An absorbing Markov chain has the following matrix P as a standard form: A B C D P = A B C D 1 0 0 0 .2 .3 .1 .4 0 .5 .3 .2 0 .1 .6 .3 I 0 R Q Let w k denote the maximum entry in Q k . Note that w 1 = .6 . (A) Find w 2 , w 4 , w 8 , w 16 , and w 32 to three decimal places. (B) Describe Q k when k is large.
An absorbing Markov chain has the following matrix P as a standard form: A B C D P = A B C D 1 0 0 0 .2 .3 .1 .4 0 .5 .3 .2 0 .1 .6 .3 I 0 R Q Let w k denote the maximum entry in Q k . Note that w 1 = .6 . (A) Find w 2 , w 4 , w 8 , w 16 , and w 32 to three decimal places. (B) Describe Q k when k is large.
Solution Summary: The author calculates the value of w_k and the transition matrix of the absorbing chain using TI-83 graphing calculator.
Task number: A1.1, A1.7
Topic: Celestial Navigation, Compass - Magnetic and Gyro
Activ Determine compass error (magnetic and gyro) using azimuth choosing a suitable
celestial body (Sun/ Stars/ Planets/ Moon). Apply variation to find the deviation of the magnetic
compass.
Minimum number of times that activity should be recorded: 6 (2 each phase)
Sample calculation (Azimuth- Planets):
On 06th May 2006 at 22h20m 10s UTC, a vessel in position 48°00'N 050°00'E observed Mars bearing
327° by compass. Find the compass error. If variation was 4.0° East, calculate the deviation.
GHA Mars (06d 22h):
Increment (20m 10s):
089° 55.7'
005° 02.5'
v (0.9):
(+)
00.3'
GHA Mars:
094° 58.5'
Longitude (E):
(+) 050° 00.0' (plus- since longitude is easterly)
LHA Mars:
144° 58.5'
Declination (06d 22h):
d (0.2):
N 024° 18.6'
(-)
00.1'
Declination Mars:
N 024° 18.5'
P=144° 58.5' (If LHA<180°, P=LHA)
A Tan Latitude/ Tan P
A Tan 48° 00' Tan 144° 58.5'
A = 1.584646985 N (A is named opposite to latitude, except when…
Task number: A1.1, A1.7
Topic: Celestial Navigation, Compass - Magnetic and Gyro
Activ Determine compass error (magnetic and gyro) using azimuth choosing a suitable
celestial body (Sun/ Stars/ Planets/ Moon). Apply variation to find the deviation of the magnetic
compass.
Minimum number of times that activity should be recorded: 6 (2 each phase)
Sample calculation (Azimuth- Planets):
On 06th May 2006 at 22h20m 10s UTC, a vessel in position 48°00'N 050°00'E observed Mars bearing
327° by compass. Find the compass error. If variation was 4.0° East, calculate the deviation.
GHA Mars (06d 22h):
Increment (20m 10s):
089° 55.7'
005° 02.5'
v (0.9):
(+)
00.3'
GHA Mars:
094° 58.5'
Longitude (E):
(+) 050° 00.0' (plus- since longitude is easterly)
LHA Mars:
144° 58.5'
Declination (06d 22h):
d (0.2):
N 024° 18.6'
(-)
00.1'
Declination Mars:
N 024° 18.5'
P=144° 58.5' (If LHA<180°, P=LHA)
A Tan Latitude/ Tan P
A Tan 48° 00' Tan 144° 58.5'
A = 1.584646985 N (A is named opposite to latitude, except when…
Activ
Determine compass error using amplitude (Sun).
Minimum number of times that activity should be performed: 3 (1 each phase)
Sample calculation (Amplitude- Sun):
On 07th May 2006 at Sunset, a vessel in position 10°00'N 010°00'W observed the Sun bearing 288°
by compass. Find the compass error.
LMT Sunset:
LIT:
(+) 00d
07d 18h
00h
13m
40m
UTC Sunset:
07d
18h
53m
(added- since longitude is westerly)
Declination (07d 18h):
N 016° 55.5'
d (0.7):
(+)
00.6'
Declination Sun:
N 016° 56.1'
Sin Amplitude = Sin Declination/Cos Latitude
= Sin 016°56.1'/ Cos 10°00'
= 0.295780189
Amplitude=W17.2N
(The prefix of amplitude is named easterly if body is rising,
and westerly if body is setting. The suffix is named same
as declination)
True Bearing=287.2°
Compass Bearing= 288.0°
Compass Error = 0.8° West
Chapter 9 Solutions
Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences (13th Edition)
University Calculus: Early Transcendentals (4th Edition)
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