We can take a greedy approach, minimum makespan. Given jobs for machines with times t₁, t₂, … , t_n, the total time from start to finish of the slowest machine is as small as possible. Example, three machines with jobs 2, 2, 2, 3, 4, 6:

M₁: 2, 2, 2
M₂: 3, 4
M₃: 6

We can write this as a decision problem. Given the number of machines and a list of job times, and a bound k, is there a way to assign jobs so that Makespan <= k? Makespan is NP-complete, by reduction from Subset Sum! Given a target W and w₁, w₂, … , w_n, split into sets W and S – W where S = sum(w_i). Take |W – (S – W)| = |2W – S| = W₀, a new constant in the list which forces the best solution where both subsets have the same sum = max(W, S – W).

Given a Subset Sum problem, create W₀ = |S – 2W| and let the w’s be the times of jobs. Ask if Makespan works on 2 machines of size <= (S + W₀) / 2.

Claim: “There is a solution to Subset Sum if and only if there is a solution to Makespan.”

Assume a Subset Sum solution. Split the w’s into subsets that sum to W and S – W. W₀ was chosen so we can make them equal, so there exists a Makespan = (S + W₀) / 2, and some machine has a subset of jobs summing to W.

Greedy Strategy: Assign each job to least busy machine, incrementally. How good is this? Call the best Makespan possible T*. The sum(t_i) = total time to be spread over m machines. Some machine must have more than sum(t_i) / m, so T* >= sum(t_i) / m. Consider M_i, the machine with worst time; it’s load is T = T_i. In general, T_j = load on M_j. Our makespan answer is T₁ by the greedy algorithm.

Let t_j be the last time assigned to M_i. At that moment, M_i’s load is T_i – t_j. That must be the smallest load, because we chose it for the last job. So T_i – t_j <= T_k for all k.

For each machine: M(T_i – t_j) <= sum(T_k) = sum(t_j). So T_i – T_j < = sum(t_j) / m &;t= T*. T_i &lt= T* + t_j so T = T_i <= T* + t_j and T* >= worst t_k >= t_j, so T <= T* + T* -> T <= 2 T*.

The greedy algorithm is always within twice the optimal value. If we presort and do the largest jobs first, we can show that T/T* <= 3/2 and T/T* <= 4/3.

Credit to Kevin Canini for the notes!