2
2λ2
Therefore, Availability = P1 + P2 + P3 = 1 − P4 = 1 − 2 1 +
2λ 2λ2
+ 2
LM OP −1
r r r N Q
11.10 λ 1 = λ 2 = λ
1 λ
r
λP1 = rP2 → P2 = P1
λ r
2 λ λ2
2r 2rP4 = λP2 → P4 = P2 = P1
2r 2r 2
λ 4
P1 +
λ
P1 +
λ2
P1 = 1 →
L λ λ OP
P = M1 + +
2 −1
r 2r 2
1
N r 2r Q 2
11-3
SOLUTIONS MANUAL
, Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.
Waveland Press, Inc., Copyright © 2009
CHAPTER 11
11.1
a)
r λ .1 .02
At2 −0 = + 1 − e − ( λ +r ) t2 → A30 = + 1 − e − (.02 +.1)30 =.8784
b
r + λ r + λ t2
2
gb g .1+.02 .1+.02 2 30 b gb g
r .1
A= → A= =.8333
r+λ .1+.02
b)
A = P1 + P2 = 1 +
LM λ1
+
λ 1λ 2 OP −1
+
λ1
N r r 2
Q r
P1
LM
A = 1+
.02 .022
+ 2
OP −1
+
.02
1+
LM
.02 .022
+ 2
OP −1
=.8065+.2 .8065 =.9678
N .1 .1 Q .1 .1 N .1 Q
c) As = 1 − (1 − Ai )2 → As = 1 − (1−.8333)2 =.9722 (note: assumes two repair crews are
available. For a single repair crew see problem 11.9)
1 − Ai
11.2 Ai = 5 .95 =.98979 and MTTRi ≤ xMTBFi =.0103 MTBFi
Ai
prop: MTBF = 10,000Γ (1 +
. ) = 892 → MTTR ≤ 9.2 hrs
avion: MTBF = 3333
. → MTTR ≤ 3.4 hrs
struc: MTBF = 2000Γ (1 +
. ) = 1771 → MTTR ≤ 18.2 hrs
elec: MTBF = 870Γ (1 +
. ) = 773 → MTTR ≤ 7.9 hrs
env: MTBF = 10,000 → MTTR ≤ 10.3 hrs
11.3 From Eq 11.19:
L λ λ λ OP
A = P + P = M1 + + 1 1 2 L λ λ λ OP = LM1 + + () OP
−1
+ M1 + +
λ1 1 1 2
−1 −1
N r r Q r N r Q N .5 Q
1 2 2 2 2
r .5
L () O
−1
.5 MN PQ =.7813+.2 .7813 =.9376
+ 1+ + 2
.5 .5
11.4 From Eq. 11.19:
A = P1 + P2 = 1 +
LM λ1
+
λ1λ 2 OP −1
+
λ1LM1 + λ + λ λ OP = LM1 + 1 + 1 OP LM1 + 1 OP = 3
1 1 2
−1 −1
N r r2 Q r N r r Q N 2 2Q N 2Q 4
2
11.5
r .5 .4
a) A = = = =.80
r + λ .5+.1 .4 + 1
.4 .1
b) A2 = + 1 − e − (.4 +.1)2 =.80+.1264 =.9264
.4+.1 .4+.1 2 2b gbg
11-1
, Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.
Waveland Press, Inc., Copyright © 2009
r λ − ( λ + r )2 .4 .1 −.5(2 )
c) A( 2) = + e = + e =.8736
r+λ r+λ .4+.1 .4+.1
d) Two components in series:
a) As = A2 =.802 =.64 , b) As = A2 2 =.92642 =.8582 , c) As = A( 2)2 =.87362 =.7632
e) Two components in parallel:
a) As = 1 − (1 − A)2 = 1 − (1−.8 )2 =.96 , b) As = 1 − (1 − A2 )2 = 1 − (1−.9264)2 =.9946
c) As = 1 − (1 − A( 2 ))2 = 1 − (1−.8736 )2 =.9840
f) A = P1 + P2 = 1 +
LM .1 .12
+
OP −1
+
LM
.1 .1 .12
1+ + 2
OP −1
=.7619+.25(.7619) =.9524
N .4 .42 Q .4N .4 .4 Q
11.6
1
λ1=2
r=3/2
Condition State
2
3 primary operating (λ1=2)
1
2
primary failed and secondary operating
λ2=4
(λ2=2)
3 both failed (r=3/2)
3 4 1
2 P1 = P3 → P3 = P1 and 2 P1 = 4 P2 → P2 = P1
2 3 2
P1 + P2 + P3 = P1 1 +
FG 1 4
+
IJ
= 1 → P1 =
6 + 3+ 8 LM OP −1
=
6
H 2 3 K 6 N Q 17
A = P1 + P2 =
6 1 6
+ =
FG IJ
9
=.529
17 2 17 17H K
11.7
State Condition
1 1 operating
1 1
2 degraded
4 3 failed
2 3
2
Steady-state equations where λ 12 = 1, λ 13 = 1, λ 23 = 2,r = 4 :
(1) rP3 = λ 12 P1 + λ 13 P1 = ( λ 12 + λ 13 ) P1 (2) λ 12 P1 = λ 23 P2
(3) λ 23 P2 + λ 13 P1 = rP3 (4) P1 + P2 + P3 = 1
λ 1
Solving (2) for P2 results in: P2 = 12 P1 = P1 =.5 P1
λ 23 2
λ P + λ 13 P1 2(.5P1 ) + P1
P3 from (3) is: P3 = 23 2 = =.5P1
r 4
1
Solving (4) for P1 results in: P1 +.5 P1 +.5 P1 = 1 → P1 = =.5
2
Therefore, P2 = P3 =.25 and Availability=P1+P2=.75 .
11-2
, Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.
Waveland Press, Inc., Copyright © 2009
1 − e − λT 1 − e −.00314 T
11.8 A(T ) = =
λ T + t1 + t2 (1 − e − λt ) .00314 T +.25 + (1 − e −.00314 T )
T in days A(T)
1 .797 8 .955 15 .9579
2 .884 10 .9576 16 .9574
3 .916 11 .9582 18 .9560
4 .932 12 .9585 20 .9544
6 .948 14 .9585
11.9
State Unit 1 Unit 2
λ 1 λ 1 operating operating
2 failed operating
r r 3
2 3 operating failed
r 4 failed failed
4 λ
λ
Steady-state equations:
(1) rP2 + rP3 = 2λP1 (2) λ 1 P1 + rP4 = rP2 + λP2 (3) λ 1 P1 = rP3 + λP3
(4) λP2 + λP3 = rP4 (5) P1 + P2 + P3 + P4 = 1
From (3): P3 =
FG λ IJ P
H λ + rK 1
Solving (1) for P2 results in: P2 =
FG IJ
2λP1 − rP3 2λ
= P1 −
λ
H K r r
P
λ +r 1
λP + λP λ λF λ I
Solving (4) for P results in: P = = P + G JP
r H λ + rK
2 3
4 4 2 1
r r
λ L 2λ F λ IJ P OP + λ FG λ IJ P = 2λ 2
= M P −G
rN r H λ + rK Q r H λ + rK r
1 1 1 2
P1
FG1 + 2λ − λ + λ + 2λ IJ P = 1 → 2
L 2λ + 2λ OP
P = M1 +
2 −1
From (5):
H r λ +r λ+r r K 2 1 1
N r r Q 2
2λ2
Therefore, Availability = P1 + P2 + P3 = 1 − P4 = 1 − 2 1 +
2λ 2λ2
+ 2
LM OP −1
r r r N Q
11.10 λ 1 = λ 2 = λ
1 λ
r
λP1 = rP2 → P2 = P1
λ r
2 λ λ2
2r 2rP4 = λP2 → P4 = P2 = P1
2r 2r 2
λ 4
P1 +
λ
P1 +
λ2
P1 = 1 →
L λ λ OP
P = M1 + +
2 −1
r 2r 2
1
N r 2r Q 2
11-3
