|CS-534: Packet Switch Architecture
|Department of Computer Science
© copyright: University of Crete, Greece
|[Lectures: 2.2 Memory Technologies]||[print version, in PDF]|
Thus, for simple, unconstrained memory operation, besides total memory throughput, the other interesting performance metric is the peak possible rate of random accesses to arbitrary, independent locations (i.e. not necessarily sequential or in the same row). This is, in other words, the peak address rate for random, independent accesses.
(a) What is this number, for some of the SRAM technologies seen in class, in millions of accesses per second (Maccesses/s)? Consider the following SRAM technologies. the SRAM blocks used in the two on-chip buffer examples in section 2.2.1: (i) in the one-port, 40 B wide buffer example, and (ii) in the two-port, 256 B wide example; the off-chip examples seen in section 2.2.2: (iii) the QDR SRAM chip example (CY7C1545V18), and (iv) the shared-bus SRAM chip example (CY7C1550V18 - this is a burst-of-2 chip, capable of receiving one address per clock cycle).
(b) Consider that we build a 64-Byte (512-bit) wide buffer memory out of each of the above technologies in (a). For technologies that provide only burst accesses, the memory "width" is the total size of the entire burst that is accessed at a time. This 64-Byte width is a customary segment size in modern networking equipment, because 64 bytes is a "round" number just above the ATM cell size or the minimum IP packet size. (In the case of 18-bit or 36-bit wide parts, the total memory width will be 64x9 = 576 bits, where the extra 64 bits per segment are usually used for parity or ECC, and/or other off-band overhead information, e.g. end-of-packet and other such mark bits).
How many blocks (on-chip) or parts (off-chip) are needed, in each case of (a), for this 64-Byte wide buffer memory to be made? What is their aggregate peak throughput in Gbits/s? What is their total power consumption at peak rate, and their consumption per Gbps of offered throughput?
(a) Find the peak access rate for trully random accesses, i.e. accesses that may fall in the same bank but in a different row relative to the previous access. Hint: this is directly linked to the (same-bank) cycle time.
(b) Assuming the peak access rate in (a), what is the chip's peak data throughput (Gb/s)? Answer separately for all-read's and for all-write's. As in the slides, assume that you may set the burst size to "full page", i.e. "very long", and that the burst goes on continuously for as long as possible for it not to adversely interfere with the next ACTIVATE command.
(a) For each SRAM technology in exercise 3.1(a), what is the peak incoming segment rate that can be supported, in Msegments/s? Hint: Each incoming segment is written into a "random" memory location (address). Thus, for each incoming segment we need to perform an (independent) write memory access. Hence, the peak incoming segment rate that can be supported is one half (50% writes - 50% reads) of the peak (independent) access rate calculated in exercise 3.1(a), except for the shared-bus SRAM that has a bus turn-around penalty where you need to derate their peak Maccesses/s by the turn-around overhead for our specific 4-write-4-read access pattern.
(b) Assume that the incoming traffic is ATM over SONET. For reasons of simplicity of memory management, each ATM cell is written into a different memory segment --hence, approximately 64-53 = 11 bytes in each segment remain unused (the exact number depends on details such as whether the header CRC is stored or just recomputed on the way out, whether any flow ID is stored together with the cell to assist in VP/VC translation in the outgoing path, etc). Thus, the peak incoming cell rate that can be supported is equal to the peak incoming segment rate that you calculated in question (a).
Translate this cell rate into an equivalent "SONET bit rate", for each SRAM technology considered in (a). Of course, SONET bit rates are strictly quantized, as listed in exercise 2.1, but, for the purposes of this exercise, assume that you can linearly scale the SONET bit rate to any number that is needed to provide the desired ATM cell rate; Assume that the percentage of SONET bit rate that is dedicated to SONET overhead (clock recovery, framing, etc) is as in exercise 2.2, i.e. 3.33 percent (3 bytes of overhead in every 90 SONET bytes). Compare the "SONET bit rate" that you find here to the buffer memory aggregate peak throughput in Gbits/s that you found in exercise 3.1(b), for each same technology. How and why do they differ?
(c) Assume, now, that the incoming traffic consists of 40-Byte (minimum sized) IP packets, which are carried in an "IP-over-SONET" technology (not IP-over-ATM-over-SONET). These minimum sized IP packets fit within one buffer memory segment (64 bytes), each. For reasons of simplicity of memory management, again, each such IP packet is written into a different memory segment --hence, approximately 64-40 = 24 bytes in each segment remain unused. Thus, the peak incoming packet rate that can be supported is equal to the peak incoming segment rate of question (a), or to the peak incoming cell rate of question (b).
Translate this packet rate into an equivalent "SONET bit rate", for each SRAM technology considered in (a). Unfortunately, I do not know the exact format of IP-over-SONET, so let us assume, for the purposes of this exercise, that the only SONET overhead, above and beyond the 40 bytes times 8 bits/byte = 320 bits of IP packet payload, is the same as for ATM over SONET, i.e. 3 bytes of overhead for every 87 payload bytes in every 90 SONET bytes (BEWARE: do not use this number in any real design of yours, because it is most probably not the real number!). Also, assume again, contrary to reality, that SONET bit rates are not quantized, and can scale linearly to provide the desired packet rate. Compare the bit rates that you find here to those of question (b) and to those of exercise 3.1(b), and explain the difference.
(d) Next, assume that the incoming traffic consists of 68-Byte IP packets. This is a "bad" size for our buffer memory, because it is just above our segment size (we assume that IP packet sizes are multiples of 4 bytes, otherwise, 65 bytes would be the worst size in this case). In this case, each IP packet needs two (2) memory segments to be written in. For reasons of simplicity of memory management, again, each such IP packet is written into two different memory segments --hence, approximately 128-68 = 60 bytes remain unused in every other segment (30 bytes per segment average fragmentation overhead). In this case, the peak incoming packet rate that can be supported is half of what it was in question (c).
Translate this packet rate into an equivalent "SONET bit rate", for each SRAM technology considered in (a), using the same IP-over-SONET assumptions used in question (c). Compare the bit rates that you find to those found earlier, and explain the difference.
(e) --Optional Question--
Assume again, as in question (c), that the incoming traffic consists of 40-Byte (minimum sized) IP packets. This time, however, the traffic arrives over a number of Gigabit Ethernet links (see also exercise 2.3). To calculate the peak packet rate of a Gigabit Ethernet link when carrying minimum sized IP packets, consider that:
(f) --Optional Question--
Answer question (e) in the case of 68-Byte IP packets, as in question (d). As in (d), two segments per packet are needed, hence two (independent) buffer memory accesses per packet. As in question (e), assume Gigabit Ethernet links; one difference, here, is that no padding is needed in the ethernet packet body, since the 68-Byte IP packet size satisfies the 46 to 1500 byte ethernet packet body requirement.
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Last updated: 12 March 2013, by M. Katevenis.