Calculating the Cost and Angle between Two Points

Solution to question 1:
Cables on land, cost: $2700/km
Cables under water, cost: $3600/km
(a)Cable length from I to E
IE¿ √(PE^2+IP^2)
=√ 422 +252
=48.877 km.
Cables laid under water hence total cost of connection:
¿ ( 48.877 km∗$ 3600 )
1 km
=$175958.63
(b) Cable length on land = 42 km
Cost, Cl= 42 km∗$ 2700
1 km
Cl = $ 113400
Cable length thro’ sea = 25 km
Cost, Cs = 25 km∗$ 3600
1 km
Cs = $90,000
Total cost of connection = Cs + Cl = $90000 + 113400
= $203400
(c) length AE = (42-x) km
Length AI = √252 + x2
= ( √625+ x2 ) km
But AI is in the Sea while AE is on land
Connection cost = ((42- x ¿ km∗$ 2700 ¿/1 km+ ¿
1

= $ (113400- 2700 x +3600( √ 625+ x2))
(d) Cost when x=20 km ;
Cost = $ (113400 – ( 2700∗20 ) +3600∗√ ( 625+400 ) ¿
= $ 174656.24
(e) Solved
Table 1: A table showing the various positions of A with the minimum connection cost
Distance
x (km)
Dist EA
(km)
Dist AI
(km)
Cost on land,
EA ($)
Cost thro'
sea, AI ($)
Total cost
($)
AI @
$3800/km
New Total cost
($)
2 40 25.0799 108000 90287.54 198287.54 95303.52 203303.52
4 38 25.3180 102600 91144.72 193744.72 96208.32 198808.32
6 36 25.7099 97200 92555.71 189755.71 97697.70 194897.70
8 34 26.2488 91800 94495.71 186295.71 99745.48 191545.48
10 32 26.9258 86400 96932.97 183332.97 102318.13 188718.13
12 30 27.7308 81000 99831.06 180831.06 105377.23 186377.23
14 28 28.6531 75600 103151.15 178751.15 108881.77 184481.77
16 26 29.6816 70200 106853.92 177053.92 112790.25 182990.25
18 24 30.8058 64800 110901.04 175701.04 117062.21 181862.21
20 22 32.0156 59400 115256.24 174656.24 121659.36 181059.36
22 20 33.3017 54000 119885.95 173885.95 126546.28 180546.28
24 18 34.6554 48600 124759.61 173359.61 131690.70 180290.70
26 16 36.0694 43200 129849.76 173049.76 137063.63 180263.63
28 14 37.5366 37800 135131.94 172931.94 142639.27 180439.27
30 12 39.0512 32400 140584.49 172984.49 148394.74 180794.74
32 10 40.6079 27000 146188.37 173188.37 154309.95 181309.95
34 8 42.2019 21600 151926.82 173526.82 160367.20 181967.20
36 6 43.8292 16200 157785.17 173985.17 166551.01 182751.01
38 4 45.4863 10800 163750.54 174550.54 172847.79 183647.79
40 2 47.1699 5400 169811.66 175211.66 179245.64 184645.64
42 0 48.8774 0 175958.63 175958.63 185734.11 185734.11
Min total cost $172931.9
4
New min
total cost $ 180263.63
(f) The cost of the cheapest route = $ 172931.94. Thus the cheapest route will be 14 km on land
and 37.537 km under water.
(g) Yes it will change the minimum connection route to:
2

16 km on land and 36.07 km under water with a new minimum cost of $180263.63
Solution to question 2.
@ y=27.5, x=0
27.5 = A + D + E
@ y=8.2, x=25
8.2 = (1.389E-11)A + 25B + 0.4226C + 0.90631D + E
@ y=13 , x=10
13 = (4.54E-05)A + 10B + 0.17365C + 0.98481D + E
@ y=15 , x=15
15 = (3.059E-07) + 15B + 0.2588C + 0.9659D + E
@ y=16 , x=16
16 = 0.006737A + 5B + 0.08715C + 0.9962D + E
Forming a matrix of these equations:
| 1
4.4 ×10−11
0.0000454
0.0006738
3.059∗10−7
0
25
10
5
15
0
0.4226
0.1737
0.8715
0.2588
1
0.9063
0.9848
0.9962
0.9659
1
1
1
1
1
|(A ,B, C) = (27.5, 8.2, 13, 16, 15)
Applying Crammer’s rule: (Thomas & Finney, 1988)
A=
|
27.5
8.2
13
16
15
0
25
10
5
15
0
0.4226
0.1737
0.8715
0.2588
1
0.9063
0.9848
0.9962
0.9659
1
1
1
1
1
|
| 1
4.4 ×10−11
0.0000454
0.0006738
3.059∗10−7
0
25
10
5
15
0
0.4226
0.1737
0.8715
0.2588
1
0.9063
0.9848
0.9962
0.9659
1
1
1
1
1
|
=29.67
3