340 Assignment 3 Solutions

Assignment 3 Solutions 1) Transition Matrix: 1 2 3 ࠵? = 1 2 3 [ 0.5 0.3 0.2 0.2 0.6 0.2 0.3 0.2 0.5 ] a) We need to calculate ࠵? 3 . ࠵? 3 = [ 0.3370 0.3850 0.2780 0.3100 0.4120 0.2780 0.3370 0.3580 0.3050 ] ࠵?(࠵? 3 = 2|࠵? 0 = 2) = ࠵? 22 (3) = 0.4120 b) We need to calculate ࠵? ࠵? , ࠵? = 1,2, . . . ,7 . ࠵? 31 (7) = ∑ ࠵? 31 (࠵?) 7 ࠵?=1 = 0.3 + 0.34 + 0.337 + 0.3316 + 0.3286 + 0.3273 + 0.3268 = 2.2913 days c) We need to calculate steady state probability ࠵? . ࠵? 100 = [ 0.3265 0.3878 0.2857 0.3265 0.3878 0.2857 0.3265 0.3878 0.2857 ] → ࠵? = [0.3265 0.3878 0.2857] ࠵? 1 = 0.3265 → 32.65% of days are good fishing days. d) ࠵? 31 = 0.3 + 0.5(1 + ࠵? 31 ) + 0.2(1 + ࠵? 21 ) → ࠵? 31 = 1 + 0.5࠵? 31 + 0.2࠵? 21 ࠵? 21 = 0.2 + 0.6(1 + ࠵? 21 ) + 0.2(1 + ࠵? 31 ) → ࠵? 21 = 1 + 0.6࠵? 21 + 0.2࠵? 31 → 0.4࠵? 21 = 1 + 0.2࠵? 31 → ࠵? 21 = 2.5 + 0.5࠵? 31 ࠵? 31 = 1 + 0.5࠵? 31 + 0.2࠵? 21 = 1 + 0.5࠵? 31 + 0.2(2.5 + 0.5࠵? 31 ) = 1.5 + 0.6࠵? 31 → 0.4࠵? 31 = 1.5 → ࠵? 31 = 3.75 days e) The question asks for mean sojourn time of state 3. ࠵?(࠵? 3 ) = 1 1−࠵? 33 = 1 1−0.5 = 1 0.5 = 2 days Medium to medium over 3 days raise matrix and add all fifty good steady store and good first purge fun bud good 2 0

2) Transition Matrix: 1 2 3 4 ࠵? = 1 2 3 4 [ 0 0.5 0.25 0.25 0.8 0 0.1 0.1 0.8 0.1 0 0.1 0.8 0.1 0.1 0 ] a) ࠵?(࠵? 2 = 1, ࠵? 1 = 2|࠵? 0 = 1) = ࠵?(࠵? 2 = 1|࠵? 1 = 2)࠵?(࠵? 1 = 2|࠵? 0 = 1) = ࠵? 12 ࠵? 21 = 0.5 × 0.8 = 0.4 b) We need to calculate ࠵? ࠵? , ࠵? = 1,2, . . . ,5 . ࠵? 14 (5) = ∑ ࠵? 14 (࠵?) 5 ࠵?=1 = 0.25 + 0.075 + 0.2125 + 0.1028 + 0.1905 = 0.8308 times c) We need to calculate steady state probability ࠵? . ࠵? 100 = [ 0.4444 0.2525 0.1515 0.1515 0.4444 0.2525 0.1515 0.1515 0.4444 0.2525 0.1515 0.1515 0.4444 0.2525 0.1515 0.1515 ] → ࠵? = [0.4444 0.2525 0.1515 0.1515] ࠵? 1 = 0.4444 d) ࠵? 21 = 0.8 + 0.1(1 + ࠵? 31 ) + 0.1(1 + ࠵? 41 ) = 1 + 0.1࠵? 31 + 0.1࠵? 41 ࠵? 31 = 0.8 + 0.1(1 + ࠵? 21 ) + 0.1(1 + ࠵? 41 ) = 1 + 0.1࠵? 21 + 0.1࠵? 41 ࠵? 41 = 0.8 + 0.1(1 + ࠵? 21 0 + 0.1(1 + ࠵? 31 ) = 1 + 0.1࠵? 21 + 0.1࠵? 31 ࠵? 31 = 1 + 0.1࠵? 21 + 0.1࠵? 41 = 1 + 0.1࠵? 21 + 0.1(1 + 0.1࠵? 21 + 0.1࠵? 31 ) = 1.1 + 0.11࠵? 21 + 0.01࠵? 31 → 0.99࠵? 31 = 1.1 + 0.11࠵? 21 → ࠵? 31 = 1.1 + 0.11࠵? 21 0.99 ࠵? 41 = 1 + 0.1࠵? 21 + 0.1࠵? 31 = 1 + 0.1࠵? 21 + 0.1 ( 1.1 + 0.11࠵? 21 0.99 ) = 1.1 + 0.11࠵? 21 0.99 ࠵? 21 = 1 + 0.1࠵? 31 + 0.1࠵? 41 = 1 + 0.1 ( 1.1 + 0.11࠵? 21 0.99 ) + 0.1 ( 1.1 + 0.11࠵? 21 0.99 ) = 1.21 + 0.022࠵? 21 0.99 → 0.99࠵? 21 = 1.21 + 0.022࠵? 21 → 0.968࠵? 21 = 1.21 → ࠵? 21 = 1.21 0.968 → ࠵? 21 = 1.25 steps Alternative solution: ࠵? = [ 1 −0.1 −0.1 −0.1 1 −0.1 −0.1 −0.1 1 ], ࠵? [ ࠵? 21 ࠵? 31 ࠵? 41 ] = [ 1 1 1 ] → [ ࠵? 21 ࠵? 31 ࠵? 41 ] = ࠵? −1 [ 1 1 1 ] [ ࠵? 21 ࠵? 31 ࠵? 41 ] = [ 1.0227 0.1136 0.1136 0.1136 1.0227 0.1136 0.1136 0.1136 1.0227 ] [ 1 1 1 ] = [ 1.25 1.25 1.25 ] → ࠵? 21 = 1.25 steps raise matrix 5 times and odd all i 74 steady state and at O first purge for flour 2 to 1

4) Transition Matrix: 1 2 3 4 ࠵? = 1 2 3 4 [ 0.1 0.3 0.4 0.2 0.15 0.3 0.5 0.05 0.265 0.588 0.147 0 0.5 0.333 0.167 0 ] a) We need to calculate steady state probability ࠵? . ࠵? 100 = [ 0.2 0.4 0.34 0.06 0.2 0.4 0.34 0.06 0.2 0.4 0.34 0.06 0.2 0.4 0.34 0.06 ] → ࠵? = [0.2 0.4 0.34 0.06] ࠵? 4 = 0.06 → 1000 × 0.06 = 60 → It is expected that 60 wine bottles would be excellent. b) We need to calculate ࠵? ࠵? , ࠵? = 1,2, . . . ,5 . ࠵? 33 (5) = ∑ ࠵? 33 (࠵?) 5 ࠵?=1 = 0.1471 + 0.421626 + 0.308376 + 0.353262 + 0.334500 = 1.564864 years ࠵? 34 (5) = ∑ ࠵? 34 (࠵?) 5 ࠵?=1 = 0 + 0.082353 + 0.047846 + 0.065115 + 0.057753 = 0.253067 years ࠵? 33 (5) + ࠵? 34 (5) = 1.564864 + 0.253067 = 1.817931 ≈ 1.8179 years c) ࠵? 34 = 0.265(1 + ࠵? 14 ) + 0.588(1 + ࠵? 24 ) + 0.147(1 + ࠵? 34 ) ࠵? 34 = 1 + 0.265࠵? 14 + 0.588࠵? 24 + 0.147࠵? 34 −0.265࠵? 14 − 0.588࠵? 24 + 0.853࠵? 34 = 1 ࠵? 14 = 0.2 + 0.1(1 + ࠵? 14 ) + 0.3(1 + ࠵? 24 ) + 0.4(1 + ࠵? 34 ) ࠵? 14 = 1 + 0.1࠵? 14 + 0.3࠵? 24 + 0.4࠵? 34 0.9࠵? 14 − 0.3࠵? 24 − 0.4࠵? 34 = 1 ࠵? 24 = 0.05 + 0.15(1 + ࠵? 14 ) + 0.3(1 + ࠵? 24 ) + 0.5(1 + ࠵? 34 ) ࠵? 24 = 1 + 0.15࠵? 14 + 0.3࠵? 24 + 0.5࠵? 34 −0.15࠵? 14 + 0.7࠵? 24 − 0.5࠵? 34 = 1 ࠵? = [ −0.265 −0.588 0.853 0.9 −0.3 −0.4 −0.15 0.7 −0.5 ] , ࠵? [ ࠵? 14 ࠵? 24 ࠵? 34 ] = [ 1 1 1 ] → [ ࠵? 14 ࠵? 24 ࠵? 34 ] = ࠵? −1 [ 1 1 1 ] [ ࠵? 14 ࠵? 24 ࠵? 34 ] = [ 5.0484 3.5586 5.7658 5.9877 3.0578 7.7687 6.8682 3.2134 7.1465 ] [ 1 1 1 ] = [ 14.377 16.819 17.233 ] → ࠵? 34 = 17.233 years steady shik 41 1000 raise to power 1 5 add all prob if 3 73 and 3 7 Art purge frm J Y furn WVU

d) ࠵?(࠵? 2 = 4|࠵? 1 = 1, ࠵? 0 = 3) = ࠵? 14 = 0.2 e) We can use the results of part (c) ࠵? 44 = 0.5(1 + ࠵? 14 ) + 0.333(1 + ࠵? 24 ) + 0.167(1 + ࠵? 34 ) = 1 + 0.5 × 14.377 + 0.333 × 16.819 + 0.167 × 17.233 = 16.667 years Alternative solution: ࠵? 44 = 1 ࠵? 4 = 1 0.06 = 16.667 years bad 7 exellent 1 s 4 Titellett extent 4 Y for n Nuu

The land is 12 million acres. 12 × 0.5170 = 6.2045 million acres are expected to be planted with corn. 12 × 0.3154 = 3.7848 million acres are expected to be planted with soybeans. 12 × 0.1177 = 1.4120 million acres are expected to be planted with other crops. 12 × 0.0498 = 0.5987 million acres are expected to be fallow. b) We need to calculate ࠵? ࠵? , ࠵? = 1,2, . . . ,5 . ࠵? 21 (5) = ∑ ࠵? 21 (࠵?) 5 ࠵?=1 = 0.3704 + 0.480788 + 0.507726 + 0.514669 + 0.516437 = 2.39 years c) Initial P: ࠵? 0 = [ 6 12 4 12 1.5 12 0.5 12 ] 12࠵? 0 ࠵? 2 = 12 [ 6 12 4 12 1.5 12 0.5 12 ] [ 0.5484 0.2883 0.1163 0.04701 0.4808 0.3487 0.1176 0.05296 0.4811 0.3421 0.1245 0.05225 0.5058 0.3226 0.1169 0.05475 ] 12࠵? 0 ࠵? 2 = 12[0.515689 0.316595 0.117745 0.049972] 12࠵? 0 ࠵? 2 = [6.1883 3.7991 1.4129 0.5997] Corn: 6.1883 acres Soybeans: 3.7991 acres Other crops: 1.4129 acres Fallow: 0.5997 acres d) Initial P: ࠵? 0 = [ 6 12 4 12 1.5 12 0.5 12 ] ࠵? 0 ࠵? 3 = [ 6 12 4 12 1.5 12 0.5 12 ] [ 0.5251 0.3083 0.1174 0.0492 0.5077 0.33237 0.1178 0.0508 0.5078 0.3232 0.1184 0.0506 0.5146 0.3178 0.1176 0.0499 ] ࠵? 0 ࠵? 3 = [0.516693 0.315705 0.117678 0.049924] The probability that will be fallow is 0.049924. e) ࠵? 44 = 1 ࠵? 4 = 1 0.0498 = 20.044 years