answer the survey Quations .short 3 answers

profileBaroosh
Chapter3-SolutionstoPracticeProblems-2.pdf

Chapter 3: Solutions to Practice

Problem

12. x1 = no. of days to operate Mill 1

x2 = no. of days to operate Mill 2

minimize Z = 6,000x1 + 7,000x2 subject to

6x1 + 2x2 ≥ 12

2x1 + 2x2 ≥ 8

4x1 + 10x2 ≥ 5

x1,x2 ≥ 0

16.

a) x1 = 320, x2 = 90

320 + 90 + s1 = 410

s1 = 0 acres uncultivated

90 + s3 = 100

s3 = 10 acres of tobacco

allotment unused

b) At point D only corn is planted. In order for

point D to be optimal the slope of the

objective function will have to be at least as

great (i.e., steep) as the slope of the constraint

line, x1 + x2 = 410, which is −1. Thus, the

profit for corn is computed as,

−c1/520 = −1

c1 = 520

The profit for corn must be greater than $520 for

the Bradleys to plant only corn.

c) If the constraint line changes from x1 + x2 =

410 to x1 + x2 = 510, it will move outward to a

location which changes the solution to the

point where 105x1 + 210x2 = 52,500 intersects

with the axis. This new point is x1 = 500, x2 =

0, Z = $150,000.

d) If the constraint line changes from x1 +

x2 = 410 to x1 + x2 = 360, it moves inward to a

location which changes the solution point to

the intersection of x1 + x2 = 360 and 105x1 +

210x2 = 52,500. At this point x1 = 260, x2 =

100, and Z = $130,000.

25.

a) Optimal solution at B: x1 = 333.3 and x2 = 166.7

(333.3) + (166.7) − s1 = 400

s1 = 100 extra gallons of

blended whiskey produced

.6(333.33) − .4(166.7) − s2 = 0

s2 = 133.3 extra gallons

of rye in the blend

(166.7) + s3 = 250

s3 = 83.3 fewer gallons of bourbon

than the maximum

(333.3) + (166.7) + s4 = 500

s4 = 100 gallons of blend

production capacity left over

b) Because the “solution space” is not really an

area, but a line instead, the objective function

coefficients can change to any positive value

and the solution point will remain the same,

i.e., point B. Observing the graph of this

model, no matter how flatter or steeper the

objective function becomes, point B will

remain optimal.

26. Z = 1,666.667 (Wait until I complete Computer-based Sensitivity Analysis)

Variable Value

x1 333.333

x2 166.667

Constraint

Slack/Surplus

Shadow

Price

c1 100.000 0.000

c2 133.333 0.000

c3 83.333 0.000

c5 0.000 3.333

Objective Coefficient ranges

Variables

Lower

Limit

Current

Values

Upper

Limit

Allowable

Increase

Allowable

Decrease

x1 −2.000 3.000 No limit No limit 5.000

x2 −6.000 4.000 No limit No limit 10.000

Right hand side ranges

Constraints

Lower Limit Current

Values

Upper

Limit

Allowable

Increase

Allowable

Decrease

c1 No limit 400.000 500.000 100.000 No limit

c2 No limit 0.000 133.333 133.333 No limit

c3 166.667 250.000 No limit No limit 83.333

c4 −250.000 0.000 500.000 500.000 250.000

c5 400.000 500.000 750.000 250.000 100.000

Z = 1,625.000

Variable Value

x1 375.000

x2 125.000

Constraint Slack/Surplus Shadow Price

c1 100.000 0.000

c2 175.000 0.000

c3 125.000 0.000

c5 0.000 3.250

Objective Coefficient ranges

Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease

x1 −1.333 3.000 No limit No limit 4.333

x2 −9.000 4.000 No limit No limit 13.000

Right hand side ranges

Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease

c1 No limit 400.000 500.000 100.000 No limit

c2 No limit 0.000 175.000 175.000 No limit

c3 125.000 250.000 No limit No limit 125.000

c4 −500.000 0.000 500.000 500.000 500.000

c5 400.000 500.000 1,000.000 500.000 100.000