WCF Slow response

daniel061974 1

Hi,
I have a WCF client . when the response is big (>5MB) , it takes more than an hour to get the response to the client , the CPU is ~40% usage
when I check in Fiddler , I see immediately the response.
How can I do it faster ?

1 answer

QiYou-MSFT 4,326 Reputation points Microsoft Vendor

2022-12-27T12:20:45.867+00:00
Hi @daniel061974 ,
When I was learning about WCF, I saw some examples of improving WCF.
1.Set ServiceHost on the server to break the maximum number of requests that the ServerHost can receive, such as:
host.Description.Behaviors.Add(new ServiceThrottlingBehavior()
{
MaxConcurrentInstances = 2000,
MaxConcurrentCalls = 2000,
MaxConcurrentSessions = 2000
});
2. Set the InstanceContextMode and ConcurrencyMode characteristics of the service by using the following features on the class that implements the service contract:

[ServiceBehavior(InstanceContextMode = InstanceContextMode.PerCall, ConcurrencyMode = ConcurrencyMode.Multiple, UseSynchronizationContext=false)] public class MyService : IInterface{}

3: Set useDefaultWebProxy="false" in APP.config

<bindings> <basicHttpBinding> <binding name="" useDefaultWebProxy="false" /> </basicHttpBinding> </bindings>

Best Regards
Qi You

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daniel061974 1 Reputation point

2022-12-27T12:42:04.877+00:00

Thnaks Qi You but the problem is in the client not in the service host

QiYou-MSFT 4,326 Reputation points Microsoft Vendor

2022-12-28T06:18:55.49+00:00

First of all, I think it may have something to do with server-side configuration, so I give you some server-side configuration. For example, if you have too many I/O interfaces on the server side, resulting in excessive resource utilization, there may be a slow problem.
Here are a few methods you can try:
The first method is to use an I/O (auto-release) WCF reuse code of the stopwatch class:

for (var i = 0; i < 10; i++) { var stopwatch = new Stopwatch(); stopwatch.Start(); using (var clienthost = new StreamServiceClient()) { var result = clienthost.Upload(new StreamRequest { Bytes = data, FileExt = ext, Passport = "" }); if (!string.IsNullOrEmpty(result.Msg)) MessageBox.Show(result.Msg); } stopwatch.Stop(); this.richTextBox1.AppendText(string.Format("{0}time：Time-consuming：{1}\r\n\r\n", i, stopwatch.ElapsedMilliseconds)); }

The second way is to customize the WCF channel according to your needs. There is no unified code for this custom channel, which needs to be completed through its own further requirements.

daniel061974 1 Reputation point

2022-12-28T08:27:17.677+00:00

Thanks Qi You
I have a client WCF , the WCF server is not on my side , it's our customer WCF server.
I received the response immediately from the server , I can see it in Fiddler but the response takes ~1 hour to load in the client
Thanks

QiYou-MSFT 4,326 Reputation points Microsoft Vendor

2022-12-29T11:58:25.96+00:00

First of all, I want to tell you how Fiddler works.
In the first step, Fiddler sends a request to the server for a handshake, obtains the CA certificate to the server, decrypts it with the root certificate public key, verifies the server data signature, and obtains the server CA certificate public key.
In the second step, Fiddler forges its own CA certificate and passes it to the client browser impersonating the server certificate, which does the same thing as Fiddler.
In the third step, the client browser generates a symmetric key for HTTPS communication, encrypts it with the public key of the certificate forged by Fiddler, and passes it to the server, which is intercepted by Fiddler.
In the fourth step, Fiddler unlocks the intercepted ciphertext with the private key of his forged certificate to obtain a symmetric key for HTTPS communication.
In the fifth step, Fiddler passes the symmetric key to the server encrypted with the server certificate public key, the server unlocks the trust with the private key, the handshake is completed, the message is encrypted with the symmetric key, and communication begins.
In the sixth step, Fiddler receives the ciphertext sent by the server, unlocks it with a symmetric key, and obtains the plaintext sent by the server. Encrypted again, sent to the client browser.
In the seventh step, the client sends a message to the server, encrypted with a symmetric key, which is intercepted by Fidller, and decrypted to obtain plaintext.

QiYou-MSFT 4,326 Reputation points Microsoft Vendor

2022-12-29T12:00:37.37+00:00

Since Fiddler has always had a symmetric key for communication, the information is transparent to it throughout the HTTPS communication process.
So you should know that Fiddler actually belongs to a proxy server with some higher privileges. His speed must be faster than the server reception, as for the reason, I think there are too many threads, your client will not be able to accept it for a while, or there are too many packets when decrypting, decryption is too slow. To sum up, I think it is necessary to check the I/O calls, and the encryption of packets during the transmission of the server is also the reason. You should publish the runtime directory so that we can better find the problem.

daniel061974 1 Reputation point

2022-12-29T14:26:03.077+00:00

Thanks,
How can I check the I/O calls, and the encryption of packets ?
What do you mean "runtime directory" ?

QiYou-MSFT 4,326 Reputation points Microsoft Vendor

2022-12-30T05:52:37.173+00:00

Each WCF server has its runtime directory, which tells you what the packet went through. I showed you the method of I/O above, but you don't have any code, and I don't know your ABC (the three elements of WCF) and the configuration of the WCF server, so I can only locate the problem here. Because Fiddler functions as a proxy server, it is definitely faster than the client. The slow data transfer of the server may also be that the server has too much data to transfer and too many threads are called, and TPL can be used on the server side to optimize the process.

QiYou-MSFT 4,326 Reputation points Microsoft Vendor

2022-12-30T05:53:39.183+00:00

The method I gave you above you also have to modify according to the configuration of the server, because you did not post any code. This is what I wrote based on a regular WCF service.
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