Question

Using the Internet Checksum algorithm, calculate the checksum which would be added to the end of the following bytes of data: 1. 39 91 8E 9D CF 85 61 AA 83 03 7C 0F A0 09 2. 21 86 CF DF 8E 67 C2 BE DB 38 8B F1 F2 F4 3. 1B F8 D9 91 02 EA 4. BA CD 7C E2 8E 7D 0F 77 AF 5D C6 48 70 0D 23 3A 5. 70 77 D2 B3 6. 44 93 6E B5 3B E7 83 36 7. 07 F0 42 CC 8. 62 AC D8 45 9. 64 88 2D 2A A9 83 3B F8 BE BC 20 8B 10. 2B 56 89 E9

Answer 1

Solution:

For the given question, I have calculate the internet checksum for the given bytes of data.

I have uploaded the images for each question.

1st Question:

solution:- According a no question, we have to to o the e the the given given internet internet alculate checksum for the following data. 29 91 8E AD CF 85 61 AA 83 03 7C OF 0 39 91 AO 09 Sol so 39.91 8 E 90 CF 85 61 AA 83 03 Te Of AO 09 First of all, we should have the knowledge about hema decimal digits. Hexa decimal values starts from 0-15 . . 10 1 2 13 14 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. Checksum at sender site:- A e carry > I+Dit 5+ A+ 3+ 4+9+0 = 56 16) 56(334 3 2 3 399! 8E9D CF 85 61 AA 8303 7C OF AO 09 to carry Apply 18+9+9+8+ A+0+0+0 to = 39 ad operation 00000 check sum (Initially) (**.**. 16)3962 em , A 98184 sum t 3 . . 987 B + EtF+1+346 =56 (sum) | 3+3+8+0+ 6+ 8+7+ Ato :57 1675113 48 (9) to get the checksum to SEND, We have to do compliment for the obtained ssum (987B). compliment means we have to Subtract sun with 15 Cre) (15-B=4 981 B 1 6 7 8 4] Tir checksum 15-1=8 115-8-7 (15-9=6

Checksum at Receiver site- Wle get the checksum for the sender sitel6984). This checksum will be added at the end of the receiver 'site lie) 399! 8E9D CF85 61 AA 8-3 63 7C OF A009 6784 Now all these values. Na 1696313 48 (157F the receiver will add 323 3991 8E9D CF 85 GIA 8303 IC OF A009 6784 (checksum) FFFC 16)60(3 48 12C 16) 47 (2 32 15F +3 . FFFF (sum) To get the checksum for receiver, we have to do compliment for the obtained sum (FFFF). Subtract with 15. G F F F F 100 00.1→ checksum (zero) It checksum at receiver site is 10000 , then it indicates that there is no Error. oooo, then If it checksum indicates at that receiver there site is not is Error.

2nd Solution:

lo Criven 2186 CF DE 8E 67 CL BE DB 38 8B FIF2 FU Similar to the above problem, Checksum at sender sites 16) 55 (3 16) 74 (4 -64 343 2186 CFDF SE 67 02 BE DB 38 8BFI & 25 oca (checksum) 9 CAT4 sum + 4 9 CAB (sum) -15 for subtract the obtained sum. Now we have to 9C AB r B 15-B = 4 16354 Llebecksum) 15-A =5 15-c. = 3. 15-9-6 Checksum at Receiver site- Add the obtained checksum at the end. les 2186 CF DE 8E67 C2 BE OB 38 8BF1F224 6354 Now add all these values.

343 2126 16)5913 1616313 ..48 TIDSB CE67 C2 BE 160-19 (4 DB38 1155sf F2 F4 6354 checksum FEFB FFFF&(sun) Non, subtract is from the sum( FF FF) F F F F checksum. 10 1oooo 0000 so No Error Here checksuon is

3rd Question:

checksum at receiver to Now, we have add obtained sum at end. IBF8 D991 02 EA 078€ A 0991 02 EA 0786 FOFF fchecksum. & subtract -15 from sum. FEFF ro00oNo Errores

4th Question:

checksum at Receiver. Весь 3 34 140 с E. Pе пр OF Hf 5) со 48 70 ор 3) 5 3 5 BAC D 1628 ( 16) 265 Т КО - 10), 3(3 ч 1с Е 2 SE 7 р Of 77 AF 5D се ч & 70 OP 23 ЗА 14 60 F2Fc (se E) + 3 F2 FE—Sum, Subtract with = 15 ERROR. F2 FE o o o o2.

5th Question:

e Given 7077 02 B3 checksum at sender 16718 4674 D2 B3 ♡ 000 432 A &sum. Subtract with 15 43 2" A BCD 5 to checksum.

Checksum at Receives- 7077 D2B3 BC D5 16) 31 7077 D2 B 3. B C D 5 - FFF F a sum. subtract with 15 .2 FF F F Toooot checksum. No Error

6th Question:

③ Given 4493 65 B5 3B E 783 36 checksum at sender;- 16)343 4493 GEB 5 38 67 32 72 65 72 66 (sum) subtract with ~15 T 2 66 180 g gh checksum. checksum at Receiver checksum 4493 6EBS BBET 8336 8D 99 221 141/2 16/30/ 16 44 9 3 GEB 5 3BE7 8336 8099 FFFE (147> E 16747182 32 Ffff Subtract with 15 . FFFF Toooot checksum. S No Error

7th Question:

1 Given on Fo 42 cc checksum at Sender 16) 210 Cu2- от рода ч 2 с об ОО чөвс Subtract with 15. че все 1959 3, checkum. checksum at Receives - отво ч2 се в543 07F. 42 cc . Ф) B5 3. FFFF Subtract with 15. F FFF Toooo checksum. No Error in data

8th Question:

(8) Given GO AC 08 45 checksum at sender 163761 62 na 0 Bs y 5 161191 SA F A-checksum. Subtract with. – 15 3A F 2 Tc 50 ot checksum. checksum at Receiver side 62 AC D845 C5OD 1G) 301 62 AC 18 4 5 Ô 15 OD FFEE - FREF asum. Subtract with 15 - FFE F Too 1o checksum I ERROR in data de

9th Question:

@ Civen 6488 2D2A A983 3B F8 BE Be 20 8B checksum at sendes: 2 333 1625263 16)5543 48 . 16)31 (2 32 1695463 4 6488 2D2A 1983 3BF8 ДЕ ВО 208 B 0000 5 6 7 4 +2 56764 (sum). subtract with 15 5616 A 989 to checksum. checksum at Receiver. 6488 2D2A 1983 3BF8 BE BC 20 SB 4989 - 162636 2 33 3 64 88 2D 2A 4983 3BF8 Ве Вс 208B A 989 1676183 48 (1370 To © F FFD F FF F <sum. Subtract with = 15 F FF Oo Ooty checksum, No Error

10th Question:

Cliven 28 56 89 E9 checksum at Sender 2B 56 169101 1992/ 89E9 6 0000 B 53 fa sum, subtract with 25 B 5 3 F 14 Acol, checksumni checksum aut Receiver 28 56 89 E9 UA.CO 16), (15) 2B 56 89€ 9. 4A CO FF FF & Sum. with ~ 15 Subtract F FFF checksum. oo 0 0 H NO ERROR