Math 340
Assignment #4
Due Wednesday March 23
1. Consider the LP:
Maximize −x1
−x1
−x1
−x2
−x2
−x2
+x3
−x3
+x3
≤ −1
≤ −2
≤ −1
x1 , x2 , x3 ≥ 0
The final dictionary is
x4
x2
x3
z
= 0 −x6
= 2
= 1 −x6
= −1 −x6
−x1
−x1
−x1
+x5
+x5
+x5
We now consider adding the constraint x1 + x2 + x3 ≤ 1. Use the method of class and use
our dual simplex method to solve. If you find an optimal solution then give the new marginal
values. If you find the dual is unbounded (and hence the primal is infeasible) then give a
parametric set of feasible solutions to the dual whose objective function goes to −∞.
2. By now you should be more comfortable with this sort of question. Assume you are given an
m × n A and a m × 1 b. Show that either
i) There exists an x ≥ 0 with Ax = b
or
ii) There exists a y with AT y ≥ 0 and b · y < 0
but not both.
3. (from an old exam) We seek a minimum cost diet selected from the following three foods.
vitamins/100gms
calories/100gms
minimum (100gms)
food 1 food 2 food 3
13.23
18.4
36
100
125
139
10
10
8
cost $/100gm
3.00
5.00
8.00
We require a diet that has at least 760 units of vitamins and at least 3500 calories. The
minimums are stated in units of 100gms. We let the variable foodi refer to the amount of
food i purchased in units of 100gms. The input to LINDO is:
min 3food1+5food2+8food3
subject to
13.23food1+18.4food2+36food3>760
100food1+125food2+139food3>3500
food1>10
food2>10
food3>8
end
The following is the output from LINDO:
OBJECTIVE FUNCTION VALUE
1) 178.6000
VARIABLE
VALUE
F OOD1
10.000000
F OOD2
10.000000
F OOD3
12.325000
REDUCED COST
0.000000
0.000000
0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2)
0.000000
−0.222222
3)
463.174988
0.000000
4)
0.000000
−0.060000
5)
0.000000
−0.911111
6)
4.325000
0.000000
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF
INCREASE
DECREASE
F OOD1
3.000000
IN F IN IT Y
0.06000
F OOD2
5.000000
IN F IN IT Y
0.911111
F OOD3
8.000000
0.163265
8.000000
RIGHTHAND SIDE RANGES
ROW
2
3
4
5
6
CURRENT ALLOWABLE ALLOWABLE
RHS
INCREASE
DECREASE
760.000000
IN F IN IT Y
119.958984
3500.000000
463.174988
IN F IN IT Y
10.000000
11.768706
9.468493
10.000000
8.461956
8.584380
8.000000
4.325000
IN F IN IT Y
a) There is a special on food 2 reducing the price to $4.10/100gms. Would this change your
purchase strategy? What about a price reduction to $3.10?
b) What is the marginal cost of 10 units of vitamins; namely what is the cost of increasing
the vitamin requirement by 10? Considering the chosen diet as a whole, what is the dollar
cost (approximately only) of the whole diet per 10 units of vitamins obtained. Which cost is
cheaper?
c) Is integrality important in diet problems such as this?
d) Give a linear inequality that expresses the requirement that at least 20% of the weight of
the purchased diet comes from food 2.
4. (from an old exam) We have a gasoline blending problem where we may mix four gasoline
products x1 , x2 , x3 , x4 from 3 types of gas (gas1,gas2,gas3). The availability of the three gases
is given as a function of a parameter p:
Maximize 12.1x1 +8.3x2 +14.2x3 18.1x4
(= z)
(gas1)
2x1
+x2
+3x3
+6x4 ≤ 37 + 2p
x1 , x2 , x3 , x4
≥
0
(gas2)
3x1
+2x3
+x4
≤ 42 + p
(gas3)
x1
+4x2
+2x3
+x4
≤ 39 − 3p
We are interested in the value of the objective function z as a function of the parameter
p near p = 1. We make p a variable (and move it to the other side of the inequalities), and
make p free (allowing it to be negative). We also impose the somewhat arbitrary condition
p ≤ 1 and then use this constraint in our analysis. We send this off to LINDO. The LINDO
output will be useful for parts b),c).
a) What happens to z in our original LP when p =
−37
?
2
What happens for p <
−37
?
2
Explain.
b) What is z for p = 1? What is the slope of the graph of z as a function of p at that point?
For what range on p is the slope valid.
c) Explain the interesting coincidence that:
2(2.525000) + 1(1.868750) − 3(1.443750) = 2.587500.
The input to LINDO is:
max 12.1x1 + 8.3x2 + 14.2x3 + 18.1x4
subject to
gas1) 2x1+x2+3x3+6x4-2p<37
gas2) 3x1+2x3+x4-p<42
gas3) x1+4x2+2x3+x4+3p<39
param)p<1
end
free p
The following is the output from LINDO:
OBJECTIVE FUNCTION VALUE
1) 230.8062
VARIABLE
VALUE
X1
11.875000
X2
4.187500
X3
3.687500
X4
0.000000
P
1.000000
REDUCED COST
0.000000
0.000000
0.000000
0.362500
0.000000
ROW
SLACK OR SURPLUS DUAL PRICES
GAS1)
0.000000
2.525000
GAS2)
0.000000
1.868750
GAS3)
0.000000
1.443750
P ARAM )
0.000000
2.587500
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF
INCREASE
DECREASE
X1
12.100000
0.161111
2.990000
X2
8.300000
0.322222
4.620000
X3
14.200000
4.271429
0.093548
X4
18.100000
0.362500
IN F IN IT Y
P
0.000000
IN F IN IT Y
2.587500
RIGHTHAND SIDE RANGES
ROW
GAS1
GAS2
GAS3
P ARAM
CURRENT ALLOWABLE ALLOWABLE
RHS
INCREASE
DECREASE
37.000000
16.750000
4.916667
42.000000
8.428572
19.000000
39.000000
19.666666
13.400000
1.000000
3.045455
2.269231
5. (from an old exam) We are running a factory and can produce products of three possible
types from four types of parts as follows.
part
part
part
part
1
2
3
4
profit $
product 1
3
4
5
4
product 2
5
6
8
7
product 3
2
2
3
4
21
35
15
available parts
286
396
440
396
We wish to choose our product mix to obtain maximum profit subject both to the limitations
on the inventory of available parts but also subject to the restriction that at most 50% of the
number of produced products can be of one type
The LINDO input/output on this page and the next page will be useful for parts a),b),c).
a) What are the marginal values of the four parts?
b) Mr. Edison visits the factory and offers to make a remarkable new part that substitutes
for one of any of the four parts and will only charge $2 for each of these new parts. Would
you buy some? How many would you buy?
c) The market for product 1 crashes and the profit drops to that of product 3. Should you
change your production?
d) Compute the marginal cost for the total parts that make up product 2 and compare with
17
for this calculation). Why aren’t they
the profit for product 2 (Please let 1.545455 be 11
equal?
e) Explain the meaning of the constraint PROD1<50 in the context of this problem.
The input to LINDO was as follows. The constraints have been labeled to aid readability:
MAX 21 PROD1 + 35 PROD2 + 15 PROD3
SUBJECT TO
PART1) 3 PROD1 + 5 PROD2 + 2 PROD3 < 286
PART2) 4 PROD1 + 6 PROD2 + 2 PROD3 < 396
PART3) 5 PROD1 + 8 PROD2 + 3 PROD3 < 440
PART4) 4 PROD1 + 7 PROD2 + 4 PROD3 < 396
PROD1<50) 0.5 PROD1 - 0.5 PROD2 - 0.5 PROD3 < 0
PROD2<50) - 0.5 PROD1 + 0.5 PROD2 - 0.5 PROD3 < 0
PROD3<50) - 0.5 PROD1 - 0.5 PROD2 + 0.5 PROD3 < 0
END
The following is the output from LINDO:
OBJECTIVE FUNCTION VALUE
1932.000
VARIABLE
VALUE
PROD1
22.000000
PROD2
36.000000
PROD3
14.000000
REDUCED COST
0.000000
0.000000
0.000000
ROW
SLACK OR SURPLUS DUAL PRICES
PART1)
12.000000
0.000000
PART2)
64.000000
0.000000
PART3)
0.000000
3.000000
PART4)
0.000000
1.545455
PROD1<50)
14.000000
0.000000
PROD2<50)
0.000000
0.363636
PROD3<50)
22.000000
0.000000
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF
INCREASE
DECREASE
PROD1
21.000000
0.363636
6.000000
PROD2
35.000000
INFINITY
0.500000
PROD3
15.000000
1.333333
2.615385
RIGHTHAND SIDE RANGES
ROW
CURRENT ALLOWABLE ALLOWABLE
RHS
INCREASE
DECREASE
PART1
286.000000
INFINITY
12.000000
PART2
396.000000
INFINITY
64.000000
PART3
440.000000
24.000000
44.000000
PART4
396.000000
44.000000
23.692308
PROD1<50
0.000000
INFINITY
14.000000
PROD2<50
0.000000
22.000000
19.250000
PROD3<50
0.000000
INFINITY
22.000000
6. (from an old exam) We are asked to choose among various carbon reduction strategies to
maximize the tons of atmospheric carbon removed subject to various constraints.
capital
labour
space
habitat damage
strategy 1
100
50
2
3
strategy 2
125
100
3
1
strategy 3
150
110
4
1
tons carbon removed
15
21
26
maximum
1000
600
23
8
There is an additional political decision that we must choose at least 1.5 units of Strategy 1.
Here is the problem formulation for LINDO.
MAX
15 STRAT1 + 21 STRAT2 + 26 STRAT3
SUBJECT TO
CAPITAL)
100 STRAT1 + 125 STRAT2 + 150 STRAT3 <=
LABOUR)
50 STRAT1 + 100 STRAT2 + 110 STRAT3 <=
SPACE)
2 STRAT1 + 3 STRAT2 + 4 STRAT3 <=
23
HABITAT)
3 STRAT1 + STRAT2 + STRAT3 <=
8
POLITICS)
STRAT1 >=
1.5
END
1000
600
Each question below is independent of each other. The LINDO input/output on this page
and the next page will be useful.
a) If tons of carbon removed per unit of strategy 1 is reduced from 15 to 14 what is now the
total tons of carbon removed by an optimal strategy?
b) We discover that the habitat damage is less crucial because new habitat for the endangered
species has been discovered on Pender Island. What is the benefit of allowing the habitat
damage to increase to 8.5 units from 8 units?
c) We are told that strategy 1 costs $5000 per unit, strategy 2 costs $6000 per unit and
strategy 3 costs $10000 per unit. Is the cost in $ per ton of carbon removed less than $400
per ton for our optimal strategy?
d) You are told of a new strategy of carbon removal using fertilizer to stimulate algae growth
that will require (per unit of new strategy) 100 units of capital, 100 units of labour, 2 units
space and 1 unit cost to habitat that will remove 19 tons of carbon. Should you investigate
further? Explain. e) A change in government has resulted in a new policy that we must
choose 2 units of strategy 1. How does this affect the number of tons of carbon removed?
The following is the output from LINDO:
OBJECTIVE FUNCTION VALUE
1)
VARIABLE
STRAT1
113.5000
VALUE
1.500000
REDUCED COST
0.000000
STRAT2
STRAT3
ROW
CAPITAL)
LABOUR)
SPACE)
HABITAT)
POLITICS)
0.000000
3.500000
SLACK OR SURPLUS
325.000000
140.000000
6.000000
0.000000
0.000000
NO. ITERATIONS=
5.000000
0.000000
DUAL PRICES
0.000000
0.000000
0.000000
26.000000
-63.000000
0
RANGES IN WHICH THE BASIS IS UNCHANGED:
VARIABLE
STRAT1
STRAT2
STRAT3
ROW
CAPITAL
LABOUR
SPACE
HABITAT
POLITICS
CURRENT
COEF
15.000000
21.000000
26.000000
OBJ COEFFICIENT RANGES
ALLOWABLE
ALLOWABLE
INCREASE
DECREASE
63.000000
INFINITY
5.000000
INFINITY
INFINITY
5.000000
CURRENT
RHS
1000.000000
600.000000
23.000000
8.000000
1.500000
RIGHTHAND SIDE RANGES
ALLOWABLE
ALLOWABLE
INCREASE
DECREASE
INFINITY
325.000000
INFINITY
140.000000
INFINITY
6.000000
1.272727
3.500000
1.166667
0.500000