Q- A group of four concentrators at located coordinates(0,3),(2,7),(4,1)and (4,4
ID: 3621060 • Letter: Q
Question
Q- A group of four concentrators at located coordinates(0,3),(2,7),(4,1)and (4,4) there are also 16 terminals at(1,1),(1,6),(2,2),(2,3),(2,5)(2,8),(3,0),(3,2),(3,4)(4,3),(4,6)(5,3),(5,5)(5,7)and (6,2) No concentrator can handle more than four termainals.A)Find an assigment that is feasible and whose total cost is less than 30 units,where the cost of a line is simply its length. Multidrop lines are not permited.
b) Use the same the same group of terimnals as above but with only a single concentrator this time, namely the (4,4), find the minimum spanning tree an it cost
c) find an MST OF COST <24 (maximum four terimnals per port)
Explanation / Answer
This question should have 16 terminals,but you given 15 terminals.
The distance from the concentrator (0, 3) to 15 terminals are
Distance from (1,1) to 2.23
Distance from (1,6) to 3.16
Distance from (2,2) to 2.23
Distance from (2,3) to 2
Distance from (2,5) to 2.8
Distance from (2,8) to 5.8
Distance from (3,0) to 4.2
Distance from (3,2) to 3.16
Distance from (3,4) to 3.16
Distance from (4,3) to 4
Distance from (4,6) to 5
Distance from (5,3) to 5
Distance from (5,5) to 5.3
Distance from (5,7) to 6.4
Distance from (6,2) to 6.08
The distance from the concentrator (2, 7) to 15 terminals are
Distance from (1,1) to 6.0
Distance from (1,6) to 1.4
Distance from (2,2) to 5
Distance from (2,3) to 4
Distance from (2,5) to 2
Distance from (2,8) to 1
Distance from (3,0) to 7
Distance from (3,2) to 5.0
Distance from (3,4) to 3.16
Distance from (4,3) to 4.47
Distance from (4,6) to 2.23
Distance from (5,3) to 5
Distance from (5,5) to 3.6
Distance from (5,7) to 3
Distance from (6,2) to 6.4
The distance from the concentrator (4, 1) to 15 terminals are
Distance from (3,0) to 1.4
Distance from (3,2) to 1.4
Distance from (3,4) to 3.16
Distance from (4,3) to 2
Distance from (5,3) to 2.23
Distance from (5,5) to 4.1
Distance from (6,2) to 2.23
The distance from the concentrator (4, 4) to 15 terminals are
Distance from (3,4) to 1
Distance from (5,3) to 1.4
Distance from (5,5) to 1.4
a)
The feasible solution is
2.23+2.23+2+2.8+1.4+1+2.23+3+1.4+1.4+2+2.23+1+1.4+1.4
=27.72
b)
The distance from the concentrator (4, 4) to 15 terminals are
Distance from (1,1) to 4.2
Distance from (1,6) to 3.6
Distance from (2,2) to 2.8
Distance from (2,3) to 2.23
Distance from (2,5) to 2.23
Distance from (2,8) to 4.4
Distance from (3,0) to 4.1
Distance from (3,2) to 2.23
Distance from (3,4) to 1
Distance from (4,3) to 2
Distance from (4,6) to 1.4
Distance from (5,3) to 1.4
Distance from (5,5) to 1.4
Distance from (5,7) to 3.16
Distance from (6,2) to 2.8
The minimum spanning tree = 3.6 +4.4+4.1+2.23+1+2+1.4+1.4+3.16
= 23.29
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