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9-1. Assume that an average SNMP response message is 100 bytes long. Assume that a manager sends 400 SNMP Get commands each second. a) What percentage of a 100 Mbps LAN link’s capacity would the resulting response traffic represent? Answer : b) What percentage of a 1 Mbps WAN link would the response messages represent? c) What are the management implications of your answers?

Answers

Answer 1

Answer:

Answer:

a) 0.32%

b) 32%

c) Mbps WAN link is of more benefit

Explanation:

Average SNMP response message = 100 bytes

Avere message for each second when manager sends 400 SNMP

: 400 SNMP per second * 100 bytes = 40000 bytes per second = 40000Bps

To convert byte to bits, We have 1 byte = 8 bits

Therefore, 40000Bps = 40000 * 8 = 320 Kbps

a) Calculating the percentage of a 100 Mbps LAN link’s capacity

100 Mbps = 100000Kbps

320 Kbps is what percent of 100000Kbps = $(320)/(100000)$ * 100 = 0.32%

So the resulting response traffic would represent 0.32% of a 100 Mbps LAN link’s capacity.

b) Calculating the percentage of a 1Mbps LAN link’s capacity

1Mbps = 1000Kbps

320 Kbps is what percent of 1000Kbps = $(320)/(1000)$ * 100 = 32%

So the resulting response messages would represent 32% of a 1Mbps LAN link’s capacity.

c) When we are using the 1 Mbps WAN link, we use 32% of its speed to response the message as opposed to 100 Mbps LAN link that uses just 0.32%. This means the 1 Mbps WAN link uses more bandwith than the 100 Mbps LAN link. Therefore the management implication is that it is better to use the 1 Mbps WAN link has it has more benefits.

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Is the willingness to put a customer’s needs above ones own needs and to go beyond a job description to achieve customer satisfaction

Answers

Answer:

customer-serious orientation

Explanation:

Customer-serious orientation can be defined as the willingness to put a customer’s needs above ones own needs and to go beyond a job description to achieve customer satisfaction.

This ultimately implies that, a customer-serious oriented business firm or company puts the needs, wants and requirements of its customers first without considering their own needs in a bid to satisfy and retain them.

Hence, customer-serious orientation requires the employees working in an organization to show and demonstrate positive attitudes and behaviors at all times.

3. Problem 5. A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. What is the maximum allowable size for memory? d. What is the largest unsigned binary number that can be accommodated in one word of memory?

Answers

The number of bits that are needed for this operation code (opcode) is 8 bits.

What is binary encoding?

Binary encoding can be defined as a standard technique that is designed and developed for converting data in plain text (source alphabets) to a form that is easily used by different operating systems (OS), especially through the use of a binary digit (bit) or two-symbol system.

How to calculate the number of bits.

The number of bits that are needed for this operation code (opcode) is given by this mathematical expression:

$2^n > 150\n\n2^8 > 150\n\n256 > 150$

n = 8 bits.

b. To determine the number of bits that are left for the address part of the instruction:

Opcode, O, I = 8 bits.

Instruction size = 24 bits.

$Address = I-O\n\nAddress =24-8$

Address = 16 bits.

c. To determine the maximum allowable size for memory:

Based on the memory unit, this is given by $2^(16)* 24$

d. The largest unsigned binary number that can be accommodated in one word of memory is given by: $2^(24)-1$

Read more on binary encoding here: brainly.com/question/5381889

Answer:

a) 8 bits b) 16 bits. c) 2¹⁶ * 24 bits d) 2²⁴ -1

Explanation:

a) In order to be able to accommodate 150 different instructions, the number of bits needed must be equal to the minimum power of 2 that satisfies this equation:

2n > 150, i.e. n=8.

b) If the total number of bits for a word is 24, and 8 are used for the op code, there are 16 bits left for the address part.

c) If the address part has 16 bits, this means that the total addressable space is just 2¹⁶, so the maximum allowable size is 2¹⁶ * 24 bits.

d) As we have 24 bits to be filled, the largest unsigned binary number is just 2²⁴ – 1.

(As we need to leave a position for all zeros).

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Answers

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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

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(IV) machines are always less expensive than human labor.

Explanation:

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