Roger K deBry
Senior Technical Staff Member
Architecture and Technology
IBM Printing Systems
email: rdebry@us.ibm.com
phone: 1-303-924-4080
owner-sdp@pwg.org on 05/07/98 04:46:54 AM
Please respond to owner-sdp@pwg.org
To: robert.herriot@Eng.Sun.COM
cc: Sdp@pwg.org
Subject: SDP>PRO Proposal for TIPSI-like protocol
Your assumption might be correct if....
1) 1284.4 was a standard and not just a work in progress
2) 1284.4 was deployed in all parallel environments
3) 1284.4 was deployed in other non-TCP/IP, direct connect environments=
- USB
- 1394
- Serial
For the small amount of overhead that a packet header provides, I think=
it
is worth the price.
Don
**********************************************
* Don Wright don@lexmark.com *
* Product Manager, Strategic Alliances *
* Lexmark International *
* 740 New Circle Rd *
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To: sdp%pwg.org@interlock.lexmark.com
cc: ipp%pwg.org@interlock.lexmark.com (bcc: Don Wright)
bcc: Don Wright
Subject: SDP, IPP>PRO Proposal for TIPSI-like protocol
I just finished scanning IEEE 1284.3 and IEEE1284.4. The most interest=
ing
part is Chapter 8 "Service Provider Interface (SPI)" in IEEE 1284.4. T=
his
chapter describes a "Berkeley Sockets-compatible interface for clients =
and
servers to access the services provided by 1284.4".
So if I understand the intent of 1284.4, it is to provide a layer that
supports sockets over parallel connections. All we need to do in IPP is=
reference sockets for TCP/IP and 1284.4 and we don't have to worry abou=
t
the
issues at that layer.
So, I conclude that we don't need to packetize IPP or do much of what i=
s
proposed in Roger and Harry's paper. Instead, we can send IPP directly =
on
sockets layered on top of TCP/IP or 1284.4. There are a few easy-to-so=
lve
dangling
issues, such as chunking for document data and intermediate acknowledge=
ment
when attributes are verified for PrintJob. But otherwise IPP stays as i=
s.
If you disagree with my conclusions, I would like to know what the
TIPSI-like packetizing layer provides that sockets don't also provide?
Bob Herriot
=