Examples of Waiting Lines
Question 1:
We know that customers seeking loans arrive on average every 45 minutes. Since we also assume that the interarrival times are exponentially distributed, we have that X, the inter arrival times between customer has an Exponential distribution with parameter$\lambda =1/45$. The cumulative distribution is computed as
\[F\left( x \right)=\Pr \left( X\le x \right)=\int\limits_{\infty }^{x}{\frac{1}{45}{{e}^{s/45}}ds}=\left\{ \begin{array}{lc} & 1{{e}^{x/45}}\text{ }x\ge 0 \\ & \,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x<0 \\ \end{array} \right. \]To simulate one interarrival time, we need to solve
\[0.353=F\left( x \right)\,\,\Leftrightarrow \,\,\,\,1{{e}^{x/45}}=0.353\,\,\,\,\Rightarrow \,\,\,\,\,{{e}^{x/45}}=0.647\,\,\,\,\]
\[\Rightarrow \,\,\,\,\frac{x}{45}=\ln \left( 0.647 \right)\,\,\,\,\,\Rightarrow \,\,\,\,\,x=45\times \ln \left( 0.647 \right)=19.593\text{ minutes}\]
Now, assume that Y represents the customers that will arrive next hour. We know that Y has a Poisson distribution with mean \(\lambda = 2\). This means that we use the following syntax in StatPro: "=POISSON_(2)"

Question 2:
Now X has an Uniform distribution on [0, 5]. This means that the cumulative distribution is
\[F\left( x \right)=\left\{ \begin{array}{lc} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,x\ge 5 \\ & x/5\,\,\,\,\,\,\,0\le x<5 \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,x<0 \\ \end{array} \right. \]This means that we need to solve:
\[0.461=F\left( x \right)\,\,\,\,\Rightarrow \,\,\,\,0.461=\frac{x}{5}\,\,\,\,\,\Rightarrow \,\,\,\,x=5\times 0.461=2.305\]
(Using StatPro we could do =UNIFORM_(0,5))
Question 3:
The distribution here is normal, so in order to simulate X, the time of pizza delivery time we use the formula
\[X=\mu +{{\Phi }^{1}}\left( 0.817 \right)\times \sigma =10+{{\Phi }^{1}}\left( 0.817 \right)\times 2.8\]
where \({{\Phi }^{1}}\) corresponds to the inverse cumulative distribution. This means that the simulated value is
\[X=10+{{\Phi }^{1}}\left( 0.817 \right)\times 2.8=10+0.903991\times 2.8=12.531\]
Note: The value of \({{\Phi }^{1}}\left( 0.817 \right)\) is calculated using NORMSINV(0.817) in Excel.
Question 4:
A simulation with 1000 replications was performed for 10,000, 10,500, 11,000, 11,500, 12,000, 12,500, 13,000, 13,500 and 14,000 tshirts (See the attach spreadsheet). The summary table of the results is shown below:
TShirts 
Expected Profit 
10,000 
$151,092.5 
10,500 
$152,890.0 
11,000 
$154,265.0 
11,500 
$153,167.5 
12,000 
$155,007.0 
12,500 
$153,335.0 
13,000 
$152,445.0 
13,500 
$149,700.0 
14,000 
$146,192.5 
The maximum profit was achieved with 12,000 Tshirts.
Yield management (revenue management) is a set of comprehensive business policies developed to maximize profit for the capacityconstrained service industries such as hotel, rentalcar, and cruise lines. In fact, airlines were the first to coin the term, yield management, after deregulation permitted airlines to set their own prices to maximize yield from each of their flights.
Discountfare policy
Julie Michaels is responsible for yield management at Tilde Airline (TA), During the recent economic slump, she discovered that airplanes on its Salt LaketoLas Vegas route have been flying with more empty seats than usual. Examining the recent data, Julie found that the demand for fullfare tickets appears to have a normal distribution, with a mean of 60 and a standard deviation of 15.
To stimulate demand, Julie has decided to offer a special, nonrefundable. 14day advancepurchase "gambler fare" for only $98 on a roundtrip ticket. Since the gambler fare is very attracting to gamblers (or riskseeking people), she is confident that all available discount gamblerfare seats will be sold 14 days prior.
The Boeing 737 used by TA has a capacity of 95 passengers in coach cabin as shown in the graph below.
Full Fan
Julie knows that she needs to limit the number of seats that are sold at the discount fare in order to sell MIfare tickets (S138 roundtrip) to passengers who have not made advance travel plans. Set up a simulation model with 1,000 replications to answer the following two questions.
Once a flight has departed, the potential revenue from an empty seat is lost forever. Because of the perishable nature of airline seats, yield management attempts to allocate the fixed capacity of seats on a flight to match the potential demand in the most profitable manner. We take a look at a common practice of the yield management in the airline industries: the discountfare policy.
Question 5
1000 replications were performed for the case of 70 fullfare ticket. Let X be the random number of fullfare clients. The random revenue is
\[R=\left\{ \begin{array}{ll} & 138X+98\times 25\text{ if }X<70 \\ & 138\times 70+98\times 25\,\,\,\,\,\,\,\,\,\text{if }X\ge 70 \\ \end{array} \right.\]Simulating the process 1000 times, we get that the expected revenue is $10,491.71.
Question 6
Now, 1000 replications were performed for the case of F fullfare ticket, for F = 30, 40, 50, 60, 70. Let X be the random number of fullfare clients. The random revenue is
\[R = \left\{ \begin{array}{ll} & 138X+98\times \left( 95F \right)\text{ if }X<F \\ & 138\times F+98\times \left( 95F \right)\,\,\,\,\,\,\,\,\,\,\,\text{if }X\ge F \\ \end{array} \right. \]Simulating the process 1000 times, we get that the following table with expected revenues:
Full Fare Reservation 
Expected Revenue 
30 
$10,493.44 
40 
$10,830.93 
50 
$11,015.65 
60 
$10,931.54 
70 
$10,494.71 
which means that the highest revenue is achieved with 50 full fare seats.
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