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NEW QUESTION: 1
Refer to the exhibit.

The junior network support staff provided the diagram as a recommended configuration for the first phase of a four-phase network expansion project. The entire network expansion will have over 1000 users on 14 network segments and has been allocated this IP address space.
192.168.1.1 through 192.168.5.255
192.168.100.1 through 192.168.100.255
What are three problems with this design? (Choose three.)
A. The broadcast domain in AREA 1 is too large for IP to function.
B. The AREA 1 IP address space is inadequate for the number of users.
C. The router-to-router connection is wasting address space.
D. The network address space that is provided requires a single network-wide mask.
E. AREA 2 could use a mask of /25 to conserve IP address space.
F. The AREA 3 IP address space is inadequate for the number of users.
Answer: B,C,E
Explanation:
The given IP addresses of areas 1 and 3 along with network masks of 24 cannot accommodate 500 users so are inadequate, while the area 2 is having over capacity so its network mask can be reduced to 25 to accommodate the only 60 users it has.

NEW QUESTION: 2
質問をドラッグアンドドロップ
IPv4ネットワークサブネットを左側から右側の選択して配置できる正しい使用可能なホスト範囲にドラッグアンドドロップします。

Answer:
Explanation:

Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet
172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or 172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.

NEW QUESTION: 3
If EV be the expected value of a firm's assets in a year, and DP be the 'default point' per the KMV approach to credit risk, and be the standard deviation of future asset returns, then the distance-to-default is given by:
A)

B)

C)

D)

A. Option B
B. Option D
C. Option C
D. Option A
Answer: B
Explanation:
Explanation
The distance to default is the number of standard deviations that expected asset values are away from the default point. The expression in Choice 'd' represents distance to default. Choice 'd' is the correct answer. The other choices are incorrect.