MATH 3170 Assignment #6 Solutions (20 marks)
Questions to be marked: 4.10.4, 4.14.4, 4.16.6 and 4.RP.19.
4.10.4 (5 marks)
Here is just one suggestion:
MODEL:
SETS:
PRODUCTS/1..3/:MADE,PROFIT;
RESOURCES/1..3/:AVAIL;
RESPRO(RESOURCES,PRODUCTS):USAGE;
ENDSETS
MAX=@SUM(PRODUCTS(I):PROFIT(I)*MADE(I));
@FOR(RESOURCES(I):@SUM(PRODUCTS(J):USAGE(I,J))* MADE(J)<=AVAIL(I));
DATA:
PROFIT= 800,1500,2500;
AVAIL=50,10 150;
USAGE= 2,3,5
0.3,0.7,0.2
10,12,20;
ENDDATA
END
4.14.4 (5 marks)
max z = z' + z''
s.t.
4x1 + x24
2x1 - x2
2x1 - 3x2 = z' - z''
All variables non-negative
First note that in any basic feasible solution, z' and z'' cannot
both be positive. Then observe that if 2x1 - 3x2>0, then z'' = 0
and the objective function will equal z' + z'' = z' = 2x1 - 3x2 =
|2x1 - 3x2| while if 2x1 - 3x2<0, z' = 0 and z'' = |2x1 - 3x2|, so
the objective function will equal z' + z'' = z'' = |2x1 - 3x2|.
4.16.6 (5 marks)
Let xip = hours partners work on job i
xis = hours seniors work on job i
xij = hours juniors work on job i
ph = partners hired
sh = seniors hired
jh = juniors hired
min P1s1- + P2s2+ + P3s3+ + P4s4+
st 160x1p + 120x2p + 110x3p + 120x1s + 90x2s +
50x2j + 40x3j + s1- - s1+ =68,000
ph + s2- - s2+ = 1
sh + s3- - s3+ = 3
jh + s4- - s4+ = 5
x1p + x2p + x3p5(40) + 40ph
x1s + x2s + x3s5(40) + 40sh
x2j + x3j5(40) + 40jh
x1p + x1s = 500
x2p + x2s + x2j = 300
x3p + x3s + x3j = 100
All variables nonnegative
70x3s +
4.16.7
Let xim = Number of students taking marketing from
professor i; xif = number of students taking finance
from professor i; xip = number of students taking
production from professor i; xis = number of students
taking statistics from professor i. Then we wish to
solve the following problem:
min s1- + s2- + s3- + s4st
7x1m + 7x2m + 3x3m + 5x4m +
5x1f + 8x2f + 5x3f + 5x4f +
8x1p + 9x2p + 7x3p + 6x4p +
2x1s + 4x2s + 9x3s + 7x4s +
x1m + x2m + x3m + x4m = 200
x1f + x2f + x3f + x4f = 200
x1p + x2p + x3p + x4p = 200
x1s + x2s + x3s + x4s = 200
x1m + x1f + x1p + x1s = 200
x2m + x2f + x2p + x2s = 200
x3m + x3f + x3p + x3s = 200
x4m + x4f + x4p + x4s = 200
All variables nonnegative
s1s2s3s4-
-
s1+
s2+
s3+
s4+
=
=
=
=
1200
1200
1200
1200
We assume that it is okay to have the average
“effectiveness” to be above 6 for any course. But since
the question does not explicitly say so, we will also
accept s1- + s2- + s3- + s4- + s1+ + s2+ + s3+ + s4+ as a correct
objective function (so that high effectiveness is also
penalized).
4.17.4
The excel file cannot be posted here. It is available upon request (See the
file “S4_17_4.xls”).
4.RP.19 (5 marks)
Let xs be the variable
leaving the basis. Then
(Coefficient
(Coefficient
(Coefficient
(Coefficient
entering the basis and xr the variable
of
of
of
of
xs
xr
xr
xs
in
in
in
in
Current Row 0) <0
Current Row 0) = 0
Pivot Row) = 1
Pivot Row) > 0
After the pivot,
(Coefficient of xr in New Row 0) =
-(Coefficient of xs in Current Row 0)
---------------------------------------- >0
(Coefficient of xs in Current Pivot Row)
Thus after the pivot, xr will have a positive coefficient in row 0
and will not reenter the basis. On a later pivot, however, it is
possible that xr may have a negative coefficient in Row 0 and again
re-enter the basis.