The wvcc case study in this chapter left some design decisions to the reader. in particular, the case study doesn’t mention actual product selections made by the network designers. making these selections is your job. read through the case study and answer the following questions.

based on the design requirements and decisions that have been made, what wireless access points would you recommend for wvcc? do some internet research to find an access point that will meet wvcc’s needs and write two or three paragraphs about the product and why you chose it.

to accomplish user authentication for the private wlan, wvcc’s it department will purchase a dedicated one-ru hardened appliance that will operate as a centralized radius server. do some internet research to find such a server that will meet wvcc’s needs and write two or three paragraphs about the product and why you chose it.

Answerthe amount of hours of driving needed to be documented is 50 hours in total; the amount of hours of nighttime is 10 hours from the 50.

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Home page for what? ? be more specific

The answer is c. 36======analyzing the for-loop, we see that the control variable, k, starts at 0 and increments by after every loop iteration. the condition  tells us that the loop will stop when k is above 7; in this case, when k is 8.0 1 2 3 4 5 6 7 are the values for k that will be used. each number is for each iteration; therefore, the for-loop will have 8 iterations in total.the code inside the loop itself goes like this: 1st iteration (k=0)starting with sum=0, val=1    means that add  to  .since sum is 0 and val is 1, 0 + 1 = 1. the new value of sum is 1.    means to increment  by 1.since val is 1, the new value of val of 2.end of iteration with  k=0; k is increased by one to become  k=12nd iteration (k=1)starting with sum=1, val=2    ======let us establish a pattern. this is moreso logic than programming: we can  rewrite so that we have the elements from the previous iteration. the 1 came from 0 + 1 before. so we can write it as      .and if we did the third iteration, by the end, sum would look like      notice how the number of plus signs is equal to the iteration you are on (which is just k+1), and what is being added is increasing by one each time.  for example, the third iteration (k=2)  ends up with  with 3 plus signs.the second iteration (k=1) ends up with  with two plus signs.the first iteration (k=0) ends up with  with one plus sign.this is a pattern. the nth iteration will have n plus signs.since we established before that there are 8 iterations of this loop in total, the value of sum at k=7 (the 8th and last iteration) will have sum with 8 plus signs.    so the value of sum at the end of the loop is 36.