Decision Models & Optimization
Solution to Make-up Exam Term II, 2016
Page 1 of 5
Indian School of Business Mohali, India Decision Models Models & Optimization Optimization Term II, 2016 Prof. Anjani Jain
Prof. Ziv Katalan
Yale School of Management Yale University
[email protected]
The Wharton School University of Pennsylvania
[email protected]
1. Transportation Safety Administration (15 points)
a) (2 points) Suppose that the total 4-day cost of training an employee on day 2 decreases by $10 and the problem is re-solved. Which one of the the following statements holds? (X) The overall cost changes but the optimal solution values stay the same. same. T2 is non-zero in the optimal solution, so the objective function value will change change with a change in the coefficient. The change change is within the the allowable decrease for the objective function coefficient for T2, so the optimal solution will not change.
b) (2 points) The allowable increase for the objective function coefficient of T1 has been hidden. Given the LP results and the rest of the sensitivity report, what can you say about the hidden value? (X) The allowable increase is infinite. T1 = 0 in the optimal solution, so it is not cost-effective to train employees on day day 1. 1. The reduced cost of $20 equals the allowable decrease for T1. Thus, a decrease of $20 makes training in day one become attractive enough so that T1 > 0 in the optimal solution, while any increase in the coefficient makes T1 remain unattractive.
c) (2 points Suppose that the number of bags to be screened on day 3 decreases by 3,000 and the problem is re-solved. Which one of the following statements statements holds? (X) The overall cost goes goes down by $420 and the solution values values change. The day 3 capacity constraint is binding, so the optimal solution changes with the right hand hand side. The change is within the allowable allowable decrease so the shadow price of 0.14 per unit holds for all 3,000 units, for a total decrease of $420.
d) (2 points) Suppose that the number of bags to be screened on day 4 increases by 1,600 and the problem is re-solved. Which one of the following statements statements holds? (X) The overall cost goes goes up by at least $384, maybe more. The change tightens the day 4 capacity constraint beyond the allowable increase, at which point the shadow price price may rise above above 0.24. Thus, $384 = 0.24 * 1,600 is a lower bound on the increase.
Decision Models & Optimization
Solution to Make-up Exam Term II, 2016
Page 2 of 5
e) (2 points) The allowable increase and decrease for Capacity on Day 2 have been hidden. Given the LP results and the rest of the sensitivity report, what is the allowable allo wable decrease? (X) The allowable decrease is infinite. This constraint constraint is not binding, so the shadow price price is zero. It is a “≥” constraint and will continue to be non-binding and have a 0 shadow price no matter matter how small the the right right hand side. Thus, the the allowable allowable decrease is infinite.
f)
(2 points) The TSA is considering transferring an additional screener from another another location. The
moving cost for the employee is $2,000, and she would be trained and paid using the same scheme, along with the 200 current current screeners. After re-running the LP, which one of the the statements below will be true? (X) Transferring and training the screener will increase increase TSA’s costs by less than than $2,650. The shadow price of $599.39 for the “total trained” constraint, indicates the cost of paying and training additional employees and holds for an allowable allowable increase increase of 6.38 > 1. The $2,000 moving cost brings the total to $2,599.39.
g) (3 points) TSA wants to ensure that, if more than 20 employees are trained on day 2, then at least 30 will be trained on day 3. Extend the original formulation to ensure this. this. Your formulation must be an integer LP. New decision variables, if any: Let Y be a 0-1 variable. If Y=0, then T2 ≤ 20, and if Y=1, then T3 ≥ 30.
Change to objective function, if any: No change.
New or modified constraints, if any: two new constraints T2 ≤ 20 + M Y, where M = 200 Note that M=200 is sufficient, since at most 200 people will be trained on day 3. T3 ≥ 30 Y.
Decision Models & Optimization
Solution to Make-up Exam Term II, 2016
Page 3 of 5
2. Princeton Review (10 points)
Princeton Review offers preparation (“prep”) classes for students taking standardized tests, and its West Philadelphia location is scheduling upcoming classes for the GMAT and GRE…. Formulate a linear program to solve Princeton Review’s problem. Decision Variables
X12,GMAT = X25,GMAT = X12,GRE = X25,GRE =
number number number number
12-seat 25-seat 12-seat 25-seat
rooms rooms rooms rooms
used used used used
for for for for
GMAT classes GMAT classes GRE classes GRE classes
All Xij’s are integer Objective Function
Min $5000 X12,GMAT + $8000 X25,GMAT + $5000 X12,GRE + $8000 X25,GRE Constraints
Enough Seats for Each Course 12 X12,GMAT + 25 X25,GMAT ≥ 212 12 X12,GRE + 25 X25,GRE ≥ 238 Enough Instructors for Course Sections X12,GMAT + X25,GMAT ≤ 14 X12,GRE + X25,GRE ≤ 15 Enough Rooms of the Assigned Sizes X12,GRE + X12,GMAT ≤ 20 X25,GRE + X25,GMAT ≤ 10 Non-Negativity (points?) Xij ≥ 0 for all i, j
Decision Models & Optimization
Solution to Make-up Exam Term II, 2016
Page 4 of 5
3. Metro Corp’s Nirvana Real Estate Development Project (25 points)
The MetroCorp Syndicate’s (MCS) flagship project Nirvana comprises 250 prestigious villas on 100 acres of land. The Nirvana project is expected to be completed in three years. a) (4 points) Using the above notation, write general expressions/formulae for the calculation of the semi-annual revenue in period t = 5, R 5 5, the inventory of unsold units at the start of period t = 6, I 6 6, and the semi-annual revenue in period t = 6, R 6. You may use any mathematical or spreadsheet functions in the formulae. R5 = min{I5, D5} x P5 + 0.03 S4 I6 = I5 - min{I5, D5}
R6 = min{I6, D6} x P6 + 0.03 S5 + max{I6 - D6, 0} x (0.3P 6) b) (3 points) Given the above notation, use the balances at the end of period t = 2 and the cash inflows and outflows during period t = 3 to write general expressions/formulae for the calculations of the cash balance, the amount borrowed, and the savings amount at the end of period t = t = 3: CB3, B3, and S3. You may use any an y mathematical or spreadsheet functions in the formulae. CB3 = R3 + 1.03S2
- C3
- 1.035B2
S3 = IF(CB3 > 0, CB3, 0) = MAX(CB 3, 0) B3 = IF(CB3 < 0, -CB 3, 0) = -MIN(CB 3, 0) c) (2 points) Based on the above simulation results, what is the approximate 95% confidence interval for the true expected NPV . (x) [26.9, 55.7] 41.3 -/+ 2x7.2
d)
results, what is the approximate approximate number of trials used (2 points) Based on the above simulation results, in the simulation run? (x) 1,900 MSE =
S /√n.
Hence, n = (S/MSE)2 = (313.8/7.2)2 = 1,900.
e) (3 points) Based on the simulation results, results, within what range lies the estimated probability that is positive? the true expected NPV is (x) between 95% and 100% The likelihood is almost 100% since the current sample mean is 41.3 and it is more than five standard errors (i.e., MSE) above zero! In other words, it is P{N(0,1) ≥ 5.7}.
Decision Models & Optimization
f)
Solution to Make-up Exam Term II, 2016
Page 5 of 5
(3 points) Based on the above simulation results, results, within what range lies the estimated probability that the actual min CB will be less than - 300mm (negative 300 million) Rs?
(x) between 10% and 30% From the cumulative frequency histogram: Prob[min CB
-300]
= 28%
g) (4 points) Suppose the original simulation simulation were re-run using 4 times as many trials. (i) The sample median of the NPV (x) is likely to stay the same The sample median is an unbiased estimate of the population median and hence is likely to stay roughly the same as the sample size increases.
(ii) A 99% confidence interval for the true expected NPV (x) will be roughly ½ of the original The width of 99% confidence interval is 2 x z0.005 x MSE. MSE is S n/√n where S n is the sample standard deviation and n is the sample size (i.e., number number of trials.) As the sample size size quadruples, quadruples, the MSE is expected to decrease by √4 = 2 since S is an unbiased estimate of the population standard deviation is likely to stay roughly the same as the sample size increases.
h) (4 points) Suppose the original simulation were re-run using the same number of trials, but with volatility of = 40% rather than the original 20%. (i) The sample range of the min CB Distribution (i.e., maximum – minimum) (x) is likely to grow With a higher value value of , the variability of semi-annual semi-annual unit prices increases and, in turn, the range semi-annual cash flows and the range of the corresponding minimum cash position distribution increase as well.
(ii) The mean standard error of the min CB (x) is likely to increase Similarly to (i), the true standard deviation of the min CB increases.
