There are five jobs to be assigned to five machines and associated cost matrix is as follows $$ \begin{matrix} \text{Machine} & 1 & 2 & 3 & 4 & 5 \\ \text{Job A} & [11, &17, &8, &16, &20] \\ \text{Job B} & [9, &7, &12, &6, &15] \\ \text{Job C} & [13, &16, &15, &12, &16] \\ \text{Job D} & [21, &24, &16, &28, &26] \\ \text{Job E} & [14, &10, &12, &11, &15] \end{matrix} $$ The question is now: Find the assignment of machines to jobs that will minimize the total cost?

I solved it using the Hungarian method but for job A and D I had only one zero that too in the same column. I don't know how to solve further if this happens.

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    $\begingroup$ Since this looks a lot like a homework question, it would be best to show you intermediate steps of your attempt to solve it $\endgroup$ Jun 18 '19 at 9:20
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    $\begingroup$ As a note, I voted to reject a tag edit of "self-study" because I think that would be a meta-tag. Not sure that we fully settled on that as a policy, but I think we were leaning that direction. (or.meta.stackexchange.com/questions/163/…) Also, welcome to OR.SE, @Tango! $\endgroup$
    – E. Tucker
    Jun 18 '19 at 14:15
  • $\begingroup$ @E. Tucke At Cross Validated stats.stackexchange.com , self-study tag is applied to all homework problems, and even for questions requesting help understanding passages in textbooks, even f being used in self-study outside any courses. $\endgroup$ Jun 18 '19 at 23:12
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    $\begingroup$ @MarkL.Stone Sounds good! The tag seems accurate; it's more that it's a meta-tag. If the community wants to go that direction, that's fine by me. $\endgroup$
    – E. Tucker
    Jun 19 '19 at 12:13

I assume you're applying the matrix version of the algorithm. When you happen to have only one $0$ for A and D the matrix is \begin{align*} \pmatrix{2&9&0&8&8\\2&1&6&0&5\\0&4&3&0&0\\4&8&0&12&6\\3&0&2&1&1} \end{align*} Now continue with Step 3: cover all zeros minimally, and adjust the weights. After that step you will find a solution.

  • $\begingroup$ That is exact matrix I got now when covering all zeros I have r=4 covering lines and since problem has n=5 since r<n. STEP3: now 1 is least uncovered element subtracting 1 from all uncover element and adding 1 to element at intersection of covering line I am left with matrix that has same problem only one zero in A and D but now r=n . Can i use same step again when I have r=n $\endgroup$
    – Tango
    Jun 18 '19 at 14:26
  • $\begingroup$ The least uncovered element will be $2$. Note that line $5$ will be unmarked after step 3. $\endgroup$ Jun 19 '19 at 5:14

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