Lab. 3. Zagadnienie dualne i algorytm simplex
2. Mając rozwiązanie zagadnienia pierwotnego - patrz tabela - skonstruuj wydruk z rozwiązaniem dla zagadnienia do mego dualnego.
Decision |
Solution |
Unit Cost |
Total |
Reduced |
Basis |
Allowable |
Allowable | |
Variable |
Value |
Profit e(j) |
Contribution |
Cost |
Status |
Mia c(j) |
Max. c(j) | |
1 |
XI |
0 |
0.9 |
0 |
-0.5667 |
at bound |
-M |
1.4667 |
2 |
X2 |
20.000 |
2.2 |
44.000 |
0 |
basie |
1.8 |
M |
3 |
X3 |
0 |
1.2 |
0 |
-0.2667 |
at boiuid |
-M |
1.4667 |
Objective |
Fiuiction |
(Max.) = |
44.000 | |||||
Constrain t |
Left Hand |
Riglit Hand |
Slack or |
Shadow |
Allowable |
Allowable | ||
Side |
Direction |
Side |
Siu plus |
Price |
Min. RHS |
Mia RHS | ||
1 |
Cl |
160.000 |
80.000.000 |
79.840.000 |
0 |
160.000 |
M | |
2 |
C2 |
600.000 |
£ |
600 000 |
0 |
0.0733 |
0 |
600 000 |
3 |
C3 |
400.000 |
600 000 |
200.000 |
0 |
400.000 |
M | |
4 |
C4 |
1.200.000 |
1.200.000 |
0 |
0 |
1.200.000 |
M | |
5 |
C5 |
60.000 |
£ |
300.000 |
240.000 |
0 |
60.000 |
M |
6 |
C6 |
20.000 |
£ |
400.000 |
380.000 |
0 |
20.000 |
M |
Decision Variable |
Solution Value |
Unit Cost or Profit c(j) |
Total Contribution |
Reduced Cost |
Basis Status |
Allowable Mia c(j) |
Allowable Max. c(j) | |
1 |
Y1 | |||||||
Objective |
Fiuiction |
(......) = | ||||||
Constrain t |
Left Hand Side |
Direction |
Riglit Hand Side |
Slack or Surplus |
Shadow Price |
Allowable Min. RHS |
Allowable Mia RHS | |
1 |
Cl |
1