HP50G SIMPLEX Version14b Max Min Pivot Algorithm, multiple/unbounded, unfeasible sol

[Gil]

Remember
as explained in
previous Note 11c above

When choosing for approximation mode (question 4/5 in OPTIONS prompted), you might get unexpected, special solutions.

Example
Go (press) DATA directory (last menu page).
Go then to (press) SPECIAL directory.
Choose last variable (last menu page) called SPEC.SPEC.
You should get
[[ 1 -1 0 0 1 0 0 'E' 3 ]
[ -3 2 9 2 0 -2 0 'E' 22 ]
[ 0 1 0 0 0 -1 0 'E' 2 ]
[ 4 3 5 0 1 0 -1 'E' 4 ]
[ -1 0 3 -1 0 0 0 0 'Max' ]]
Press then —>GO.

a) Press 5 times ENTER.
The full solution is then given by
{lambda × [ 0 2 2 0 5 0 17 ]
+ (1-lambda) [ 5 2 '11/3' 0 0 0 '121/3' ]
+ mu1 × [ 1 1 '1/3' 0 0 1 '26/3' ]
+ mu2 × [ 0 1 0 0 1 1 4 ]},
with 0<=lambda<=1
& mu i >= 0

b) Press now INPUT.last (at 1st menu page)
& launch (press) a 2nd time the main program —>GO.
Be careful and, to the questions/prompts that appear:
- answer 1 for 1st question (MAX option)
- answer 0 for 2nd question (no multiple solution)
- answer 1 for 3rd question here (automatic mode)
- answer 0 for 4th question here (no fractions —> approximation)
- answer 1 for 5th question (full stacks & details).
The full and unrecognisable solution is then given as:
[ 0 8.28947368439 2 0 11.2894736843 6.28947368439 42.1578947369 ].

c) In fact, the above solution b) is a special case of a).
Indeed:

Solution b) [ 0 8.28947368439 2 0 11.2894736843 6.28947368439 42.1578947369 ]
=?
solution a)
lambda × [ 0 2 2 0 5 0 17 ]
+ (1-lambda) × [ 5 2 '11/3' 0 0 0 '121/3' ]
+ mu1× [ 1 1 '1/3' 0 0 1 '26/3' ]
+ mu2 × [ 0 1 0 0 1 1 4 ]

Set lambda = 1, mu1 =0, mu2 =6.28947368439.
Then solution b)
[ 0 8.28947368439 2 0 11.2894736843 6.28947368439 42.1578947369 ]
=?
solution a)
1 × [ 0 2 2 0 5 0 17 ]
+ 6.28947368439 × [ 0 1 0 0 1 1 4 ]

Or [ 0 8.28947368439 2 0 11.2894736843 6.28947368439 42.1578947369 ]
=?
[ 0 2 2 0 5 0 17 ]
+ [ 0 6.28947368439 0 0 6.28947368439 6.28947368439 25.1578947376 ]

And we see that the expected (or unexpected!) equality is, to the rounding errors, indeed fully fulfilled.