cancel
Showing results for 
Show  only  | Search instead for 
Did you mean: 
nanson
Explorer
Explorer
1,019 Views
Registered: ‎08-31-2017

Clarification of the time to perform one complete execution of a function against II

Jump to solution

Hi,

In Figure 1-3: Latency and Initiation Interval Example in UG 902, it explains the latency, II, loop iteration latency, Loop II, Loop latency well. However, as shown in the following figure, the note under the II describes that it takes 11 clock cycles.

Note: The time to perform one complete execution of a function is referred to as one transaction. In this example, it takes 11 clock cycles before the function can accept data for the next transaction.

 

Q1: I don't quite understand how it calculates the 11 clock cycles. Can anyone help to elaborate the definition in you know?

Q2: Regarding throughput, how is it normally be defined in HLS? Or what's your experience in the calculation of throughput in HLS? 

Thanks

 

ask2.PNG
0 Kudos
1 Solution

Accepted Solutions
nupurs
Moderator
Moderator
979 Views
Registered: ‎06-24-2015

@nanson 

Q1: I don't quite understand how it calculates the 11 clock cycles. Can anyone help to elaborate the definition in you know?

The next new data is read after second C0 is completed(and not when second C0 starts), which is after 11 cycles. Here the data being talked about is the one read through Addr.

Q2: Regarding throughput, how is it normally be defined in HLS? Or what's your experience in the calculation of throughput in HLS? 

Throughput is same as II in HLS.

Thanks,
Nupur
--------------------------------------------------------------------------------------------
Google your question before posting. If someone's post answers your question, mark the post as answer with "Accept as solution". If you see a particularly good and informative post, consider giving it Kudos (click on the 'thumbs-up' button).

View solution in original post

0 Kudos
1 Reply
nupurs
Moderator
Moderator
980 Views
Registered: ‎06-24-2015

@nanson 

Q1: I don't quite understand how it calculates the 11 clock cycles. Can anyone help to elaborate the definition in you know?

The next new data is read after second C0 is completed(and not when second C0 starts), which is after 11 cycles. Here the data being talked about is the one read through Addr.

Q2: Regarding throughput, how is it normally be defined in HLS? Or what's your experience in the calculation of throughput in HLS? 

Throughput is same as II in HLS.

Thanks,
Nupur
--------------------------------------------------------------------------------------------
Google your question before posting. If someone's post answers your question, mark the post as answer with "Accept as solution". If you see a particularly good and informative post, consider giving it Kudos (click on the 'thumbs-up' button).

View solution in original post

0 Kudos