4th Ed-SG# 4, Chapter 3, Problem 14

[soaringAlone]

Can someone explain how did we arrive at 107 minutes if partial loading were allowed ?

[mithunsam]

Assuming that you understand how we arrive at 80 mins & 64 mins for the two machines...

Machine A takes 80 mins for 1 unit of work
Machine B takes 64 mins for 1 unit of work

How much time would the machines together take for 1 unit of work?

1/80 + 1/64 = 1/h
9/320 = 1/h
320/9 = h

How much time would the machines together take for 3 units of work?

h = (320/9) * 3
h = 320/3 ~ 107 mins

[mithunsam]

One more point...

Book says to divide the load between the machines as 4/9 and 5/9 respectively. However, we can avoid this computation.

We just have to remember that when two machines work together, the fastest they could finish a job is when none of the machines sit idle. That means, both the machines have to work for the same amount of time. In other words, the fastest completion time will depend on the average rate of the machines together. That is all I have done above.

[tim]

thanks..

[soaringAlone]

Thanks mithunsam, You have explained it from a different angle and I get it now.

[tim]

glad to hear it!