Transcript Handling Multiple Loaders

Handling Multiple Loaders
©Dr. B. C. Paul 2000
Revised 2008
The Binomial Formula is widely published in works on
statistics. These slides contain screen shots of binomial
calculations worked in Excel (a Microsoft Product) and
Caterpillar’s FPC program.
Availability

Truck and Loader Systems
require two things to work at
the same time
• If you have one truck available
85% of the time
• One loader available 85% of the
time
• Combination is available
0.85*0.85= 0.7225 or 72.25%
Problem Comes with
Multiples

If I have 5 trucks and 1 loader (like fleet
#3)
•
•
•
•
•
•

I could have 5 trucks running
4 trucks running
3 trucks running
2 trucks running
1 truck running
0 trucks running
The number of trucks running raises or
lowers the extend of over or undertrucking
• Changes the amount of wait time for a truck to
load
Getting More Complex

With just one loader
• The truck fleet either can work or
cannot


Gets more complicated with
two or more loaders
Key to truck production is
Binomial Distribution
Binomial Distributions


If a result is 0 or 1 and a certain
proportion are 1s and a certain 0’s
called a binomial distribution
Used to model truck and loader
systems
• Let 1 mean truck or loader available
• Let 0 mean not available
• Let the proportion of 1’s be equal to the
proportion of time a truck or loader is
available.
Back to the Truck Setting

Our trucks have 85% availability

We have 6 trucks (5 for the Cable shovel
case)
• We set p for our binomial distribution equal to
0.85
• We set q (not available) to 0.15
•
•
•
•
•
•
•
6
5
4
3
2
1
0
trucks running
trucks running
trucks running
trucks running
trucks running
truck running
trucks running
What is the Probability of
having 5 trucks running?

Formula is
n
• N is total number
of trucks
• n is the number
running

p *q
N n
Plug and Chug
0.85 * 0.15  0.0665557
5
1
Of Course There is More
Than One Configuration for
5 Trucks Running


For simple cases we can count the
number of ways it could happen
For more involved cases we can use
the binomial coefficient
N!
( N  n)!
n!
N! means N factor
For N=6, N! =
6*5*4*3*2*1
For Our 5 Trucks Running
Case


Plug and Chug
6!
1!
5!
6
Probability of a situation with 5 trucks
running is 0.0665557 with 6 ways it could
happen
• 0.066557 * 6 = 0.3993 or 39.93% of the time
Can Use the Binomial
Formula to Tell How Much
of Time We’ll have what
Available
Trucks
0
1
2
3
4
5
6
0.85
6
1.13906E-05
6.45469E-05
0.000365766
0.002072672
0.011745141
0.066555797
0.377149516
1
6
15
20
15
6
1
1.13906E-05
0.000387281
0.005486484
0.041453438
0.176177109
0.399334781
0.377149516
½% for 2
4.14% for 3
17.62% for 4
39.93% for 5
37.71% for 6
If You Suspect I Can Do
This for Two Loaders Your
Right
Available
Loaders
0
1
2
0.85
2
0.0225
0.1275
0.7225
1
2
1
0.0225
0.255
0.7225
2.25% all out
25.5% just 1
72.25% all
So How Do I Deal With
Both a Once?
Trucks
Loaders
0
1
2
0.0225
0.255
0.7225
0
1.13906E-05
2.56289E-07
2.90461E-06
8.22973E-06
1
0.000387281
8.71383E-06
9.87567E-05
0.000279811
2
0.005486484
0.000123446
0.001399054
0.003963985
3
0.041453438
0.000932702
0.010570627
0.029950109
4
0.176177109
0.003963985
0.044925163
0.127287962
5
0.399334781
0.008985033
0.101830369
0.288519379
Cross Multiply it Out to get ratio of time you will
Have any number of trucks or loaders.
6
0.377149516
0.008485864
0.096173126
0.272490525
Checking FPC
You can see FPC gets
The production with
6 trucks and 2 loaders
Good Program – Bad Screw
Up
Fleet Availability is
0.85 * 0.85 = 0.7225
(it is not true that all or
None of the trucks are
Available or all or none
Of the loaders)
At 100% availability
Production is 7,658,623
•0.7225 = 5,533,354
•WOOPS
How Bad is the Error
Proportion of time in that State
Trucks
Loaders
0
1
2
0.0225
0.255
0.7225
0
1.13906E-05
2.56289E-07
2.90461E-06
8.22973E-06
1
0.000387281
8.71383E-06
9.87567E-05
0.000279811
2
0.005486484
0.000123446
0.001399054
0.003963985
3
0.041453438
0.000932702
0.010570627
0.029950109
4
0.176177109
0.003963985
0.044925163
0.127287962
5
0.399334781
0.008985033
0.101830369
0.288519379
6
0.377149516
0.008485864
0.096173126
0.272490525
X Annual Production achievable in that state
Trucks
Loaders
0
0
0
0
0
1
2
1
0
1,450,167.10
1,450,167.10
2
0
2,758,797.90
2,900,334.21
3
4
5
6
0
0
0
0
3,829,311.25 4,493,774.70 4,782,294.08 4,881,884.52
4,305,691.15 5,517,595.79 6,625,813.49 7,658,622.50
= Production Achieved by that State
0
0
0
0
143.2137444
405.7722759
0
3859.705901
11496.88119
0
0
0
0
40478.21922 201883.5604 486982.7719 469506.0974
128955.9175 702323.5206 1911675.597 2086902.066
Sum of Which is Annual Production 6,045,000 tons
Commentary

We Suffered a 10% Error
• Not negligible

Why did the answer change
• Because we neglected partial
credit for part of the fleet being
available
• Equipment ratio and bunching is
non-linear
Coping


Some of Caterpillar’s Cycle
times are optimistic
Program gets closer because
under-estimating availability
covers part of over-estimating
production
• Over-estimated production leads
to more TMPH problems being
predicted
How to Correct FPC





Use a Binomial Probability
Spreadsheet (as I did)
Run FPC at 100% availability in
each state
Enter them in a table
Multiply the production table by
the probability table
Sum up the answer.