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Notes on Storage Server Operation
We recommend that you dedicate the Storage Server computer exclusively to running this software.
Be sure to avoid the following on the Storage Server computer:
• Running Web server software on the computer
You cannot operate the Storage Server on a computer where Web server software such as Windows standard IIS
(Internet Information Server) or Apache is already installed.
The Storage Server uses HTTP protocol to communicate with cameras. Therefore, the Storage Server may not work
normally if other Web service applications exist on the same computer.
The HTTP port used by this software is 80 and cannot be changed.
• Running database server software (e.g., SQL Server, Oracle) on the computer
Database processing can place a load on the hard disk that may affect recording performance. Conversely, this
software may affect database server performance when it performs tasks that place a load on the hard disk.
• Using the computer as a file server
The hard disk is frequently accessed when the computer is used as a file server, and this may affect recording
performance. In addition, the Storage Server is more likely to run out of disk capacity if the computer is operated as a file
server.
• Running other applications on the computer
The additional processing load from running applications other than this software may affect recording performance.
This software uses the TCP port numbers 80, 443, 11080 and 2380. These ports should not be used by other
applications.
Calculating Network Bandwidth
You need to consider the network bandwidth if you want to use a network camera for both recording and live view.
You must calculate the network bandwidth both for the Storage Server and the Viewer.
When using the VB-M40 for example, calculate the bandwidth as follows if you use one Storage Server (320 x 240 1 fps)
and two Viewers (320 x 240 5 fps):
• Bandwidth for recording
10.4 (KB/frame) x 1 (frames/sec.) x 8 (bits) = 83.2 Kbps
• Bandwidth for live view (per Viewer)
10.4 (KB/frame) x 5 (frames/sec.) x 8 (bits) = 416 Kbps
Therefore, the required network bandwidth is calculated as follows:
83.2 Kbps x 1 + 416 Kbps x 2 = 915.2 Kbps (approx. 0.9 Mbps)
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