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Readme [11.4K]
3 years ago
13

Sushant is a new manager and he wants to share his ideas and working protocol with his team.Compare the advantages and disadvant

ages of oral and written communication and suggest him which form of communication would be more effective.
Computers and Technology
1 answer:
saw5 [17]3 years ago
3 0

Answer:

Advantages of Oral Communication

  1. Use of visual aids like PowerPoint presentation while explaining his ideas and working protocols would help his team to understand him better.
  2. There is a spontaneous response from the staff which is more genuine than written response.

Disadvantages of Oral Communication

  1. It could prove to be a problem if Sushant has stage fright or is not an effective communicator, which could form an impression of him by the staff.
  2. Staff may forget some of Sushant's words as no memory is as reliable as written information.

Advantages of Written Communication

  1. A well written memo explains the ideas and working protocols that Sushant wants to relay to the staff.
  2. There could be a questionnaire for the staff to give feedback.
  3. Aims, objectives are clearly stated.

Disadvantages of Written Communication

  1. If the written material is too lengthy, it could be a problem to learn or memorise.

It is recommended that Sushant should use written form of communication to convey his ideas and working protocols as it is more effective.

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What technique is used when setup times at a workstation are sequence dependent?
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3 years ago
1. How many bits would you need to address a 2M × 32 memory if:
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Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

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