Question

A slightly more complicated net present value problem The equation for net present value (NPV) is: NPV = {=o8+14 where 8 is the discount factor and It is income in period t. Let S = 0.9. Suppose a risk-neutral individual has two potential career paths, A (career straight from high school) and B (career after college), each which give different levels of income over 5 periods. If the individual chooses path A, their income (I) in each of the 5 periods is: 10 = 25,000, 11 = 25,000, 12 = 30,000, 13 = 35,000, 14 = 40,000. If the individual chooses path B, there is some probability, p, that the individual will drop out of college and work in the job that only requires a high school education, and a probability, (1 – p), that the individual will graduate college and work in the job that requires a college education. Let p = 0.1. If the individual drops out of college, their income is: 1. = –10,000, 11 = 25,000, 12 = 30,000, 13 = 35,000, 14 = 40,000. If the individual graduates college, their income is: 1o = -10,000, 11 = -10,000, 12 = 55,000, 13 = 65,000, 14 = 70,000. Calculate the NPV for career paths A and B and explain which path the individual should take if they base their decision solely on NPV. (5 points)

Answer 1

There are 2 career paths A and B,

In Career Path A:-

Here he starts his career straight from high school,

The equation for NPV in the case of career A is,

NPV = 5*14 = 3°10 + 5+ 11 +82 I2 + 5° 13 + 5414

and value of delta given 0.9

So,

NPV = (0.9)°10 + (0.9) 11 + (0.9) 12 + (0.9) 13 + (0.9) 14

= 1 x 25000+0.9 x 25000+0.81 x 30000+0.729 x 35000+ 0.6561 x 40000

= 25000 + 22500 + 24300 + 255154 26244

= 123.559

In Career Path B:-

In career path B there are two ways to go.

• No.1 is to drop out of college and work, where probability is p = 0.1
• No.2 is to graduate from college and then work, where probability is (1-p) = (1-0.1) = 0.9

...................No1 Case,............................................

In this case, he will drop out of college and join for the work;

'I, Xp= (5°o X p)+(5', X 2) +(8°I Xp)+(8°IA X ) + NPV = \ (5414 xp)

[Here, p= probability= 0.1 ]

  

= (0.9) 10 * p +(0.9)'11 *p+(0.9) 12 *p+(0.9) 13 * p +(0.9) *14*p

= (1 x (-10000) x 0.1) + (0.9 x 25000 x 0.1) + (0.81 x 30000 x 0.1) + (0.729 x 35000 x 0.1) + (0.6561 x 40000 x 0.1)

  

= -1000 + 2250 + 2430 + 2551.5+ 2624.4

     

= 88559

...........For No2 Case,....................................................

In this case, he will complete his graduation and then join for work;

NPV =ΣΙ, x (1 - p)

= (5°10 x (1 –p))+(51 11 (1 - p)) +(8412 x (1 – p)) +(8813 x (1 –p)) + (5414 x (1 - p))

  

= ((0.9)°1, *(1-p))+((0.9)'11*(1-p))+((0.9)12(1-p))+((0.9)313 x (1 – p))+((0.9)*14 x (1 - p))

  

= ((0.9) 10 x (1 – 0.1))+((0.9)'11 x (1 – 0.1))+((0.9)-12 x (1 – 0.1))+ (0.9) 13 x (1 – 0.1))+((0.9)414 x (1 – 0.1))

(= ((0.9)°1, 0.9) + ((0.9))11 0.9) + ((0.9)1, x 0.9) + ((0.9)* 13 * 0.9) + ((0.9) 414 x 0.9)

   

= (1 + (-10000) x 0.9) + (0.9 x (-10000) x 0.9) +(0.81 x 55000 x 0.9) + (0.729 x 65000 X 0.9) + (0.6561 x 70000 x 0.9)

  

=-9000-8100 + 40095 + 42646.5-41334.3

  

= 106.975.8

............................................................................................................

NPV for A = 123,559

NPV for B in College drop out case = 8855.9

NPV for B in College graduation case = 106,975.8

..................................................................................................................

As NPV for career path A is the highest, that's why the person should choose the Career path A for his future.