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Nt1310 Unit 9 Case Study

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Phase 1 is that, BTC of the node j parent node field. After broadcastingADV1 message, if a node receives any of the ADV1 message from any type of other node, then the all node compare its own ID with the parent node ID stored in the received broadcast the message. If its own ID is equal to that the parent node ID in received ADV1 message, the node declares itself as an internal node, If the node does not receive any type of the ADV1 message where its own ID is equal to the parent node ID stored in that broadcast message, after then the node declare itself as an leaf node. The algorithm to add second parent to each of the node,the tree constructed in phase 1 is given in the procedure BTC phase2 in which performs its all task as follows. At the starting of this phase, sink node broadcasts anADV2 message to all its neighbours,the node executes the following steps
1. …show more content…
If the node receives the ADV2 message from the sink node, then it computes the new cost by adding reciprocal of its left over power to the received cost, and sets its two cost fields to new cost and its stores the sink node ID in its both parent node fields.
2. If both the parent node fields of the receiving node are equal, then it stores that new cost value as computed in step 1 in the second cost field and stores the received node ID in the second parent node field.
3. If both the parent node fields of the receiving node are not equal, then it compares the new cost with that cost are stored in the second cost field, and if the new cost is less than the value stored in the second cost field, then it stores the new cost value in the second cost field and stores the received node ID in the second parent node

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