如何在AMPL+CPLEX中用二进制变量建模机器效率阶跃下降？

I’ll walk you through exactly how to model this 10% efficiency drop after 100 hours of machine operation using binary variables. We’ll build on your existing AMPL setup (3 machines, 5 parts) with a complete, runnable example.

Core Idea #

Each machine has two operational states:

• High-efficiency state: Total runtime ≤ 100 hours, producing at full capacity
• Low-efficiency state: Total runtime > 100 hours, producing at 90% capacity

We use a binary variable y[m] to track if machine m has entered the low-efficiency state (y[m] = 1 if yes, 0 otherwise). Then split production time into two variables: one for high-efficiency hours, one for low-efficiency hours, and link them to y[m] via constraints to enforce the state switch.

Complete AMPL Code #

Model File (production.mod) #

# Define parameters
param M;  # Number of machines
param N;  # Number of parts
param r{M, N};  # Base production rate (parts per hour) for machine m making part n
param d{N};  # Demand for each part
param T{M};  # Total available hours for each machine

# Define variables
var x1{M, N} >= 0;  # High-efficiency hours: machine m making part n (max 100 total per machine)
var x2{M, N} >= 0;  # Low-efficiency hours: machine m making part n (only after 100 hours)
var y{M} binary;    # 1 if machine m has used >100 hours, 0 otherwise

# Objective: Minimize total machine runtime (adjust to your goal, e.g., minimize cost)
minimize Total_Runtime: sum{m in 1..M, n in 1..N} (x1[m,n] + x2[m,n]);

# Constraints
subject to Demand_Satisfaction{n in 1..N}:
    sum{m in 1..M} (r[m,n] * x1[m,n] + 0.9 * r[m,n] * x2[m,n]) >= d[n];

subject to Max_High_Efficiency_Time{m in 1..M}:
    sum{n in 1..N} x1[m,n] <= 100;  # High-efficiency can't exceed 100 hours per machine

subject to Low_Efficiency_Only_If_Over_100{m in 1..M}:
    sum{n in 1..N} x2[m,n] <= (T[m] - 100) * y[m];  # Can't use low-efficiency hours unless y[m]=1

subject to Trigger_Low_Efficiency{m in 1..M}:
    sum{n in 1..N} (x1[m,n] + x2[m,n]) >= 100 * y[m];  # If y[m]=1, total time must hit at least 100

subject to Total_Available_Time{m in 1..M}:
    sum{n in 1..N} (x1[m,n] + x2[m,n]) <= T[m];  # Don't exceed machine's total available hours

Data File (production.dat) #

data;
param M := 3;  # Your existing machine count
param N := 5;  # Your existing part count

# Example production rates (replace with your actual data)
param r:
    1   2   3   4   5 :=
1   10  15  8   12  9
2   12  10  14  7   11
3   9   13  11  15  8;

# Example part demands
param d :=
1   500
2   700
3   600
4   800
5   650;

# Example total available hours per machine
param T :=
1   150
2   200
3   180;

Key Explanations #

• Binary Variable y[m]: Acts as a state switch. When y[m] = 0, the Low_Efficiency_Only_If_Over_100 constraint forces all x2[m,n] to 0 (no low-efficiency hours allowed). When y[m] = 1, this constraint unlocks low-efficiency hours up to the machine’s remaining available time after 100 hours.
• Efficiency Calculation: The Demand_Satisfaction constraint uses the full production rate for x1 (high-efficiency hours) and a 90% rate for x2 (low-efficiency hours) to reflect the step drop.
• Trigger Constraint: Trigger_Low_Efficiency prevents the solver from setting y[m] = 1 unnecessarily—it ensures the machine only switches to low efficiency if it actually uses at least 100 hours.

How to Run #

1. Save the model and data files
2. In AMPL, execute these commands:

   model production.mod;
   data production.dat;
   option solver cplex;
   solve;
   display x1, x2, y;
   

This will output the optimal production plan that accounts for the efficiency drop after 100 hours of machine operation.

内容的提问来源于stack exchange，提问作者Madras