1. Notes set #1:
Salary multiplication factor = 2.5 ( rather a low number for American standards)
Ex: A salary of $80K means an expense of 80x2.5 = $200K to the company!
Productivity of an engineer is 50 LOC /day (LOC = Lines of code)
1 year = 12 months
1 month = 30 days
1 month = 4 weeks
Cost of an engineer
Salary - $80,000
Burden - 2.5
80,000 * 2.5 = $200,000

2. Notes set #2: Model to be used for Money and for Time:
Requirements 24%
Design 16%
Subtotal 40%
Coding 12%
Testing 48%
Subtotal 60%
Total 100%
EX: A project is estimated to be 450KLOC. There are 18 engineers available
Evaluate Effort (People) Time and Money per stage ($90K/year-engineer)
450000LOC /50(LOC/day) = 9000 days-engineer
9000 days-eng/18 (eng) = 500 days = months = 1.39 years
Cost of one engineer = 90x2.5 = $225K/year … etc.

3. Notes set #3: Was Brooks right?
Consider the following problem: S/W is being produced by 10 engineers at the standard rate for 12 months (linear monthly productivity of 1.5*T [KLOC/person], where T is given in months). Management thinks the project is late and decides to bring in 4 more developers.
With training and relocation the productivity becomes 2*(T^0.5) [KLOC/person], T still in months for the following 9 months. After that the productivity becomes again the standard linear. When does this solution become effective? Any comments?
Answer
For the first 10 months the team produced 1.5*12*10 KLOC = 180 KLOC
Starting at this point the team produces 2*3*14 = 84 KLOC
THIS TOTAL = 264 KLOC PER THE FIRST 12+9 = 21 MONTHS.
What rests is to solve 10 engineers at linear productivity for 21 + x months versus
14 engineers for x months at linear productivity starting at the point (21, 264)
The equation is 15(21+x) = x => 6x = 51 => x = 8.5 months.
Therefore: if the project lasts for more than = 29.5 months the trade is worth it!