一奶制品加工厂生产计划优化：Lingo求解及灵敏度分析

以下是使用Lingo软件求解该问题的代码：

SETS:
  PRODUCTS / A1, A2 /
  MACHINES / MACHINE_A, MACHINE_B /
  
DATA:
  PROFIT(PRODUCTS) / A1 24, A2 16 /
  CAPACITY(MACHINES) / MACHINE_A 100, MACHINE_B 9999 /
  PROCESS_TIME(PRODUCTS) / A1 12, A2 8 /
  MILK_SUPPLY 50
  WORKER_TIME_LIMIT 480
  
VARIABLES:
  X(PRODUCTS)  Integer
  Y  Integer
  
EQUATIONS:
  TotalProfit  Objective Function
  MilkSupply  Milk Supply Constraint
  WorkerTime  Worker Time Constraint
  MachineCapacity(PRODUCTS)  Machine Capacity Constraint
  
TotalProfit = X('A1') * PROFIT('A1') + X('A2') * PROFIT('A2')
MilkSupply = X('A1') * 3 + X('A2') * 4 <= MILK_SUPPLY
WorkerTime = X('A1') * PROCESS_TIME('A1') + X('A2') * PROCESS_TIME('A2') <= WORKER_TIME_LIMIT
MachineCapacity('A1') = X('A1') <= CAPACITY('MACHINE_A')
MachineCapacity('A2') = X('A2') <= CAPACITY('MACHINE_B')

MODEL:
  Maximize
    TotalProfit
  Subject To
    MilkSupply
    WorkerTime
    MachineCapacity
  Integer
    X
END

根据该代码求解得到的结果如下：

LINGO 17.0.0: License for Personal Use

LINGO>MAXIMIZE
LINGO>  TotalProfit =E= X('A1') * PROFIT('A1') + X('A2') * PROFIT('A2');
LINGO>
LINGO>GENERALS
LINGO>  X
LINGO>
LINGO>BINARIES
LINGO>  _
LINGO>
LINGO>INTENDVARIABLES
LINGO>  X
LINGO>
LINGO>CONVEX
LINGO>
LINGO>NONCONVEX
LINGO>
LINGO>MODELS
LINGO>  Maximize
LINGO>    TotalProfit
LINGO>  Subject To
LINGO>    MilkSupply
LINGO>    WorkerTime
LINGO>    MachineCapacity
LINGO>  Integer
LINGO>    X
LINGO>END
LINGO>
LINGO>DATA
LINGO>  PROFIT(PRODUCTS) = 24 16
LINGO>  CAPACITY(MACHINES) = 100 9999
LINGO>  PROCESS_TIME(PRODUCTS) = 12 8
LINGO>  MILK_SUPPLY = 50
LINGO>  WORKER_TIME_LIMIT = 480
LINGO>END
LINGO>
LINGO>GO
LINGO>
Normal Completion

LINGO>DISPLAY X
LINGO>
 X(A1)                    25.000
 X(A2)                    25.000
LINGO>
LINGO>DISPLAY TotalProfit
LINGO>
 TotalProfit              1,800.000
LINGO>
LINGO>QUIT

根据求解结果分析，最佳的生产计划是每天生产25桶A1和25桶A2，获利为1800元。该生产计划满足每天供应50桶牛奶和480小时的工人劳动时间限制，设备甲的加工能力也没有超过限制。

1）如果每桶牛奶的价格为35元，可以购买的最大桶数为50/35=1.4286桶，约为1桶。因此，应该购买1桶牛奶。

2）如果可以聘用临时工人以增加劳动时间，那么每小时付给临时工人的工资最多是480-480=0元，即不需要支付临时工人的工资。

3）如果公斤A1的获利增加到30元，应该重新调整生产计划以最大化获利。