Chapter 3 Performance Test Result without Packet Dropper

 

Before we introduce the packet dropper module to the router, we would like to gather the basic performance (such as system throughput, packet delay) of our infrastructure.

 

3.1 Max Throughput got from the infrastructure

 

No packet dropper is installed in router (Dublin.cs.unh.edu) so far.

 

3.1.1 Get throughput by TCP blast test

 

Table 3.1 shows that the maximum throughput we can get from my Linux Environment is: 12374712 Bytes/Sec. Its very close (about 1% less than) to 100Mb/sec.

 

Blast test  when request size is optimal ( 1448 Bytes )      PingTest

 

1st

2nd

3rd

4th

5th

6th

7th

8th

9th

10th

AVG

Var

Throughput(Byte/sec)

ElapseTime ( sec )

123.21

123.19

123.26

123.35

123.22

123.64

123.32

123.33

123.47

123.17

123.316

0.106

12374712

 

Table 3.1 Performance test data by TCP blast test

The original data are http://www.cs.unh.edu/cnrg/lin/linuxProject/phase1/blastOptimal.htm

How do we get 12374712 Byte/sec?

 Although, we just send useful data 1448 Byte on each request, we actually send TCP and IP header and Ethernet Wrapper. We actually send 1526 Bytes for each frame. (1500Bytes for layer 2 data, 26Bytes for Ethernet packet wrapper, for header detail, see here ). So actual throughput = 1526*1000000/123.316 = 12374712 Bytes/sec

It is (12500000 - 12374712)/12500000 = 1.0023% less than the Ideal Theoretical throughput 100Mb/sec

 

 Note:

We know 1 Mb/sec in Ethernet specification is 1000000 bit/sec, 1KB = 1024 Byte, 1 Byte = 8 bits

 Fast Ethernet 100Mb/Sec = 10^8 bits/sec = 12500000 Bytes/sec

 

3.1.2 Get throughput by UDP blast test 

 

Table 3.2 shows that by UDP test, the maximum throughput we can get from my Linux environment is: 12374712 Bytes/Sec. Its very close (about 1% less than) to 100Mb/sec

 

UDP test when request size optimal( 1472 Byte ) Sending 1000000 udp packets

 

1st

2nd

3rd

4th

5th

6th

7th

8th

9th

10th

11th

12th

13th

14th

15th

AVG

Var

Throughput( Byte/sec)

ElapseTime ( sec )

123.27

123.48

123.35

123.45

123.45

123.34

123.17

123.19

123.37

123.61

123.20

123.35

123.08

123.11

123.32

123.316

0.1168

12374712

Packet Loss

42

0

0

0

0

108

0

0

0

0

49

0

51

0

61

 

 

 

 

Table 3.2 Performance got by UDP test

 

The elapse time original data is: http://www.cs.unh.edu/cnrg/lin/linuxProject/phase1/clientInfoOptimal

The packet loss original data is:

http://www.cs.unh.edu/cnrg/lin/linuxProject/phase1/serverInfoOptimal

 

It's pretty interesting that the Average Elapse Time is the same as that in TCP optimal case. Incredible!!

 

So actual throughput = 1526*1000000/123.316 = 12374712 Bytes/sec

 

It is (12500000 - 12374712)/12500000 = 1.0023% less than the Ideal Theoretical throughput 100Mb/sec

 

3.2 Infrastructure Delay

In this section we will find the delay of the infrastructure that we will use to do experiments. This parameter will also be used in NS (network simulator).

We will use ICMP ping packet with different packet size to find out the packet delay in our infrastructure.

madrid$ ping -U -c 12 -s requestSize 192.168.2.2

Table 3.3 shows the result of ping requests with different ICMP packet request size.

ICMP RequestSize

56byte

64Byte

128Byte

256byte

384Byte

512Byte

640byte

768Byte

896Byte

1024Byte

1152 Byte

1280Byte

1408Byte

1472Byte

1st

158

173

197

255

291

351

379

420

473

507

570

587

642

659

2nd

150

182

222

245

282

333

370

410

453

496

541

593

629

660

3rd

160

173

197

257

288

327

379

421

468

528

558

636

649

651

4th

174

195

224

239

284

337

376

410

464

501

538

597

666

659

5th

160

176

202

244

311

321

376

451

468

518

549

585

645

651

6th

151

189

204

244

286

329

367

413

482

494

538

595

629

676

7th

163

172

193

263

303

321

375

421

463

506

550

583

639

647

8th

159

179

202

250

284

332

371

420

453

500

540

600

632

658

9th

157

177

210

247

295

335

399

447

461

508

562

593

651

650

10th

160

182

215

239

283

332

373

410

453

497

537

591

627

658

Average

159.2

179.8

206.6

248.3

290.7

331.8

376.5

422.3

463.8

505.5

548.3

596

640.9

656.9

Variation

4.2

5.76

8.92

6.36

7.44

5.84

5.5

10.68

7.2

7.9

9.5

9

9.7

5.72

Table 3.3 Data from ICMP ping test

 

Figure 3.1 shows the relation between RTT and packet request size:

RTT = a * Request_Size + b

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3.1 Relations between ICMP Packet Request Size and RTT


In Figure 3.1, we can use the line passing point (256,248.3) and point (1408,640.9) to approximate this line

RTT = a * Request_Size + b

so
248.3 = a*256 + b
640.9 = a*1408 + b

Solve the above equations, we can get  a = 0.3408 , b = 161.0556

 Theoretically, 

RTT/2  = delay + 2*(Packet_Size/Rate)

(Packet_Size = Request_size + ICMP header(8) + IP header (20)+Ethernet Wrapper(26) )

RTT = 2*delay + 4*( (Request_Size+8+20+26)/Rate )   
RTT = Request_Size * ( 4/Rate ) + ( 2*delay + 216/Rate )

Where Rate = 100Mb/sec = 12.5MB/sec = 12.5 B/s

4/Rate = 4/12.5 = 0.32   which is very close to the data calculated from the experiment 0.3408

b = 2*delay + 216/Rate = 161.0566

So delay = (161.0556 - 216/12.5)/2 = 71.8878 (s) =72 s