Clark-Wright Algorithm
1. You will be given the following information (in diagram or in table format).
3
2
1
4
0
0
1
2
3
1
3
2
3
3
3
4
4
5
4
3
6
8
2
This table above is called the _________________________________
2. Develop the Net Savings Matrix
0
1
2
3
1
…..
2
3
4
….. ….. …..
3
Net Savings of going 0 to 1 to 2 to 0 (rather than 0 to 1 and 0 to 2 and back to 0) is:
= D0i + D0j – Dij
= 3+3–3
=3
Distance between 0 and 1 = D0i = D01 = 3
Distance between 0 and 2 = D0j = D02 = 3
Distance between 1 and 2 = Dij = D12 = 3
0
1
2
3
1
…..
2
3
….. …..
3
3
2
4
…..
0
0
5
Net Savings per Route
1 to 3
1 to 4
2 to 3
2 to 4
3 to 4
3+4–4=3
3+3–6=0
3+4–5=2
3 + 3 – 8 = -2 or 0
4 + 3 -2 = 5
This table above is called the _________________________________
3. Run the Optimization
A)
Start with the most inefficient route 3 + 3 + 3 + 3 + 4 + 4 + 3 + 3 = 26 miles
3
2
1
4
3
4
0
3
3
a. Original Net Savings Matrix with the bold values being the original “T”
values.
0
1
2
3
1
2
3
4
2
2
3
2
3
2
2
0
0
5
b. Find the highest number on the Net Savings Matrix and see if the route
satisfies our assumptions.
 Route 3 to 4 has a savings of 5 miles if we don’t go back to the warehouse.
 Does it meet the assumptions?

1st Assumption is that the D0i and D0j do not have a “T” value
of 0
Answer: This is true. In this model, D03 currently has T = 2 and
D04 has T = 2
 2nd Assumption is that D0i and D0j are not on the same path
Answer: This is true.

Reroute the original trip from: 0 – 3 – 0 – 4 - 0 to: 0 – 3 – 4 – 0
3
2
1
4
0
c. Update the Net Savings table with the new T value
0
1
2
3
1
2
3
4
2
2
3
2
3
2
2
0
0
1
Go to the next highest value on the Net Savings Matrix. It is 3 (doesn’t matter
which 3 you choose)
B)
0
1
2
3
1
2
3
4
2
2
3
2
3
2
2
0
0
1
 Does it meet the assumptions?

1st Assumption is that the D0i and D0j do not have a “T” value
of 0
Answer: This is true. In this model, D01 currently has T = 2 and
D02 has T = 2
 2nd Assumption is that D01 and D02 are not on the same path
Answer: This is true.

Reroute the original trip from: 0 – 1 – 0 – 2 - 0 to: 0 – 1 – 2 – 0
3
1
2
4
0
a. Update the Net Savings table with the new T value
0
1
2
3
C)
1
2
3
4
2
2
1
2
3
2
2
0
0
1
Go to the next highest value on the Net Savings Matrix. It is 3 (trips between
1 and 3).
0
1
2
3
1
2
3
4
2
2
1
2
3
2
2
0
0
1
 Does it meet the assumptions?

1st Assumption is that the D0i and D0j do not have a “T” value
of 0
Answer: This is true. In this model, D01 currently has T = 1 and
D03 has T = 1
 2nd Assumption is that D01 and D02 are not on the same path
Answer: This is true.

Reroute the original trip from: 0 – 1 – 0 – 3 - 0 to: 0 – 1 – 3 – 0
3
2
1
4
0
a. Update the Net Savings table with the new T value
0
1
2
3
1
2
3
4
2
2
1
2
1
2
2
0
0
1
Now, take a look at the model. We can eliminate the trip between 0 and 3 because it
serves no purpose. The most efficient route becomes 0 – 2 – 1 – 3 – 4 – 0 or
3 + 3 + 4 + 2 + 3 = 15 miles
3
1
2
4
0