Question

Problem 16-13 (Algorithmic)

The wedding date for a couple is quickly approaching, and the wedding planner must provide the caterer an estimate of how many people will attend the reception so that the appropriate quantity of food is prepared for the buffet. The following table contains information on the number of RSVP guests for the 145 invitations. Unfortunately, the number of guests does not always correspond to the number of RSVPed guests.

Based on her experience, the wedding planner knows it is extremely rare for guests to attend a wedding if they notified that they will not be attending. Therefore, the wedding planner will assume that no one from these 50 invitations will attend. The wedding planner estimates that the each of the 25 guests planning to come solo has a 75% chance of attending alone, a 20% chance of not attending, and a 5% chance of bringing a companion. For each of the 60 RSVPs who plan to bring a companion, there is a 90% chance that she or he will attend with a companion, a 5% chance of attending solo, and a 5% chance of not attending at all. For the 10 people who have not responded, the wedding planner assumes that there is an 80% chance that each will not attend, a 15% chance each will attend alone, and a 5% chance each will attend with a companion.

RSVped Guests	Number of invitations
0	50
1	25
2	60
No response	10

1. Assist the wedding planner by constructing a spreadsheet simulation model to determine the expected number of guests who will attend the reception. Round your answer to 2 decimal places.
   
   _________ guests
   
2. To be accommodating hosts, the couple has instructed the wedding planner to use the Monte Carlo simulation model to determine X, the minimum number of guests for which the caterer should prepare the meal, so that there is at least a 90% chance that the actual attendance is less than or equal to X. What is the best estimate for the value of X? Round your answer to the nearest whole number.
   
   __________ guests

Answer 1

a)

You have a probability distribution for each of the 4 sets of invitations. The 4 sets are 50,25,60,10.

Ignore the set of 50. Nobody is going to turn up.

For the set of 25 the probability distribution is

1	0.75
-	0.20
2	0.05

Generate a sample of 25 numbers. This is one trial. do 1000 trials (or simulations) in this way. Add the sum at the end of each row

Generate a sample of 60 numbers using the same way of 25 set in a seperate sheet. This is one trial. do 1000 trials. Probability distribution for 60 is

2	0.90
1	0.05
-	0.05

Generate a sample of 10 numbers using the same way of 25 set in a seperate sheet. This is one trial. do 1000 trials.

-	0.80
1	0.15
2	0.05

Accumulate all the 1000 trials' sums for each of the 3 sets as follows.

Trial	50	25	60	10	Total
1	0	19	106	4	129
2	0	20	117	2	139
3	0	21	115	0	136
4	0	20	111	5	136
5	0	26	108	1	135
6	0	20	109	0	129

Now you have 1000 rows with total ranging from a, b. For my trial, the min was 117 and max was 149, average was 135

b)

To find out the number of guests X or below which has 90% chance

P(guests <=X) = 90%

Forumulate in excel using the formula

norm.inv(0.9,135,4.8684)

as the mean is 135 and the std devn is 4.8684 for 1000 trials.

X is 141 (approx)

best estimate for the value of x is

P (x<=141) = 93% and P (x<=140) =89%. this comes closer as P (x<=141) is 90%

All the other options

P(x>=140) = 93% and P(x>=141) = 93% is wrong, as the greater than sign is wrong, cumulative probability means till that point and hence lesser than sign.

P(x>=140) = 89% and P(x>=141) = 93% is wrong, as P(x>=140) being 89% is wrong.

P(x<=140) = 89% and P (x>=141)=93% is wrong.as p(x>=141) being 93% is wrong