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 LC 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.

PARAMETER	​$$\bf{L_1}$$​​	​$$\bf{L_2}$$​​
Man-year needed for development	LOC/1000	LOC/1000
Development cost	Rs. 70,000	Rs. 90,000
Cost of Maintenance per year	Rs. 1,00,000	Rs. 40,000

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?

Correct option is C

The company is considering two languages $$L_1$$​ and $$L_2$$​ for developing software and wants to compare the total cost of development and maintenance for each language. The goal is to find the LOC (Lines of Code) that makes the total costs of developing software in both languages equal.
Given Data:
• For $$\bf{L_1}$$​:
o Development cost per 1000 LOC = Rs. 70,000
o Maintenance cost per year per 1000 LOC = Rs. 1,00,000
• For $$\bf{L_2}$$​:
o Development cost per 1000 LOC = Rs. 90,000
o Maintenance cost per year per 1000 LOC = Rs. 40,000
Let the number of LOC for language$$\bf{L_1}$$​ be x.
• For $$L_2$$​, the number of LOC is three times that of $$L_1$$​, so the number of LOC for $$L_2$$​ will be 3x.
Total Cost Calculation:
• Total cost for$$\bf{L_1}$$​:
​$$\text{Total cost for } L_1 = \text{Development cost for } L_1 + \text{Maintenance cost for } L_1$$​​
​$$\text{Total cost for } L_1 = 70,000 \times \left(\frac{x}{1000}\right) + 100,000 \times \left(\frac{x}{1000}\right) \times 10$$​​
​$$= 70x + 1,000x = 1,070x$$​​
• Total cost for$$\bf{L_2}$$​:
​$$\text{Total cost for } L_2 = \text{Development cost for } L_2 + \text{Maintenance cost for } L_2$$​​
​$$\text{Total cost for } L_2 = 90,000 \times \left(\frac{3x}{1000}\right) + 40,000 \times \left(\frac{3x}{1000}\right) \times 10$$​​
​$$= 270x + 1200x = 1,470x$$​​
Now, set the total costs equal to each other to find x:
​$$1,070x = 1,470x$$​​
Solving for $$x$$​, we get:
​$$x = 3,000$$​​
Thus, the LOC for $$L_1$$​ that makes the total cost equal for both languages is 3000.
Information Booster:
1. Development Cost Calculation: Development costs are calculated based on the number of LOC, multiplied by the cost per 1000 LOC for each language. The cost of development is generally a one-time cost associated with building the application.
2. Maintenance Cost: Maintenance cost is calculated annually based on the number of LOC, and it is multiplied by 10 (for the next 10 years). The maintenance cost includes updates, patches and running the system over time, which is an ongoing cost over the software's lifespan.
3. Equating Total Costs: To find the equal cost point, both development and maintenance costs for both languages must be set equal. This results in the equation used to find the value of x.
4. Comparing Costs: Language $$L_1$$​ is cheaper for development, but language $$L_2$$​ has a lower maintenance cost.
Additional Knowledge:
LOC in Software Projects: LOC is a measure of the size of the software, and higher LOC often leads to higher development and maintenance costs.