Exam 200-301 topic 1 question 181 discussion

C is correct. '0C:0E:15:1A' is the smallest MAC. It is assumed that priority is the same for all 4 switches.

Root bridge > switch with lowest BID. The BID (bridge ID) is composed by the switch priority and by its MAC address. Since the question doesn’t mention any priority, just focus on the MAC addresses and compare them “hex value- to-hex value”, starting from left to right: Switch 1 – 0C:E0:38:41:86:07 Switch 2 – 0C:0E:15:22:05:97 Switch 3 – 0C:0E:15:1A:3C:9D Switch 4 – 0C:E0:18:A1:B3:19 The first two values are equal for all MAC addresses. Starting from the 3rd value, Switch 2 and Switch 3 have a MAC address which is lower than Switch 1 and Switch 4. Starting from the 7th value, you can see that Switch 3 is lower than Switch 4. Answer C is correct.

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Answer is C. Useful link (in spanish) to enforce the concept: https://learningnetwork.cisco.com/s/article/practica-eleccion-de-roles-y-estados-de-puertos-protocolo-rapid-pvst

C - It has the lowest MAC address.

the decimal equivalent of the hexadecimal number 1A is 26. the decimal equivalent of the hexadecimal number 22 is 34.

not sure why i was thinking 0(zero) is latter O

C correct answer

Place all the MAC addresses in Excel then sort them from smallest to largest (A-Z). You will find the answer and how to sort in Excel. You will also notice in the OCGs the numbers are listed first in the glossary and index. C is the answer.

the Switch which has the lowewst Mac address which is SW3 correct answer is C

D is the answer - SW4 will become root bridge Assuming bridge priorities are equal, tie breaker will be the lower mac address wins SW1 and SW2 are eliminated due to high value of MAC address SW1: 0C:E0:38:41:86:07 - SW2: 0C:0E:15:22:05:97 - Comparing SW3 and SW4's mac addresses, they are fairly similar until the 5th and 6th quartet , 3C=15 is higher than B3=14 and 9D=22 is higher than 19, so SW4 becomes the root bridge SW3: 0C:0E:15:1A:3C:9D - SW4: 0C:E0:18:A1:B3:19 - we need to consider checking the entire mac address not just half

C is the Answer. 1A= (1 × 16¹) + (10 × 16⁰) = 26 < A1 = (10 × 16¹) + (1 × 16⁰) = 161

maybe correct answer - B? 1a = 26

C is correct Least MAC value is selected incase of all the interface - RootBridge priorities are same

C is correct Assuming Bridge Priority is the same, the lowest MAC will be elected as the root bridge. 0 is lower than E which equals 14. Eliminating SW1 and SW4. 1 is lower than 2 which equals 10. Eliminating SW2. SW3 will be elected as the Root Bridge. Chart: https://ptgmedia.pearsoncmg.com/images/chap7_9780136633662/elementLinks/07fig05_alt.jpg

i understand that the smallest mac but what would be lower 29 or 2a? cant find a clear answer of this. how does the address increase

D is the correct answer