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4 years ago

Why is it difficult to convince top management to commit funds to develop and implement SIS

2 answers:

5 0

ANSWER: It is difficult to convince top management to commit funds to develop and implement SIS because these management are only committed to increase profit and reduce cost in the organization. Since SIS does not add to the profit of the organization, instead it increase cost, these managers finds it difficult to commit funds to such department.

EXPLANATION: SIS means Special Information System. This information systems are the department, responsible for creating medium for transferring Information within the organization and outside the organization. They also control the information technology of the company. These department enhances the communication pathways of the organization, thereby making the job of each department more easier.

But it is always very costly to develop and maintain such department in an organization. It is seen that such department incur more cost for an organization and does not bring in profit for the organization, this is the perspectives that such top managements see the SIS department.

Viefleur [7K]4 years ago

3 0

Answer:

It is difficult to convince top management to commit funds to develop and implement SIS because the system is not simply reducing costs or increasing output, it might also change the way an organization does business and because many fundamental business changes are involved, measuring the financial impact gets more difficult which leads to management's questioning to commit funds to develop and implement SIS.

Explanation:

An SIS means strategic information system, it is an information system that helps organizations to seize opportunities. It requires top management to be involved and must be part of the overall organizational strategic plan.

Whether to develop a new SIS and estimating the financial benefits is extremely difficult, as many fundamental business changes may be involved.

This is why it is difficult to convince top management to implement SIS.

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Write a function get_initials(the name) that takes a string of a name and returns the initials of the name, in upper case, separ

spayn [35]

Answer:

Following are the code to this question:

def get_initials(the_name): #defining a method get_initials

name = the_name.split(" ") #defining name variable that splits value

val = ''#defining a string variable val

for n in name: #defining loop to convert value into upper case and store first character

val = val + n[0].upper() + "." # use val variable to hold value

return val # return val

print(get_initials("Gloria kind of a big deal Tropalogos"))#call method and print return value

print(get_initials("Homer J Simpson"))#call method and print return value

print(get_initials("John Q Public")) #call method and print return value

Output:

J.Q.P.

H.J.S.

G.K.O.A.B.D.T.

Explanation:

In the given code, the last is incorrect because if we pass the value that is "Gloria kind of a big deal Tropalogos" so, will not give G.O.O.D. instead of that it will return G.K.O.A.B.D.T. So, the program description can be given as follows:

In the given python code, a method "get_initials" is declared, that accepts "the_name" variable as a parameter, inside the method "name" variable is declared, which uses the split method that splits all values.
In the next step, a string variable "val" and for loop is used, in which the "val" variable converts the first character of the string into the upper case and stores its character with "." symbol in "val" variable and return its value.
In the last step, the print method is used, that passes the string value and prints its sort value.

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4 years ago

A bookstore owner wants to allow customers to pay for books using their mobile phones. Which version of iOS will support the boo

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3 years ago

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Write a recursive method called printNumPattern() to output the following number pattern. Given a positive integer as input (Ex:

TiliK225 [7]

Answer:

See Explanation Below

Explanation:

// C++ printNumPattern() recursive method

// Comments are used for explanatory purpose

// Only the recursive is submitted

// Method starts here

void printPattern(int num1, int num2, bool dec)

{

// Print num2.

cout << num2 << " ";

//Printing to num1

if (dec == false && num1 ==num2) {

return; }

// Printing to 0 or negative.

if (dec)

{

// If num2 is greater than num2

if (num2-num1 > 0)

printPattern(num1, num2-num1, true);

else // recur with false dec

printPattern(num1, num2-num1, false);

}

else // If dec is false.

printPattern(num1, num2+num1, false);

}

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4 years ago

Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has0.020 defects /cm2Assume a 20 cm diameter wafer has a c

Vitek1552 [10]

Answer:

1. yield_1=0.959 and yield_2=0.909

2. Cost_1=0.148 and Cost_2=0.165

3. New area per die=1.912 cm^2 and yield_1=0.957

New area per die=2.85 cm^2 and yield_2=0.905

4. defects=0.042 per cm^2 and defects=0.026 per cm^2

Explanation:

1. Find the yield for both wafers.

yield= 1/(1+(defects per unit area*dies per unit area/2))^2

Wafer 1:

Radius=Diameter/2=15/2=7.5 cm

Total Area=pi*r^2=pi(7.5)^2=176.71 cm^2

Area per die= 176.71/84=2.1 cm^2

yield_1= 1/(1+(0.020*2.1/2))^2

yield_1=1/1.04244=0.959

Wafer 2:

Radius=Diameter/2=20/2=10 cm

Total Area=pi*r^2=pi(10)^2=314.159 cm^2

Area per die= 314.159/100=3.14 cm^2

yield_2= 1/(1+(0.031*3.14/2))^2

yield_2=1/1.0997=0.909

2. Find the cost per die for both wafers.

Cost per die= cost per wafer/Dies per wafer*yield

Wafer 1:

Cost_1=12/84*0.959=0.148

Wafer 2:

Cost_2=15/100*0.909=0.165

3. If the number of dies per wafer is increased by 10% and the defects per area unit increases by 15%, find the die area and yield.

Wafer 1:

There is a 10% increase in the number of dies

10% of 84 =8.4

New number of dies=84.4+8=92.4

There is a 15% increase in the defects per cm^2

15% of 0.020=0.003

New defects per area= 0.020 + 0.003=0.023 defects per cm^2

New area per die= 176.71/92.4=1.912 cm^2

yield_1= 1/(1+(0.023*1.912/2))^2=0.957

Wafer 2:

There is a 10% increase in the number of dies

10% of 100=10

New number of dies=100+10=110

There is a 15% increase in the defects per cm^2

15% of 0.031=0.0046

New defects per area= 0.031 + 0.00465=0.0356 defects per cm^2

New area per die= 314.159/110=2.85 cm^2

yield_2= 1/(1+(0.0356*2.85/2))^2=0.905

4. Assume a fabrication process improves the yield from 0.92 to 0.95. Find the defects per area unit for each version of the technology given a die area of

Assuming a die area of 2cm^2

We have to find the defects per unit area for a yield of 0.92 and 0.95

Rearranging the yield equation,

yield= 1/(1+(defects*die area/2))^2

defects=2*(1/sqrt(yield) - 1)/die area

For 0.92 technology

defects=2*(1/sqrt(0.92) - 1)/2

defects=0.042 per cm^2

For 0.95 technology

defects=2*(1/sqrt(0.95) - 1)/2

defects=0.026 per cm^2

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4 years ago