(sorry beforehand I know putting scanned diagrams may seem not-so-professional but this problem is sticking for long and its interesting too)The language corresponding to given regex seems to accepts all strings of 1's with length in multiple of 2 or 3 (i.e.4,6,8,9,10,12,...)I prepared following NFA first:Figure 1Then I followed steps given here to prepare DFA.First I prepared below table to get equivalent DFA steps:Figure 2Then I prepared below DFA, which seems to be quite correct.Figure 3Next to get equivalent states I followed table filling algorithm as explained [here](http://books.google.co.in/books?id=tzttuN4gsVgC&lpg=PP1&pg=PA144#v=onepage&q&f=false)and formed below:(cross between every intersecting column and row indicates two intersecting states are distinguishable / not equivalent)Figure 4But I dont get from above table how can I get below minimized DFA as given in the book solution:Figure 5So in the solution of the textbook, their is one less state (total 6) than my dfa in figure 3 (which has 7 states). I should be able to derive the same (equivalent 6-state dfa) from above triangular table. 
Getting minimum DFA for regular expression (11)*+(111)*
automata;finite automata
The problem is that you did not make state $A$ accepting during $\epsilon$ transition elimination. Since it has $\epsilon$ transitions to accepting states, you should have done so. As a result, the automaton you obtain doesn't accept the empty word as it should. If you fix this, minimization then merges $A$ with $B,D$.