This preview shows pages 1–3. Sign up to view the full content.
SOLUTIONS TO MP CHAPTER 5 PROBLEMS
SECTION 5.1
1. Typical isoprofit line is 3x
1
+c
2
x
2
=z. This has slope
3/c
2
. If slope of isoprofit line is <2, then Point C
is optimal. Thus if 3/c
2
<2 or c
2
<1.5 the current basis
is no longer optimal. Also if the slope of the
isoprofit line is >1 Point A will be optimal. Thus if
3/c
2
>1 or c
2
>3 the current basis is no longer optimal.
Thus for 1.5
≤
c
2
≤
3 the current basis remains optimal.
For c
2
= 2.5 x
1
= 20, x
2
= 60, but z = 3(20) + 2.5(60) =
$210.
2. Currently Number of Available Carpentry Hours = b
2
=
80. If we reduce the number of available carpentry
hours
we see that when the carpentry constraint moves
past the point (40, 20) the carpentry and finishing
hours constraints will be binding at a point where
x
1
>40. In this situation b
2
<40 + 20 = 60.
Thus for
b
2
<60 the current basis is no longer optimal. If we
increase the number of available carpentry hours we see
that when the carpentry constraint moves past (0, 100)
the carpentry and finishing hours constraints will both
be binding at a point where x
1
<0. In this situation
b
2
>100. Thus if b
2
>100 the current basis is no longer
optimal. Thus the current basis remains optimal for
60
≤
b
2
≤
100. If 60
≤
b
2
≤
100, the number of soldiers and
trains produced will change.
3. If b
3
, the demand for soldiers, is increased then
the current basis remains feasible and therefore
optimal. If, however, b
3
<20 then the point where the
finishing and carpentry constraints are binding is no
longer feasible (it has s
3
<0). Thus for b
3
>20 the
current basis remains optimal. For b
3
>20 the point
where the carpentry and finishing constraints are still
binding remains at (20, 60) so producing 20 soldiers
and 60 trains remains optimal.
4a. If isocost line is flatter than HIM constraint,
point C is optimal. If isocost line is steeper than
HIW, point B is optimal. Thus current basis remains
optimal if 7/2
≤
c
1
/100
≤
1/6 or 50/3
≤
c
1
≤
350.
4b. Current basis remains optimal if 7/2
≤
50/c
2
≤
1/6 or
100/7
≤
c
2
≤
300.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document4c. If we decrease HIW requirement optimal solution
moves towards D = (0, 2). We lose feasibility if HIW
requirement is 2(2) = 4. Increase HIW requirement and
optimal solution moves towards C = (12, 0). We lose
feasibility when HIW requirement = 7(12) = 84. Thus
current basis remains optimal for 4,000,000
≤
HIW
Requirement
≤
84,000,000. If HIW = 28 +
∆
, the optimal
solution is where 7x
1
+ 2x
2
= 28 +
∆
and 2x
1
+ 12x
2
= 24.
This yields x
1
= 3.6 + .15
∆
and x
2
= 1.4 .025
∆
.
4d. If we decrease HIM requirement optimal solution
moves towards A = (4, 0). We lose feasibility if HIM
requirement is 2(4) = 8. Increase HIM requirement and
optimal solution moves towards B = (0, 14). We lose
feasibility when HIM requirement = 14(12) = 168. Thus
current basis remains optimal for 8,000,000
≤
HIW
Requirement
≤
168,000,000. If HIM = 28 +
∆
, the optimal
solution is where 7x
1
+ 2x
2
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '07
 HOCHBAUM

Click to edit the document details