Lecture 12: Exercises

a. A discrete distribution, with possible outcomes $1, 2, 3, 4, 5, 6$.

b. Still a discrete distribution in principle (there are only finitely many possible averages of 100 die-max outcomes), but by the central limit theorem it will look close to a normal (continuous) distribution.

c. The confidence interval is $\left[4.5 - \frac{2 \times 1.4}{\sqrt{2500}}, 4.5 + \frac{2 \times 1.4}{\sqrt{2500}}\right] = [4.444, 4.556]$. We are 95% confident that the true expected outcome lies in this interval. More precisely: if we repeated this experiment many times and built a 95%-CI each time in the same way, about 95% of those intervals would contain the true expected value. Any single realized interval either does or does not contain it; no probabilistic statement applies to it on its own.

d. All 36 equally likely outcomes of (die 1, die 2) and their maximum:

Each outcome has probability $1/36$, and the value $k$ appears $2k - 1$ times (1, 3, 5, 7, 9, 11 times for $k = 1, \dots, 6$), so

a. $X$ takes countably many values, so it is discrete.

b. $E[X] = 0 \cdot \tfrac14 + 2 \cdot \tfrac12 + 6 \cdot \tfrac18 + 10 \cdot \tfrac18 = \tfrac{24}{8} = 3$.

c. Using $E[X + Y] = E[X] + E[Y]$, $E[b] = b$ for a constant $b$, and $E[aX] = aE[X]$ for a constant $a$:

using part b for the last step.

d. $\text{Var}[X] = E[(X - E[X])^2] = (0-3)^2 \tfrac14 + (2-3)^2 \tfrac12 + (6-3)^2 \tfrac18 + (10-3)^2 \tfrac18 = 2.25 + 0.5 + 1.125 + 6.125 = 10$.

e. Split $[0, 1]$ into consecutive intervals with widths equal to the probabilities: $[0, \tfrac14)$ for $X = 0$, $[\tfrac14, \tfrac34)$ for $X = 2$, $[\tfrac34, \tfrac78)$ for $X = 6$, $[\tfrac78, 1]$ for $X = 10$. Draw $u \sim \text{Uniform}[0, 1]$ and return the $X$-value whose interval contains $u$.

f. Using the intervals from part e: $u_1 = 0.11$ falls in $[0, \tfrac14)$, so $X_1 = 0$; $u_2 = 0.98$ falls in $[\tfrac78, 1]$, so $X_2 = 10$; $u_3 = 0.47$ falls in $[\tfrac14, \tfrac34)$, so $X_3 = 2$. Then $g(X_1), g(X_2), g(X_3) = 0, 100, 4$, and the sample average is $(0 + 100 + 4)/3 = 104/3 \approx 34.7$.

g. Only 3 samples is far too few for the law of large numbers to give a reliable approximation of $E[g(X)]$; the estimate in part f should be treated with strong suspicion.

a. $r(X, \pi) = 0.6 \min(X, \pi) - 0.2\pi + 0.1 \max(\pi - X, 0)$: revenue from what is sold, minus the purchase cost of everything ordered, plus salvage revenue on leftovers ($\pi - X$ leftover units when $\pi \ge X$, 0 otherwise).

b. For $\pi = 45$: demand 50 gives profit $0.6 \times 45 - 0.2 \times 45 - 0.1 \times 0 = 18$ euro; demand 45 gives $0.6 \times 45 - 0.2 \times 45 - 0.1 \times 0 = 18$ euro; demand 39 gives $0.6 \times 39 - 0.2 \times 45 + 0.1 \times 6 = 15$ euro. The sample average is $(18 + 18 + 15)/3 = 17$ euro, and the sample variance (dividing by $n-1$) is $s^2 = \big((18-17)^2 + (18-17)^2 + (15-17)^2\big) / 2 = 6/2 = 3$.

a. A and B are uniform on $[1, 5]$, with expected duration 3; C is uniform on $[1, 3]$, with expected duration 2.

b. np.maximum(a, b) compares the two arrays run by run, giving [3.5, 4.5, 2.5, 4.0], so finish is [4.5, 7.0, 4.0, 6.0]. The four statements print:

• the array [4.5, 7.0, 4.0, 6.0];

• the average $(4.5 + 7.0 + 4.0 + 6.0)/4 = 5.375$;

• finish > 5 is [False, True, False, True], and its mean is the fraction of True, 0.5;

• np.max(finish) is the largest value in the array, 7.0.

So np.maximum takes the maximum of two arrays per run and returns an array, while np.max returns one number: the largest value of a single array.

c. $\max(3, 3) + 2 = 5$. The true expected finish time is larger (flaw of averages): the project waits for the slower of A and B, and $E[\max(X_A, X_B)] > \max(EX_A, EX_B)$. A delay of either activity delays the project, while an early finish of one of them does not help if the other is late. (In fact $E[\max(X_A, X_B)] = 1 + 4 \times \tfrac23 \approx 3.67$, so the expected finish time is about 5.67.)

d. The deviations from the mean are $-0.875, 1.625, -1.375, 0.625$, so $s^2 = (0.766 + 2.641 + 1.891 + 0.391)/3 = 5.6875/3 \approx 1.90$ and $s \approx 1.38$. The CI is $5.375 \pm 2 \times 1.38/\sqrt 4 = 5.375 \pm 1.38$, so about $[4.00, 6.75]$. With only four runs this CI is not reliable: the normal approximation needs many more runs. The width decreases with $\sqrt n$, so a ten times narrower CI needs $10^2 = 100$ times as many runs: n_runs = 400.

e. (a > b).mean(). For the four runs, a > b is [False, True, False, False], which gives 0.25. (The true probability is 0.5, by symmetry.)

a. Inverse-CDF sampling for an exponential with mean 2 (rate $\lambda = 1/2$) is $-\ln(u) \times 2$. This gives inter-arrival times $0.89, 0.45, 2.16, 0.50$ hours.

b. Inverse-CDF sampling for $\text{Uniform}[1.1, 2]$ is $1.1 + 0.9u$. This gives service times $1.73, 1.99, 1.59, 1.89$ hours.

c. A job can only start once it has arrived and the machine is idle. Using parts a and b:

The number of jobs at the machine over time is then a step function: 0 until 0.89, up to 1 when job 1 arrives, up to 2 when job 2 arrives (at 1.34) while job 1 is still in service, down to 1 when job 1 departs (2.62), up to 2 when job 3 arrives (3.50), up to 3 when job 4 arrives (3.99) while job 2 is still in service, down to 2 when job 2 departs (4.61), down to 1 when job 3 departs (6.20), and down to 0 when job 4 departs (8.09).

d. Integrating the step function from part c and dividing by the 10-hour horizon:

a. The state is the stock (stock), together with whether an order is on its way. There are two types of events: a demand (a customer arrives) and a delivery. Their times are next_demand and next_delivery; next_delivery is np.inf when no order is on its way.

b. It makes sure that the shop does not place a new order while an earlier order is still on its way. Without it, every demand that leaves the stock at or below the reorder point would schedule a new delivery, overwriting the time of the earlier one.

c. Following the events:

The next demand would be at $2.7 + 0.5 = 3.2 > 3$, so the loop stops. The code prints:

time 0.4: stock 4
time 0.9: stock 2
time 1.2: stock 0
time 1.8: stock 0
time 1.9: stock 8
time 2.7: stock 3
stockouts: 2 out of 5 demands

d. In the delivery event, replace stock = order_up_to by stock = stock + 6 (or stock += 6).

e. Add lost_units = 0 before the loop, and in the stockout case add lost_units += demand_size - stock before stock = 0 (after it, stock is 0 and the result would be wrong). Print lost_units at the end.