1. MBA 201A Section 1

2. Overview  Introduction  Section Agenda -Math Review -Class Concepts (briefly) -Problem Set #1 -Answer Additional Questions

3. Introduction  Objective: teach the how to and to learn by doing  Office Hours – By email appointment  sharat_raghavan@haas.berkeley.edu sharat_raghavan@haas.berkeley.edu  My Job:  Get you through 201A with as little stress as possible  Go through problem sets; review for exams  Problem Set Questions  Exam Questions  Your Job:  Ask Questions and suggest improvements  Work though problem sets  Section is open ended, so don’t be afraid to speak up

4. Math Review  Can you solve this problem?  Do you understand the basic ideas?  What is a function and how do I solve one for an unknown “x” (Algebra)  How / why is the slope (derivative) of a function associated with its maximum? (Simple Calculus) a) What value of x maximizes profits (  in the following function? b) What is the value of the profit function at its maximum?  (x) = (10 – x)(x – 4)

5. What is a function?  Formal definition  A function is a mathematical expression of the relationship between two variables. Changes to the independent variable (right hand side) cause changes in the dependent variable (left hand side). Examples: DEPENDENT VAR. (F(x)) INDEPENDENT VAR (X) Profits for a firmQuantity produced Demand for mortgagesInterest rate Batting averageHits in a season

6. Algebra  Step 1: Simply the left side 10X – X – 2 + 3X – 3 = 19 12X – 5 = 19  Add 5 to both sides 12X – 5 + 5 = 19 + 5 12X = 24  Divide 12 by both sides 12X/12 = 24/12 X = 2 10X – (X + 2) + 3(X – 1) = 19 What is X?

7. Lines and Slope  Slope  “Steepness of a line”  “rise / run”  What are signs of the slopes of these lines?

8. Derivatives  Derivative  “Slope” for a curve at a point (ie the slope of the tangent line) X (x = 3) dy /dx = y’(x) Y (note that: y’(3) > 0) y(x)

9. Derivative Formulas  If f(x) = 5x, then df/dx = 5  If f(x) = 5x 2, then df/dx = 10x  For this class, if f(x) = ax b, then df/dx = (ba)x (b-1)  *Note: typically, we will only use equations where b=1 or 2 in this class  If f(x) = 5x 2 + 5x, then df/dx = 10x + 5  Product Rule: d[f(x)g(x)]/dx = f(x)[dg/dx]+ [df/dx]g(x)

10. Derivatives and Maximizing – Some Intuition  Remember that the derivative of f(x) w.r.t. X evaluated at any X is the slope of the curve at X.  Notice that at the top of the curve the curve is flat i.e. the slope is zero.  This is why to find the value of X that maximizes f(x) we set the derivative equal to zero and solve for X.  In other words, we find the value of X where the slope of the curve is zero.

11. Derivative Example f(X) = 16X + 2 – 2X 2 What is the value of X that maximizes f(X)?  Step 1: Take the derivative of f(x) w.r.t. X df(x)/dX = 16 – 4X  Step 2: Set the derivative of f(x) equal to zero df(x)/dX = 16 – 4X = 0  Step 3: Solve for X X = 4

12. Derivative Example  Now let’s solve that problem:  Step 1: Use Algebra to simplify  (x) = (10 – x)(x – 4)  (x) = 10x – x 2 – 40 + 4x  (x) = – x 2 + 14x - 40 a) What value of x maximizes profits (  in the following function? b) What is the value of the profit function at its maximum?  (x) = (10 – x)(x – 4)

13. Derivative Example (cont’d)  Step 2: Use Calculus to find maximum  (x) = – x 2 + 14x – 40 d  dx= -2x + 14 At max, -2x+14 = 0 -2x = -14 x * = 7  Step 3: Evaluate function at Maximum (use original function) at x=7  *  (x) = -49 + 98 – 40 = 9 a) What value of x maximizes profits (  in the following function? b) What is the value of the profit function at its maximum?

14. Derivative Example (cont’d)  Step 4: Check for Consistency (Optional)  (x) = (10-x)(x-4)  (7) = 3*3 = 9  (8) = 2*4 = 8  (6) = 4*2 = 8  Note: You can also use the product rule of derivates to answer the question, this would save you the effort of simplifying √√√√

15. Class Concepts – Setting up Decision Trees  Node Types : Chance Node : Decision Node :Terminal Node  Setting up a Decision Tree -Timeline -What has happened already?  Sunk Costs: ignore but record? Expected value vs. sunk decision- but equivalence with sunk cost -Is a given node a decision, chance, or terminal node? -How many branches from the node? Are you sure? Continue until every branch has come to a dead end (i.e., a terminal node)

16. Class Concepts – Solving Decision Trees  Backward Induction  Start from each terminal node and work towards the beginning  Decision Nodes -Choose the best of the options (and mark it!)  Chance Nodes -Find the expected value of the outcomes Finally, calculate the EV of the entire tree: each POSSIBLE terminal node X probability of each POSSIBLE terminal node (note all probabilities added together better equal 100% and generally NOT all terminal nodes are possible) EV = 5*0.4 + -10*0.6 EV = 2 – 6 EV = -4

17. Problem 3 from Problem Set 1  Should AK steel undertake a multi-step R&D Project?  3 sequential steps, each with a.8 probability of success (.2 probability of failure)  Each step costs $500k  If all three are successful, then save $4,000k  Need all three steps to have any savings.  Can decide whether to do the next step after seeing the result of the previous step.

18. Problem 3 from Problem Set 1: Decision Tree part a

19. Problem 3 from Problem Set 1: Understanding the Tree b) Probability of Success? 0.8^3 = 0.512 c) Expected gross return:.512*4mm= $ 2,048,000 d) Should they do it? Expected net return: 0.8*0.8*0.8 * (4,000,000 – 500,000 – 500,000 – 500,000) + 0.8*0.8*0.2 * (-500,000 – 500,000 – 500,000) + 0.8*0.2 * (-500,000 – 500,000) + 0.2 * (-500,000) = $ 828,000 e)Should it quit if there are no failures? Continue even if there is a failure? No and No (value is 0 if have one step that fails) f) What if there is an alternative that AK steel leans about before the third step? With it, AK steel can spend $150k on a technology that saves $1,000k with certainty. The technologies cannot be used together.

20. Problem 3 from Problem Set 1: Decision Tree part f 2,750,000

21. Problem 3 from Problem Set 1: Decision Tree g) What is the chance the alternative is valuable?.2 chance valuable (i.e., if step 3 of original fails) h) How much is it worth? CONDITIONIAL on having the first 2 step’s succeed, the option value is: EV[|option value] = 0.2 * 1,000,000 = 200k i) What should it do? Firm should pursue both simultaneously (highest EV- follow the arrow) j) What to do if it knew about the alternative technology to begin with? (careful here) Only the 3-step project: EV= $828,000 Only the alternative project: EV=(1,000,000-150,000)=$850,000 Do Both? EV[marginal value of 3 step] = 3,000,000 * 0.8^3 = 1,536,000 EV[costs] = 0.2 * 500,000 + 0.8* 0.2 * 1,000,000 + 0.8 * 0.8 * 1,500,000 = 1,220,000 THUS, do both still… What if risk averse?

22. Questions on anything else?