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miskamm [114]
3 years ago
8

int) You are the head of a division of a big Silicon Valley company and have assigned one of your engineers, Jim, the job of dev

ising an algorithm to sort through an English text of n words and convert it into an Esperanto document. Jim comes up with an algorithm which takes 2n2+2n bit operations to handle an input text with n words. Suppose the computers in your business can handle one bit operation every nanosecond (1 nanosecond =10−9 seconds). How many nanoseconds would it take Jim's algorithm to convert a text with 10 wor
Computers and Technology
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

Correct option is E

Explanation:

a) 2n^2+2^n operations are required for a text with n words

Thus, number of operations for a text with n=10 words is 2\cdot 10^2+2^{10}=1224 operation

Each operation takes one nanosecond, so we need 1224 nanoseconds for Jim's algorithm

b) If n=50, number of operations required is 2\cdot 50^2+2^{50}\approx 1.12589990681\times 10^{15}

To amount of times required is 1.12589990681\times 10^{15} nanoseconds which is

1125899.90685 seconds (we divided by 10^{9}

As 1$day$=24$hours$=24\times 60$minutes$=24\times 60\times 60$seconds$

The time in seconds, our algortihm runs is \frac{1125899.90685}{24\cdot 60\cdot 60}=13.0312 days

Number of days is {\color{Red} 13.0312}

c) In this case, computing order of number of years is more important than number of years itself

We note that n=100 so that 2(100)^2+2^{100}\approx 1.267650600210\times 10^{30} operation (=time in nanosecond)

Which is 1.267650600210\times 10^{21} seconds

So that the time required is 1.4671881947\times 10^{16} days

Each year comprises of 365 days so the number of years it takes is

\frac{1.4671881947\times 10^{16}}{365}=4.0197\times 10^{13} years

That is, 40.197\times 10^{12}=$Slightly more than $40$ trillion years$

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For the (pseudo) assembly code below, replace X, Y, P, and Q with the smallest set of instructions to save/restore values on the
Dimas [21]

Answer:

Explanation:

Let us first consider the procedure procA; the caller in the example given.

  Some results: $s0,$s1,$s2, $t0,$t1 and $t2 are being stored by procA. Out of these registers, few registers are accessing by procA after a call to procB. But, procB might over-write these registers.

       Thus, procA need to save some registers into stack first before calling procB, .

      only $s1,$t0 and $t1 are being used after return from procB in the given example,

       Caller saves and restores only values in $t0-$t9, according to MIPS guidelines for caller-saved and callee-saved registers, .

       Thus, procA needs to save only $t0 and $t1.

    jal instruction overwrites the register $ra by writing the address, to which the control should jump back, after completing the instructions of procB, when procB is called,.

       Therefore, procA also need to save $ra into stack.

 ProcA is writing new values into $a0,$a2, procA must save $a0 and $a1 first before calling procB, .

     In the given example, after return from procB, only $a0 is being used. It is therefore enough to save $a0.

   Also, procA needs to save frame pointer, which points the start of the stack space for each procedure.

       Generally, as soon as the procedure begins, frame pointer is set to the current value of the stack pointer,.

Let us consider the procedure procB; the callee in the given example.

 The callee is responsible for saving values in $s0-$s7 and restoring them before returning to caller, this is according to MIPS guidelines for caller-saved and callee-saved registers,

   procB is expected to over-write the registers $s2 and $t0. Nonetheless, in the first two lines, procB might over-write the registers $s0 and $s1.

   Thus, procB is responsible for saving and restoring $s0,$s1 and $s2.

X:

We need to create space for 5 values on the stack since procA needs to save $a0,$ra,$t0,$t1 and $fp(frame pointer), . Each value(word) takes 4 bytes.

$sp = $sp – 20 # on the stack, create space for 5 values

sw $a0, 16($sp) # store the result in $a0 into the memory address

               # indicated by $sp+20

sw $ra, 12($sp) # save the second value on stack

sw $t0, 8($sp) # save the third value on stack

sw $t1, 4($sp) # save the fourth value on stack

sw $fp, 0($sp) #  To the stack pointer, save the frame pointer

$fp = $sp      #  To the stack pointer, set the frame pointer

Y:

lw $fp, 0($sp) #  from stack, start restoring values

lw $t1, 4($sp)

lw $t0, 8($sp)

lw $ra, 12($sp)

lw $a0, 16($sp)

$sp = $sp + 20 # decrease the size of the stack

P:

$sp = $sp – 12 #  for three values, create space on the stack

sw $s0, 0($sp) # save the value in $s0

sw $s1, 0($sp) # save the value in $s1

sw $s2, 0($sp) # save the value in $s2

Q:

lw $s0, 0($sp) #  from the stack, restore the value of $s0

lw $s1, 0($sp) #  from the stack, restore the value of $s1

lw $s2, 0($sp) #  from the stack, restore the value of $s2

$sp = $sp + 12 # decrease the stack size

8 0
2 years ago
A dietician wants you to write a program that will calculate the number of calories a person can lose by walking at a slow pace
gregori [183]

Answer:

The pseudocode is as follows

1. Input Steps

2. Input Day

3. Miles = Steps/2000

4. Calories = 65 * Miles

5. Print Calories

6. Stop

Explanation:

This line gets the number of steps for the day

1. Input Steps

This line gets the current day

2. Input Day

The line calculates number of miles

3. Miles = Steps/2000

This line calculates the calories lost

4. Calories = 65 * Miles

This line prints the calories lost

5. Print Calories

The pseudocode ends here

6. Stop

6 0
2 years ago
Users need to be able to make use of _____ , such as rumors, unconfirmed reports, and stories, when solving problems.
jekas [21]

Answer:

Informal information

Explanation:

The informal information system is employee based system design to meet personnel and vocational needs and to help in the solution of work-related problems. it also funnels information upward through indirect channels.

However, one of the most important reasons for why informal communication is critical to businesses is that it allows employees to give feedback to their superiors.

Thus, the informal or grapevine communication promotes social relationship among the participants. It helps to build up unity, integrity and solidarity among them and boosts up their morale. Grapevine or informal communication is faster than the formal communication.

8 0
2 years ago
Explain an application of a data communication system, including issues encountered with implementing the application.
aleksandrvk [35]

Answer:

Data communication system is referred to the exchange of information between the sender and receiver by use of transmission media.

Explanation:

The data communication system is referred to as the exchange of information between the sender and receiver by the use of transmission media.

some application of data communication system are;

videoconferencing,  

instant messaging

Telnet, etc.

In the data communication system, one information is transferred is transmission circuit then the media transfer the information to the receiver into the desired format.

the main issue is:

- data privacy

- cost of implementation

- use of the system for personal used during working hour

6 0
2 years ago
Assume we have a computer where the clocks per instruction (CPI) is 1.0 when all memory accesses hit in the cache. The only data
umka21 [38]

Answer:

6.6 times faster considering I-cache

Explanation:

<em>Given data </em>:

CPI  = 1

Data accesses ( loads and stores ) = 40% of instructions

Miss penalty = 200 clock cycles

Miss rate = 2% for I-cache ,  5% for D-cache

<u>Determine how much faster the computer will be if all instructions were Cache hits </u>

In this condition the memory stall = 0

hence: CPU ideal time = ( Ic * CP1  + memory stall ) * clock cycle time --- ( 1 )

                                     = ( Ic * 1 + 0 ) * clock cycle time

<em>Note : Memory stall = Ic * ( 1 + load fraction ) * miss rate * miss penalty  --- ( 2)</em>

back to equation 1

Memory stall ( for I-cache ) = Ic * ( 1 + 40% ) * 2% * 200

                                              = 5.6 Ic

Input value into equation 1

CPU ideal time = Ic * 1 + 5.6Ic * clock cycle time

                         = ( 6.6 Ic ) * clock cycle time

To determine how much faster we take the ratio of the CPU ideal time

=   6.6 Ic * clock cycle time / 1  Ic * clock cycle time

= 6.6 times faster

4 0
2 years ago
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