Download Simulation Formulas Worksheet with Crystal Ball Answers

Spreadsheet Formulas for Simulation Worksheet
Give a formula or a spreadsheet section and a formula to simulate:
1. The cost for a repair is uniformly distributed between $50 and $1000.
2. The number of customers averages 35, but could be 15 above or below that.
(Symmetrical triangular distribution)
3. Customer Johnson either orders 50, 100, or 150 units. Twenty-five percent of the time,
he orders 50, 35 percent of the time, 100, and the rest of the time, 150.
4. One fifth of the time, the machine breaks down. (Let 0 = no breakdown, 1= a
breakdown)
5. If the machine is broken down, the cost to repair it is normally distributed with a mean
of $350 and a standard deviation of $100.
6. The number of units demanded is between 1 and 20. (symmetrical triangular)
7. The amount of rainfall is normally distributed with a mean of 27.5 inches and a
standard deviation of 5.2 inches.
8. Half of the employees can perform the task in the regular amount of time. One fourth
take twice as long and one fourth take three times as long.
9. Historically, the interest rate has varied uniformly from 5% to 19%.
10. The number of customers is negative-exponentially distributed with a mean of 117
customers per hour.
Spreadsheet Formulas for Simulation Worksheet answers with Crystal Ball
1. The cost for a repair is uniformly distributed between $50 and $1000.
=CB.Uniform(50,100)
2. The number of customers averages 35, but could be 15 above or below that.
=CB.Triangular(20,35,50)
3. Customer Johnson either orders 50, 100, or 150 units. Twenty-five percent of the time,
he orders 50, 35 percent of the time, 100, and the rest of the time, 150.
A
1
2
3
4
B
ordered
50
100
150
C
probability
.25
.35
.40
=CB.Custom(B2:C4)
4. One fifth of the time, the machine breaks down. (Let 0 = no breakdown, 1= a
breakdown)
A
1
2
3
B
breakdown
0
1
C
probability
.80
.20
=CB.Custom(B2:C3)
5. If the machine is broken down, the cost to repair it is normally distributed with a mean
of $350 and a standard deviation of $100.
=CB.Normal(350,100)
6. The number of units demanded is between 1 and 20. (symmetrical triangular)
=INT(CB.Triangular(1,10.5,20))
7. The amount of rainfall is normally distributed with a mean of 27.5 inches and a
standard deviation of 5.2 inches.
=CB.Normal(27.5,5.2)
8. Half of the employees can perform the task in the regular amount of time. One fourth
take twice as long and one fourth take three times as long.
A
1
2
3
4
B
time multiplier
1
2
3
C
probability
.50
.25
.25
=CB.Custom(B2:C4)
9. Historically, the interest rate has varied uniformly from 5% to 19%.
=CB.Uniform(.5,.19)
10. The number of customers is negative-exponentially distributed with a mean of 117
customers per hour.
=CB.Exponential(117)