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- Exam Name : MKT-101
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- Update Time: 2019-01-10
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NEW QUESTION: 1
Your route map includes a collaborative step. What does a collaborative route map step allow users in the
step to do?
Please choose the correct answer.
A. Allows the employees to add a user to the route map to collaborate on the form.
B. Allows all user roles to see the form in their inbox at the same time and enter data at almost the same
time.
C. Allows the employee to send the form to another user to collaborate on the form.
D. Allows all user roles to simultaneously view and edit the form at the same time.
Answer: D
NEW QUESTION: 2
利用可能なウェビナーコネクタを選択します
A. デミオ
B. WebinarJam
C. ClickMeeting
D. GoToWebinar
E. ReadyTalk
F. Webex
Answer: D,E,F
NEW QUESTION: 3
What is the 1-day VaR at the 99% confidence interval for a cash flow of $10m due in 6 months time? The risk free interest rate is 5% per annum and its annual volatility is 15%. Assume a 250 day year.
A. 0
B. 1
C. 2
D. 3
Answer: A
Explanation:
Explanation
The $10m cash flow due in 6 months is equivalent to a bond with a present value of 10m/(1.05)
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Exams
=$9,759,000. Essentially, the question requires us to calculate the VaR of a bond.
The VaR of a fixed income instrument is given by Duration x Interest Rate x Volatility of the interest rate x z-factor corresponding to the confidence level.
In this case, since the question requires us to calculate the value "closest to" the correct answer, we can use an estimate for the modified duration of the bond as equal to 0.5 years/(1+r) = 0.5/1.05 = 0.47 years. The VaR would be given by 0.5 * 5% * 15% * 2.326 * sqrt(1/250) * 9,759,000 = $5,384 which is closest to $5,500.
Therefore Choice 'a' is the correct answer. Note that we have to multiply by sqrt(1/250) as the given volatility is annual and the question is asking for daily VaR. All other answers are incorrect.
NEW QUESTION: 4
A. Option A
B. Option C
C. Option D
D. Option B
Answer: D
Explanation:
Explanation
NAT operation is typically transparent to both the internal and external hosts. Typically the internal host is aware of the true IP address and TCP or UDP port of the external host. Typically the NAT device may function as the default gateway for the internal host. However the external host is only aware of the public IP address for the NAT device and the particular port being used to communicate on behalf of a specific internal host.
NAT and TCP/UDP
"Pure NAT", operating on IP alone, may or may not correctly parse protocols that are totally concerned with IP information, such as ICMP, depending on whether the payload is interpreted by a host on the "inside" or
"outside" of translation. As soon as the protocol stack is traversed, even with such basic protocols as TCP and UDP, the protocols will break unless NAT takes action beyond the network layer.
IP packets have a checksum in each packet header, which provides error detection only for the header. IP datagrams may become fragmented and it is necessary for a NAT to reassemble these fragments to allow correct recalculation of higher-level checksums and correct tracking of which packets belong to which connection.
The major transport layer protocols, TCP and UDP, have a checksum that covers all the data they carry, as well as the TCP/UDP header, plus a "pseudo-header" that contains the source and destination IP addresses of the packet carrying the TCP/UDP header. For an originating NAT to pass TCP or UDP successfully, it must recompute the TCP/UDP header checksum based on the translated IP addresses, not the original ones, and put that checksum into the TCP/UDP header of the first packet of the fragmented set of packets. The receiving NAT must recompute the IP checksum on every packet it passes to the destination host, and also recognize and recompute the TCP/UDP header using the retranslated addresses and pseudo-header. This is not a completely solved problem. One solution is for the receiving NAT to reassemble the entire segment and then recompute a checksum calculated across all packets.
The originating host may perform Maximum transmission unit (MTU) path discovery to determine the packet size that can be transmitted without fragmentation, and then set the t fragment (DF) bit in the appropriate packet header field. Of course, this is only a one-way solution, because the responding host can send packets of any size, which may be fragmented before reaching the NAT.