Question

The domain of this question is scheduling operations on constrained hardware. The resolution of the result is the number of clock cycles the schedule fits within. The search space grows very rapidly where early decisions constrain future decisions and the total number of possible schedules grows rapidly and exponentially. A lot of the possible schedules are equivalent because just swapping the order of two instructions usually result in the same timing constraint.

Basically the question is what is a good strategy for exploring the vast search space without spending too much time. I expect to search only a small fraction but would like to explore different parts of the search space while doing so.

The current greedy algorithm tend to make stupid decisions early on sometimes and the attempt at branch and bound was beyond slow.

Edit:
Want to point out that the result is very binary with perhaps the greedy algorithm ending up using 8 cycles while there exists a solution using only 7 cycles using branch and bound.

Second point is that there are significant restrictions in data routing between instructions and dependencies between instructions that limits the amount of commonality between solutions. Look at it as a knapsack problem with a lot of ordering constraints as well as some solutions completely failing because of routing congestion.

Clarification:
In each cycle there is a limit to how many operations of each type and some operations have two possible types. There are a set of routing constraints which can be varied to be either fairly tight or pretty forgiving and the limit depends on routing congestion.

Answer 1

Integer linear optimization for NP-hard problems

Depending on your side constraints, you may be able to use the critical path method or
(as suggested in a previous answer) dynamic programming. But many scheduling problems are NP-hard just like the classical traveling sales man — a precise solution has a worst case of exponential search time, just as you describe in your problem.

It’s important to know that while NP-hard problems still have a very bad worst case solution time there is an approach that very often produces exact answers with very short computations (the average case is acceptable and you often don’t see the worst case).

This approach is to convert your problem to a linear optimization problem with integer variables. There are free-software packages (such as lp-solve) that can solve such problems efficiently.

The advantage of this approach is that it may give you exact answers to NP-hard problems in acceptable time. I used this approach in a few projects.

As your problem statement does not include more details about the side constraints, I cannot go into more detail how to apply the method.

Edit/addition: Sample implementation

Here are some details about how to implement this method in your case (of course, I make some assumptions that may not apply to your actual problem — I only know the details form your question):

Let’s assume that you have 50 instructions cmd(i) (i=1..50) to be scheduled in 10 or less cycles cycle(t) (t=1..10). We introduce 500 binary variables v(i,t) (i=1..50; t=1..10) which indicate whether instruction cmd(i) is executed at cycle(t) or not. This basic setup gives the following linear constraints:

v_it integer variables
0<=v_it; v_it<=1;       # 1000 constraints: i=1..50; t=1..10
sum(v_it: t=1..10)==1   # 50 constraints:   i=1..50

Now, we have to specify your side conditions. Let’s assume that operations cmd(1)…cmd(5) are multiplication operations and that you have exactly two multipliers — in any cycle, you may perform at most two of these operations in parallel:

sum(v_it: i=1..5)<=2    # 10 constraints: t=1..10

For each of your resources, you need to add the corresponding constraints.

Also, let’s assume that operation cmd(7) depends on operation cmd(2) and needs to be executed after it. To make the equation a little bit more interesting, lets also require a two cycle gap between them:

sum(t*v(2,t): t=1..10) + 3 <= sum(t*v(7,t): t=1..10)   # one constraint

Note: sum(t*v(2,t): t=1..10) is the cycle t where v(2,t) is equal to one.

Finally, we want to minimize the number of cycles. This is somewhat tricky because you get quite big numbers in the way that I propose: We give assign each v(i,t) a price that grows exponentially with time: pushing off operations into the future is much more expensive than performing them early:

sum(6^t * v(i,t): i=1..50; t=1..10) –> minimum. # one target function

I choose 6 to be bigger than 5 to ensure that adding one cycle to the system makes it more expensive than squeezing everything into less cycles. A side-effect is that the program will go out of it’s way to schedule operations as early as possible. You may avoid this by performing a two-step optimization: First, use this target function to find the minimal number of necessary cycles. Then, ask the same problem again with a different target function — limiting the number of available cycles at the outset and imposing a more moderate price penalty for later operations. You have to play with this, I hope you got the idea.

Hopefully, you can express all your requirements as such linear constraints in your binary variables. Of course, there may be many opportunities to exploit your insight into your specific problem to do with less constraints or less variables.

Then, hand your problem off to lp-solve or cplex and let them find the best solution!