Bookmark The one-year lease for a machine is $400,000 with no option for early c

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The one-year lease for a machine is $400,000 with no option for early cancellation. So if you aren’t breaking even, you are still stuck with it for the rest of the year. You are considering signing the contract because you think the more advanced device will save some labor and raw materials and because you think the maintenance cost will be lower than the existing process.

You elicit experts to give their 95% confidence interval estimates for the random variables representing savings(materials, labor, and production) and production level:

MaintSavings(prodLevel) = 14 + prodLevel / 2500 + epsilon  

where epsilon follow a normal distribution with mean = 0 and standard deviation = 3

Labor savings (LS): -$2 to $8 per unit

Raw materials savings (RMS): $3 to $9 per unit

Production level (PL): 15,000 to 35,000 units per year

Thus, your annual savings will be random variable that is equal to:

(MS + LS + RMS) × PL - $400,000.

Use R to create a representative sample (n = 100,000) of savings outcomes (you should use normal distributions when modeling the experts opinions). What is your estimate for the 5% "Value at Risk" estimate (i.e. use the quantile function with probs = 0.05)? This represents the amount of savings that you expect to exceed 95% of the time. Round your answer to the nearest whole dollar (note: if your answer is negative, be sure to include the negative sign).

Explanation / Answer

The complete R snippet is as follows :

set.seed(1234)

n<-100000
# create the variables
epsilon <- rnorm(n,0,3)
prodlevel <- runif(n,min =15000 ,max=35000)

MS <- 14+ (prodlevel/2500) + epsilon
LS <- runif(n,2,8)
RMS<- runif(n,3,9)

# find the savings
savings <- (MS + LS + RMS)*prodlevel -400000

# find the var at 5%
round(quantile(savings,probs = 0.05),0)

The results are

> round(quantile(savings,probs = 0.05),0)
5%
83734

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