modbsoln

B Q U A N T I T A T I V E M O D U L E

Linear Programming

DISCUSSION QUESTIONS

1. Students may select from eight LP applications given in the introduction. These include school bus scheduling, police patrol allocation, scheduling bank tellers, selecting product mix, picking blends to minimize cost, minimizing shipping cost, developing production schedules, and allocating space.

2. LP theory states that the optimum lies on a corner. All three solution techniques make use of the “corner point” feature.

3. The feasible region is the area bounded by the set of problem constraints. A feasible solution is any combination of x, y coordinates (or x1, x2 coordinates) that is in or on the feasible region.

4. Each LP problem that has been formulated correctly does have an infinite number of possible solutions. Any point within the feasible region is a solution that satisfies all constraints (although it is not necessarily optimal). In addition, for any problem in which the optimal solution lies on a constraint that is parallel to the objective function, all points along that constraint are also both feasible and optimal.

5. The objective function contains the profit or cost informationthat enables us to determine whether one solution is better than an-other solution. Our choice of best depends only on the objective.

6. Before activity values can be placed into the objective, they must meet the constraints. Notice that the objective function has nominimum-required profit level unless it is included as a constraint.

7. As long as the costs do not change, the diet problem always provides the same answer. In other words, the diet is the same every day. Unlike animals, people enjoy variety, and variety cannot be included as a linear constraint.

8. The number of feasible solutions is infinite. We only need to consider extreme points—corner points—to find the optimal solution. If we use isoprofit lines, we only need to examine one corner point to determine the optimal solution.

9. Shadow price or dual: the value of one additional unit of a resource, such as one more hour of a scarce labor resource or one more dollar to invest.

10. The isocost line is moved down in a minimization problem until it no longer intersects with any constraint equation. The last

point in the feasible region that the line touches is the optimalcorner point.

11. The corner point method examines the profit at every corner point, whereas the isoprofit line method draws a series of parallel profit lines until one line finally touches the last tip (corner point) of the feasible region. That last point touched is the optimal solution, so other corner points need not be tested.

12. When two constraints do not cross at an axis, we use simultaneous equations—there is only one point where two linear equations (constraints) cross.

13. (a) Adding a new constraint will reduce the size of the feasible region unless it is a redundant constraint. It can never make the feasible region any larger.

(b) A new constraint can only reduce the size of the feasible region; therefore the value of the objective function will either decrease or remain the same. If the original solution is still feasible, it will remain the optimal solution.

ACTIVE MODEL EXERCISE

ACTIVE MODEL B.1: LP Graph

1. By how much does the profit on x-pods need to rise to make it the only product manufactured?If the profit per x-pod is more than $10 per unit, thenit is the only product that should be manufactured.

2. By how much does the profit on x-pods need to fall to stop manufacturing it?At $6.66 and below, we should not manufacture any x-pods.

3. What happens to the profit as the number of assembly hours increases by 1 hour at a time? For how many hours does this hold true?The profit rises by $.50 per hour until we reach 120 hours, at which point the rise stops.

4. What happens if we can reduce the electronics time for Blue-berrys to 2.5 hours?The profit rises by $70.

228

229 QUANTITATIVE MODULE B L INEAR PROGRAMMING

END-OF-MODULE PROBLEMS

Feasible corner points (x, y): (0,3), (0,10), (2.4,8.8), (6.75,3).Maximum profit 100 at (0,10).

B.3 We solve this problem by the isocost line method:

Unique optimal solution is (0, 4) with z = 4.

B.4 (a) Corner points (0,50), (50,50), (0,200), (75,75), (50,150).

(b) Optimal solutions: (75,75) and (50,150). Both yield profit of $3,000.

X Y Z 4X 6Y

0 0 04 0 160 4 24

1.33 3.33 25.33 (optimal)

B.1

B.2

QUANTITATIVE MODULE B L INEAR PROGRAMMING 230

B.5

X Y Z 24X 15Y

0 20 30011 0 264

3.86 4.54 160.86 (optimal)

B.6 (a) Let x1 number of liver-flavored biscuits in a packagex2 number of chicken-flavored biscuits in a package

Minimize x1 2x2Subject to x1 x2 40

2x1 4x2 60x1 15

x1, x2 0

(b) Corner points are (0,40) and (15,25). Optimal solution is(15,25) with cost of 65 cents.

(c) Minimum cost 65 cents.B.7

Let x1 number of air conditioners to be produced

x2 number of fans to be produced

Maximize 25x1 15x2Subject to 3x1 2x2 240 (wiring)

2x1 1x2 140 (drilling)x1, x2 0 (nonnegativity)

Profit:@a: (x1 0, x2 0) Obj $0@b: (x1 0, x2 120) Obj 25 0 15 120 $1,800@c: (x1 40, x2 60) Obj 25 40 15 60 $1,900*@d: (x1 70, x2 0) Obj 25 70 15 0 $1,750

* The optimal solution is to produce 40 air conditioners and 60 fans each period. Profit will be $1,900.

B.8

Let x1 number of Model A tubs producedx2 number of Model B tubs produced

Maximize 90x1 70x2Subject to 125x1 100x2 25,000 (steel)

20x1 30x2 6,000 (zinc)


x1, x2 0 (nonnegativity)Profit:@a: (x1 0, x2 200) Obj 90 0 70 200

$14,000.00@b: (x1 85.71, x2 142.86) Obj 90 85.71 70 142.86

$17,714.10@c: (x1 200, x2 0) Obj 90 200 70 0

$18,000.00** The optimal solution is to produce 200 Model A tubs, and 0 Model B tubs. Profit will be $18,000.

B.9 (a) Let T = number of trucks to produce per day

C = number of cars to produce per day

Maximize z = 300T + 220C

such that:

(b) Graph feasible region:

(c)

(d) Produce 20 trucks and 30 cars daily for a profit of $12,600 per day.

B.10

Let x1 number of Alpha-4 computersx2 number of Beta-5 computers

Maximize 1200x1 1800x2Subject to 20x1 25x2 800 (total hours)

x1 10 (Alpha-4s) x2 15 (Beta-5s)

x1, x2 0 (nonnegativity)Profit:@ a: (x1 10, x2 24) Obj 1200 10 1800 24

$55,200*@ b: (x1 21.25, x2 15) Obj 1200 21.25 1800 15

$52,500* The optimal solution is to produce 10 Alpha-4 and 24 Beta-5 computers per period. Profit is $55,200.

B.11 Let: X1 number of pounds of compost in each bag X2 number of pounds of sewage waste in each bag

Minimize cost 5X1 4X2 (in cents)Subject to X1 X2 60 (pounds per bag)

X1 30 (pounds compost per bag)X2 40 (pounds sewage per bag)

Corner point a:(X1 30, X2 40) cost 5(30) (4)(40) $3.10Corner point b (which is optimal):(X1 30, X2 30) cost 5(30) (4)(30) $2.70Corner point c:(X1 60, X2 0) cost 5(60) (4)(0) $3.00

Point Coordinates z Value

O (0, 0) 0

A (0, 50) 11,000

C (40 ,0) 12,000B Solve 2 equations in 2 unknowns

derived from constraints (1) and (2) to obtain (20, 30)

12,600


B.12

The optimal point, a, lies at the intersection of the constraints:3x1 2x2 120x1 3x2 90

To solve these equations simultaneously, begin by writing them in the form shown below:

3x1 2x2 120x1 3x2 90

Multiply the second equation by –3, and add it to the first:3x1 2x2 120 → 3x1 2x2 120

–3(x1 3x2 90) → 3x1 – 9x2 –270

–7x2 –150

Therefore, x2 150/7 21.43. Given: 3x1 2x2 120 then3x1 120 – 2x2 120 – 2 21.43

and

Thus, the optimal solution is x1 25.71, x2 21.43.The cost is given by:

C x1 2x2 25.71 2 21.43 $68.57

B.13 The last constraint is not linear because it contains the square root of x and the objective function and first constraint are not because of the x1x2 term.

B.14 (a) Using software, we find that the optimal solution is:x1 7.95x2 5.95

x3 12.60Profit $143.76 (rounded)

(b) There is no unused time available on any of the threemachines.

(c) An additional hour of time on the third machine would be worth $0.26.

(d) Additional time on the second machine would be worth $0.786 per hour for a total of $7.86 for the additional 10 hours.

(c) If X1’s profit coefficient was overestimated, but should only have been $1.25, it is easy to see graphically thatthe solution at point b remains optimal.

B.16 (a) Let Xij number of students bused from sector i to school j. Objective:

(b)

B.15 (a)


Subject to:

(b) Solution: XAB 400

XAE 300

XBB 500

XCC 100

XDC 800

XEE 400

Distance 5,400 “student miles”

X1 $ invested in Treasury notesX2 $ invested in bondsMaximize ROI 0.08X1 0.09X2

X1 $125,000X2 $100,000

X1 X2 $250,000X1, X2 0

Point a (X1 150,000, X2 100,000),ROI $21,000 (optimal solution)Point b (X1 250,000, X2 0), ROI $20,000

B.17

B.18 Problem Data Solution

Period

Time PeriodWorkersRequired

HireSolution

1

HireSolution 2

HireSolution 3

1 3 A.M.–7 A.M. 3 0 3 32 7 A.M.–11 A.M. 12 16 9 143 11 A.M.–3 P.M. 16 0 7 24 3 P.M.–7 P.M. 9 9 2 75 7 P.M.–11 P.M. 11 2 9 46 11 P.M.–3 A.M. 4 3 0 0

30


Let xi number of workers reporting for the start of workin period i, where i 1, 2, 3, 4, 5, or 6. The equations become:

Objective:x1 x2 x3 x4 x5 x6 (Minimize staff size)Subject to:

Note that three alternate optimal solutions are provided to this problem. All solutions could be implemented using only 30 staff members.

B.19 (a) Let X1 wren housesX2 bluebird houses

Maximize profit 6X1 15X2

4X1 2X2 604X1 12X2 120

Corner Points

X1 X2 Profit

0 0 015 0 90 0 10 15012 6 162 (Optimal)

The maximum value of the objective is $162, obtained by producing 12 wren houses and 6 bluebird houses.

B.20 The original equations are:Objective: 9x1 12x2 (maximize)Subject to: x1 x2 10 (gallons, varnish)

x1 2x2 12 (lengths, redwood)where: x1 number of coffee tables/weekx2 number of bookcases/weekOptimal: x1 8, x2 2, Profit $96

B.21 (a, b) Let S = number of standard bags to produce per week

D = number of deluxe bags to produce per week

Maximum: z = 10S + 8D

(b)


B.22

The original equations are:Objective: 4x1 5x2 (minimize)Subject to: x1 2x2 80

3x1 x2 75The optimal solution is found at the intersection of the two constraints:

x1 2x2 803x1 x2 75

To solve these equations simultaneously, begin by writing them in the form shown below:

x1 2x2 803x1 x2 75

Multiply the second equation by –2 and add it to the first:

Thus, x1 70/5 14. Given: x1 2x2 80,2x2 80 – x1 80 – 14 or x2 66/2 33.The cost is given by:

C 4x1 5x2 (4 14) (5 33) 221

B.23 Let x1 number of class A containers to be usedx2 number of class K containers to be usedx3 number of class T containers to be used

The appropriate equations are:Maximize: 9x1 7x2 15x3Subject to: 2x1 x2 3x3 130 (material)

2x1 6x2 4x3 240 (time)x1, x2, x3 0 (nonnegativity)

Using software we find that the optimal solution is:x1 0, x2 14.29, x3 38.57

and

Profit $678.57

B.24

There are 5 corner (extreme) points.

Let: X1 number of TV spotsX2 number of newspaper ads

Maximize exposures 35,000X1 20,000X2

Extreme (Corner) Points

Point Profit

(0, 0) $0

(0, 300) $2,400

(360, 0) $3,600

(240, 180) $3,840 Optimal solution and answer

x 4 (line 1)

–x 2 (line 2)

x +2y 6 (line 3)

–x +2y 8 (line 4)

y 0 (line 5)

B.25


B.26 (a) Minimize:

(b) Solution:

An alternate solution, at the same $219 ($219,000) cost, is

B.27 (a) V1 fertilizer shipped to Customer A from Warehouse W1V2 fertilizer shipped to Customer A from Warehouse W2V3 fertilizer shipped to Customer A from Warehouse W3V4 fertilizer shipped to Customer B from Warehouse W1V5 fertilizer shipped to Customer B from Warehouse W2V6 fertilizer shipped to Customer B from Warehouse W3

(b) Z 7.5V1 6.25V2 6.5V3 6.75V4 7V5 8V6

(d) How many of the constraints are binding? 4 (tell by non-zero shadow prices)

(e) How much slack/surplus is there with the non-binding constraint(s)? 50 tons (600–550)

(f) The range of optimality on Variable V3? $6.25 to $7.25 (to get this look at Allowable Increase (.75) and Decrease (.25) for V3. Add/subtract from C3 (6.5): 6.25 6.5 – .25 & 7.25 6.5 .75.

(g) If we could ship 10 tons less to Customer A, we might be able to save: $65 (shadow price $6.5/ton). Cust B Shadow price is higher for filling Customer B’s demand $7.25 > $6.5 and we are trying to minimize costs.

B.28 Let x1 number of medical patientsx2 number of surgical patients

The appropriate equations are:Maximize 2280x1 1515x2

Optimal: x1 2790, x2 2104, Profit $9,551,659or 2,790 medical patients, 2,104 surgical patients,Profit $9,551,659 (all numbers rounded)Beds required:

Medical uses:

Surgical uses:

Here is an alternative approach that solves directly for the number of beds:

Maximize revenues 104,025x1 110,595x2

where x1 no. of medical beds 61.17x2 no. of surgical beds 28.83Revenue is $9,551,659, as before.

B.29 (a) Let X1 no. of units of internal modems produced per week

X2 no. of units of external modems produced per week

X3 no. of units of circuit boards produced per weekX4 no. of units of jump drives produced per weekX5 no. of units of hard drives produced per weekX6 no. of units of memory boards produced per week

Objective function analysis: First find the time used on each test device:Hours on test device 1:

(c) Description Variables and Coefficients?

What Type? RHS?

C1: Cust A’s demand formula: V1 V2 V3 >| 650 (either > or okay)

C2: Cust B’s demand formula: V4 V5 V6 >| 800 (either > or okay)

C3: W1’s supply formula: V1 V4 < 400


Hours on test device 2:

Hours on test device 3:

Thus, the objective function isMaximize contribution per unit Revenue – Material cost – Test cost

This can be rewritten as

Subject to:

(b) The solution is

X1 496.55 internal modems

X2 1,241.38 external modems

X3 through X6 0

Profit $195,504.80

(c) The shadow prices, as explained in Module B, for addi-tional time on the three test devices are $21.41, $5.75, and $0, respectively, per minute.


B.30

FoodsCost/

ServingCalories/Serving

Percent Pro-tein

Percent Carbs

Percent Fat

Fruit/Vegetable

Apple sauce(AS) $0.30 100 0% 100% 0% 1Canned corn (CC)

$0.40 150 20% 80% 0% 1

Fried chicken (FC)

$0.90 250 55% 5% 40% 0

French fries (FF) $0.20 400 5% 35% 60% 0

Mac & cheese (MC)

$0.50 430 20% 30% 50% 0

Turkey breast (TB)

$1.50 300 67% 0% 33% 0

Garden salad (GS)

$0.90 100 15% 40% 45% 1

AS CC FC FF MC TB GS

Cost 0.3 0.4 0.9 0.2 0.5 1.5 0.9Servings 0 1.333 0.457 0 1.130 0 0 $1.51

Constraints AS CC FC FF MC TB GS LHS RHS

Cals min 100 150 250 400 430 300 100 800 500

Cals max 100 150 250 400 430 300 100 800 800

Protein min 0 30 137.5 20 86 201 15 200 200

Carb min 100 120 12.5 140 129 0 40 311.43 200

Fat max 0 0 100 240 215 99 45 288.57 400Fruit/veg. min. 100 150 0 0 0 0 100 200 200

Target cell (Min.) Answer Report (Relevant Section)Cell Name Original Value Final Value$I$14 servings $2.91 $1.51

Adjustable cellsCell Name Original Value Final Value$B$14 serving AS 1.50 0.00$C$14 serving CC 0.00 1.33$D$14 serving FC 1.33 0.46$E$14 serving FF 0.00 0.00$F$14 serving MC 0.00 1.13$G$14 serving TB 0.00 0.00$H$14 serving GS 1.40 0.00

ConstraintsCell Name Cell Value Formula Status Slack$I$17 Cals min LI 800 $I$17 $J$1 Not binding 300$I$18 Cals max L 800 $I$18 $J$1 Binding 0$I$19 Protein min 200 $I$19 $J$1 Binding 0$I$20 Carb min L 311.43 $I$20 $J$2 Not binding 111.4286$I$21 Fat max LI 288.57 $I$21 $J$2 Not binding 111.4286$I$22 Fruit + Veg I 200 $I$22 $J$2 Binding 0


Adjustable cells

Sensitivity Report (Relevant Section)

Cell NameFinal Value

Reduced Cost

Objective Coefficient

Allowable Increase

Allowable Decrease

$B$14 serving AS 0 0.1726 0.3 1E + 30 0.1726$C$14 serving CC 1.333 0 0.4 0.2589 0.2256$D$14 serving FC 0.457 0 0.9 0.1051 0.1006$E$14 serving FF 0 0.1527 0.2 1E + 30 0.1529$F$14 serving M 1.130 0 0.5 0.0629 0.7078$G$14 serving TB 0 0.1693 1.5 1E + 30 0.1694$H$14 serving GS 0 0.6661 0.9 1E + 30 0.6882

Constraints

Cell NameFinal Value

Shadow Price

Constraint R.H. Side

Allowable Increase

Allowable Decrease

$I$17 Cals min LI 800 0 500 300 1E + 30$I$18 Cals max L 800 – 0.00023 800 200 251.6129$I$19 Protein min 200 0.008983 200 155 40$I$20 Carb min L 311.43 0 200 111.4285 1E + 30$I$21 Fat max LI 288.57 0 400 1E + 30 111.4286$I$22 Fruit + Veg I 200 0.0015 200 485.7143 200

modbsoln

Documents

optimal corner point

feasible region

feasible corner points

corner point method

number of feasible solutions

tip corner point

corner point feature

solution techniques