Transportation Problem Optimal Solution with MODI and ZQ (Total Cost)

Transportation Video 6 of 7
Example 2

Two examples of the Linear Programming Transportation Problem are given. The first is videos 1, 2, and 3. The second is videos 4, 5, and 6.
* The Least Cost Method (sometimes called the Low Cost Method or the Minimum Cost Method) is use to find the initial feasible solution. This method typically gets you closer to a solution than the Northwest Corner Method.
* The Stepping Stone Method is the second video in the series and is a good teaching tool to aid in the understanding of how to reach an optimal solution.
* The 3rd video in each Series is the MODI or Modified Distribution Method and is the best way to solve a Transportation Problem.

How to solve a transportation problem using the MODI (Modified Distribution) method and finding the optimal solution. The total shipping cost, ZQ, is calculated.

All rights reserved, copyright 2012 Ed Dansereau

[EdDansereau]

In cell CE, I wrote 150 on board and should have been 50 (200-150 = 50). The calculation for ZQ is correct.
From To Cost Qty Sub Total
A D 6 150 900
A F 9 50 450
B E 7 200 1400
C F 12 150 1800
C E 10 50 500
Total 5050

Taking a guess at your sequence you numbers are cells AD + CF + BE + CE + AF. I agree with cells AD, CF, and AF. Cell BE has a cost of $7 per unit and CE has a cost of $10 per unit and not $12.

Thank you very much. I enjoy the intellectual exchange.

[driphard]

In germany we call people like you "Ehrenmann"

[exzellentz]

For some reason that was more understandable than our MODI lesson in german :D
Thanks a lot! :)

[ayannahwoods]

So Amazing! Helped me through a rough study session. Great clear and relatable examples. Thanks so much.

[riveroaguilar]

I believe there are some errors. CORRECT ME IF I AM WRONG

Once you have calculated all the U and V, and you analyze that you should add more units in C-F (since = -1) . THEN IN C-E you need to ad 50 and NOT 150.
THIS AFFECTS THE FINAL MODI COST = 150x6 + 150x12 + 200x12 +50x12 +50x9 = 6050,
Therefore the TOTAL SAVINGS ARE 6050-5450= 600 $

[utukufunkombezi]

I'm writing today, this has been more helpful to me, thanks

[ShEng]

So it is just like this? Why did my lecturer made a big deal out of it

[EdDansereau]

Yes, It appears I wrote the number incorrectly on the board but used the correct number (50 and 150) in calculating ZQ. I will correct in the coming weeks. Thank you. I cracked under pressure! LOL.

[shamsudeenakodu]

I feel cell B to F should be blank and C to F should be 150. I dont know if i am correct.
Please guide. Just learning this.

[robertkafulu]

But when you have more than one negative values in the NET EVAL do you create loops around all of them ?

[EdDansereau]

I will review and get back to you. Thank you for your input. Sorry for the delay, I have been away.

[aimanameen]

The best explanation ever. so much thanks to you.

[ammarsaneej]

in this case there is only one minus @ C-F i.e (-1). how to do optimal solution if there two minus(-) values ?

[kelvinsaah]

Thanks
Please can you use Degeneracy ??

[dzakybima]

oh god thanks your lecture is so easy to understand

[lindzillao]

How do we determine where to take units from and how much units to take? Also, we have homework about this but one U value, specifically U3, ended up being negative. Is that allowed? Thank you!

[robertcarricaburujr.]

Thank you so much for an easy to understand lecture.

[rto]

I do not understand when the evaluation. why the capacity exceeds the limit?

[eglishenmaredi]

THANK YOU VERY MUCH...NOW IT MAKES MORE SENSE. :-)

[niler]

Really nice, lets hope i remember all this for my test on Friday :D