Chapter 1, Problem 6P

6.3A-3. Multiple Access protocols (3). Consider the figure below, which shows the arrival of 6 messages for transmission at different multiple access wireless nodes at times t=0.1, 1.4, 1.8, 3.2, 3.3, 4.1. Each transmission requires exactly one time unit. 1 t=0.0 2 3 45 t=1.0 t-2.0 t-3.0 6 t=4.0 t-5.0 For the CSMA protocol (without collision detection), indicate which packets are successfully transmitted. You should assume that it takes .2 time units for a signal to propagate from one node to each of the other nodes. You can assume that if a packet experiences a collision or senses the channel busy, then that node will not attempt a retransmission of that packet until sometime after t=5. Hint: consider propagation times carefully here. (Note: You can find more examples of problems similar to this here B.] ☐ U ப 5 - 3 1 4 6 2

Just wanted to know, if you had a scene graph, how do you get multiple components from a specific scene node within a scene graph? Like if I wanted to get a component from wheel from the scene graph, does that require traversing still?   Like if a physics component requires a transform component and these two component are part of the same scene node. How does the physics component knows how to get the scene object's transform it is attached to, this being in a scene graph?

How to develop a C program that receives the message sent by the provided program and displays the name and email included in the message on the screen?Here is the code of the program that sends the message for reference: typedef struct {    long tipo;    struct {        char nome[50];        char email[40];    } dados;} MsgStruct; int main() {    int msg_id, status;    msg_id = msgget(1000, 0600 | IPC_CREAT);    exit_on_error(msg_id, "Creation/Connection");    MsgStruct msg;    msg.tipo = 5;    strcpy(msg.dados.nome, "Pedro Silva");    strcpy(msg.dados.email, "pedro@sapo.pt");    status = msgsnd(msg_id, &msg, sizeof(msg.dados), 0);    exit_on_error(status, "Send");    printf("Message sent!\n");}