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MBA 201A Section 1

MBA 201A Section 1. Overview. Introduction Section Agenda Math Review Class Concepts (briefly) Problem Set #1 Answer Additional Questions. Introduction. Objective: teach the how to and to learn by doing Office Hours – By email appointment sharat_raghavan@haas.berkeley.edu My Job:

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MBA 201A Section 1

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  1. MBA 201ASection 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 • 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 (P) in the following function? b) What is the value of the profit function at its maximum? P(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 firm Quantity produced • Demand for mortgages Interest rate • Batting average Hits 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) = 19What 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) dy /dx = y’(x) Y y(x) (note that: y’(3) > 0) (x = 3) X

  9. Derivative Formulas • If f(x) = 5x, then df/dx = 5 • If f(x) = 5x2, then df/dx = 10x • For this class, if f(x) = axb, 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) = 5x2 + 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 – 2X2What 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 • P(x) = (10 – x)(x – 4) • P(x) = 10x – x2 – 40 + 4x • P(x) = – x2 + 14x - 40 a) What value of x maximizes profits (P) in the following function? b) What is the value of the profit function at its maximum? P(x) = (10 – x)(x – 4)

  13. Derivative Example (cont’d) • Step 2: Use Calculus to find maximum • P(x) = – x2 + 14x – 40 • dP/dx= -2x + 14 • At max, -2x+14 = 0 • -2x = -14 • x* = 7 • Step 3: Evaluate function at Maximum (use original function) • at x=7 P *(x) = -49 + 98 – 40 = 9 a) What value of x maximizes profits (P) 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) • P(x) = (10-x)(x-4) • P(7) = 3*3 = 9 • P(8) = 2*4 = 8 • P(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?

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