A company has a choice of two languages \(𝐿_1\) and \(𝐿_2\) to develop a…

2020

A company has a choice of two languages \(𝐿_1\) and \(𝐿_2\) to develop a software for their client. Number of LOC required to develop an application in \(𝐿_2\) is thrice the LOC in language \(𝐿_1\). Also, software has to be maintained for next 10 years. Various parameters for two languages are given below to decide which language should be preferred for development.

\(\begin{array}{|l|l|l|} \hline \text{PARAMETER} & L_1 & L_2 \\ \hline \text{Man-year needed for development} & \text{LOC}/1000 & \text{LOC}/1000 \\ \hline \text{Development cost} & \text{Rs}. 70,000 & \text{Rs}. 90,000 \\ \hline \text{Cost of Maintenance per year} & \text{Rs.} 1,00,000 & \text{Rs.} 40,000 \\ \hline \end{array}\)

Total cost of project include cost of development and maintenance. What is the LOC for \(𝐿_1\) for which cost of developing the software with both languages must be same?

  1. A.

    2000

  2. B.

    6000

  3. C.

    3000

  4. D.

    5000

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Correct answer: C

Find LOC for language L1 so that total cost (development + 10-year maintenance) is the same for both languages.

  1. Let LOC for L1 be x. Then LOC for L2 = 3x.

  2. Development cost: for L1 = (x/1000) * 70,000 = 70x. For L2 = (3x/1000) * 90,000 = 270x.

  3. Maintenance for 10 years: for L1 = 100,000 * 10 = 1,000,000. For L2 = 40,000 * 10 = 400,000.

  4. Total cost expressions: L1 = 70x + 1,000,000. L2 = 270x + 400,000.

  5. Set equal and solve: 70x + 1,000,000 = 270x + 400,000

  6. Rearrange: 1,000,000 - 400,000 = 270x - 70x => 600,000 = 200x => x = 600,000 / 200 = 3000.

Answer: LOC for L1 = 3000

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