Take the 2-minute tour ×
Arqade is a question and answer site for passionate videogamers on all platforms. It's 100% free, no registration required.

Does the SCV repair scale proportionally for time and resources?

This question suggests that repair speed varies from building to building, but what about multiple SCVs repairing the same building?

Assuming I have 2 Bunkers that have 50% life and 2 SCVs to repair them. Is there a time or resources difference between:

  • 1st SCV repairs 1st Bunker and 2nd SCV repairs 2nd Bunker.
  • Both SCVs repair 1st Bunker then 2nd Bunker.
share|improve this question

2 Answers 2

up vote 5 down vote accepted

Using more SCV's to repair a building only increases the repair speed, the actual resources used is the same as if you had used only one SCV.

However, there is an opportunity cost in using more than one SCV for the job, because those extra SCV's would have been mining otherwise. You're basically paying those extra minerals to repair your building faster. Whether or not that's worth it is up to you.

share|improve this answer
1  
Since they also repair faster, this means each SCV is off the line for less time. The opportunity cost is just the travel time from the line to the repair site. If you're repairing something you can easily move near your line, you can save some of this by having the travel time taken care of by the damaged unit/building. –  PeterL Aug 30 '12 at 15:24
1  
Plus the danger of the SCV dying I guess, think of SCV's mass repairing a planetary fortress under siege tank fire. –  Blindy Aug 30 '12 at 15:29

Each worker repairs at a fixed HP/resource rate. Applying more workers to one building repairs them faster. The resources spent on the repair depend on the original price of the building and the amount of HP that needs to be restored.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.