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  • Exam Number/Code : NSE7_SDW-7.0
  • Exam Name : Fortinet NSE 7 - SD-WAN 7.0
  • Questions and Answers : 260 Q&As
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NEW QUESTION: 1
Instructions
This item contains several questions that you must answer. You can view these questions by clicking on the corresponding button to the left. Changing questions can be accomplished by clicking the numbers to the left question. In order to complete the questions, you will need to refer to the topology.
To gain access to the topology, click on the topology button of the screen. When you have finished viewing the topology, you can return to your questions by clicking on the Questions button to the left.
Each of the windows can be minimized by clicking on the [-]. You can also reposition a window by dragging it by the title bar.
Scenario
Refer to the topology. The diagram represents a small network with a single connection to the Internet.

When a packet is sent from Host 1 to Server 1, in how many different frames will the packet be encapsulated as it is sent across the internetwork?
A. 0
B. 1
C. 2
D. 3
E. 4
Answer: A
Explanation:
Explanation/Reference:
Explanation:
We believe the correct answer is 3 because the packet will be encapsulated in one more frame sent between routers R1 and R2. Source MAC is interface S0/0/0 on router R1 and destination is the serialinterface on router R2.

NEW QUESTION: 2
View the Exhibit; examine the structure of the PROMOTIONS table.

Each promotion has a duration of at least seven days.
Your manager has asked you to generate a report, which provides the weekly cost for each promotion done to l date.
Which query would achieve the required result?
A. SELECT promo_name, promo_cost/(promo_end_date-promo_begin_date/7) FROM promotions;
B. SELECT promo_name, promo_cost/((promo_end_date-promo_begin_date)/7) FROM promotions;
C. SELECT promo_name, promo_cost/promo_end_date-promo_begin_date/7 FROM promotions;
D. SELECT promo_name, (promo_cost/promo_end_date-promo_begin_date)/7 FROM promotions;
Answer: B

NEW QUESTION: 3
When evaluating the memory subsystem, the Pages/Sec counter is higher than what is acceptable. What should you do to rectify this situation?
A. Add more memory.
B. Use memory RAID.
C. Use disk drives with a faster transfer rate to keep up with memory.
D. Use an online spare memory.
Answer: A
Explanation:
Pages/Sec - Displays the number of pages read from or written to disk to resolve hard page faults. Hard page faults occur when a process requires code or data that is not in its working set or elsewhere in physical memory, and must be retrieved from disk.
This counter was designed as a primary indicator of the kinds of faults that cause systemwide delays. It is the sum of Memory: Pages Input/sec and Memory: Pages Output/sec. It is counted in numbers of pages, so it can be compared to other counts of pages, such as Memory: Page Faults/sec, without conversion.
The counter includes pages retrieved to satisfy faults in the file system cache (usually requested by applications) and non-cached mapped memory files. This counter displays the difference between the values observed in the last two samples, divided by the duration of the sample interval.
On most servers, if this value is consistently greater than 5, it indicates an excessive amount of paging. Try to identify the application that is creating the paging condition. If this is not normal behavior for the application, adding memory might increase system performance.

 

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