Introduction to Operations Research Answers to selected problems · 2014. 8. 14. · Introduction to Operations Research Answers to selected problems Prof. dr. E-H. Aghezzaf, assistant

Introduction to Operations ResearchAnswers to selected problems

Prof. dr. E-H. Aghezzaf, assistant D. VerleyeDepartment of Industrial Management (IR18)

Exercise session 1

1. Decision variables:

x1 = number of barrels of beer produced

x2 = number of barrels of ale produced

LP formulation:

Maximize Z = 5x1 + 2x2,subject to

5x1 + 2x2 ≤ 60 (Availability of corn)2x1 + x2 ≤ 25 (Availability of hops)

and x1, x2 ≥ 0.


x1 = ton of alloy 1 produced

x2 = ton of alloy 2 produced

Since x1 + x2 = 1, these are also the percentages.

LP formulation:

Minimize Z = 190x1 + 200x2,subject to

x1 + x2 = 1 (Producing 1 ton)3.2 ≤ 3x1 + 4x2 ≤ 3.5 (Carbon requirement)

1.8 ≤ 2x1 + 2.5x2 ≤ 2.5 (Silicon requirement)0.9 ≤ x1 + 1.5x2 ≤ 1.2 (Nickel requirement)

42000x1 + 50000x2 ≥ 45000 (Tensile strength requirement)and x1, x2 ≥ 0.

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x11 = amount of steel 1 produced at mill 1 (ton)






LP formulation:

Minimize Z = 10x11 + 12x12 + 14x13+11x21 + 9x22 + 10x23,

subject tox11 + x12 + x13 ≥ 500 (Minimum production steel 1)x21 + x22 + x23 ≥ 600 (Minimum production steel 2)

20x11 + 22x21 ≤ 12000 (Available furnace time at mill 1)24x12 + 18x22 ≤ 12000 (Available furnace time at mill 2)28x13 + 30x23 ≤ 12000 (Available furnace time at mill 3)

and xij ≥ 0, i = 1, 2; j = 1, 2, 3.

4. A graphical representation of the problem:

Decision variables:

xLA−H = amount of barrels shipped from LA to Houston (million barrels/year)

xLA−NY = amount of barrels shipped from LA to NY (million barrels/year)

xC−H = amount of barrels shipped from Chicago to Houston (million barrels/year)

xC−NY = amount of barrels shipped from Chicago to NY (million barrels/year)

yLA = added refining capacity for LA (million barrels)

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yC = added refining capacity for Chicago (million barrels)

LP formulation:

Maximize Z = 10 (20xLA−H + 15xLA−NY + 18xC−H + 17xC−NY )− 120yLA − 150yCsubject to

xLA−H + xLA−NY ≤ 2 + yLA (Don’t ship more from LA than available)xC−H + xC−NY ≤ 3 + yC (Don’t ship more from Chicago than available)xLA−H + xC−H ≤ 5 (Amount shipped to Houston ≤ Houston Demand)xLA−NY + xC−NY ≤ 5 (Amount shipped to NY ≤ NY Demand)and all variables≥ 0.


xi = number of hours workers spend on machine i each week, i = 1, 2, 3

pi = units of product i produced each week, i = 1, 2, 3

LP formulation:

Maximize Z = 6p1 + 8p2 + 10p3,subject to

2p1 + 3p2 + 4p3 ≤ x1 (process time machine 1 ≤ assigned time machine 1)3p1 + 5p2 + 6p3 ≤ x2 (process time machine 2 ≤ assigned time machine 2)4p1 + 7p2 + 9p3 ≤ x3 (process time machine 3 ≤ assigned time machine 3)

x1 ≤ 200 (assigned time machine 1 ≤ available time)x2 ≤ 120 (assigned time machine 2 ≤ available time)x3 ≤ 160 (assigned time machine 3 ≤ available time)

x1 + x2 + x3 ≤ 350 (total workers time)and xi, pi ≥ 0, i = 1, 2, 3.


Sij = type i trucks sold during year j

Pij = type i trucks produced during year j

Iij = number of type i trucks in inventory at end of year j

LP formulation:

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Maximize Z = 20S11 + 20S12 + 20S13 + 17S21

+17S22 + 17S23 − 15P11 − 15P12

−15P13 − 14P21 − 14P22 − 14P23

−2I11 − 2I12 − 2I13−2I21 − 2I22 − 2I23,

subject toS11 − P11 + I11 = 0 (Year 1 production type 1)S12 − P12 − I11 + I12 = 0 (Year 2 production type 1)S13 − P13 − I12 + I13 = 0 (Year 3 production type 1)S21 − P21 + I21 = 0 (Year 1 production type 2)S22 − P22 − I21 + I22 = 0 (Year 2 production type 2)S23 − P23 − I22 + I23 = 0 (Year 3 production type 2)P11 + P21 ≤ 320 (Year 1 capacity limit)P12 + P22 ≤ 320 (Year 2 capacity limit)P13 + P23 ≤ 320 (Year 3 capacity limit)S11 ≤ 100 (Year 1 maximum type 1 sold)S12 ≤ 200 (Year 2 maximum type 1 sold)S13 ≤ 300 (Year 3 maximum type 1 sold)S21 ≤ 200 (Year 1 maximum type 2 sold)S22 ≤ 100 (Year 2 maximum type 2 sold)S23 ≤ 150 (Year 3 maximum type 2 sold)5P11 + 5P12 + 5P13 − 5P21 − 5P22 − 5P23 ≤ 0 (pollution emission limit)


x11 = units of product 1 produced on machine 1




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LP formulation:

Minimize Z = 1.5x11 + 0.4x21 + 2.2x12 + 4x22subject to

x11 + x12 ≥ 200 (Minimum production product 1)x21 + x22 ≥ 240 (Minimum production product 2)

12(x11 + x12) ≤ x11 → 1

2(x12 − x11) ≤ 0 (Half of product 1 on machine 1)

12(x21 − x22) ≤ 0 (Half of product 2 on machine 2)

0.7x11 + 0.75x21 ≤ 200 (Machine 1 availability)0.8x12 + 0.9x22 ≤ 200 (Machine 2 availability)

0.75x11 + 0.75x21 + 1.2x12 + x22 ≤ 400 (Labor availability)and xij ≥ 0, i = 1, 2; j = 1, 2.


xij = number of officers who get off days i and j of the week (1 = Saturday, ..., 7 =Friday). Note that for example on Saturday, 30 - 28 = 2 officers have to take a dayoff. This reasoning leads us to the following LP formulation:

LP formulation:

Maximize Z = x12 + x23 + x34 + x45 + x56 + x67 + x17subject to

x12 + x13 + x14 + x15 + x16 + x17 = 2x12 + x23 + x24 + x25 + x26 + x27 = 12x13 + x23 + x34 + x35 + x36 + x37 = 12x14 + x24 + x34 + x45 + x46 + x47 = 6x15 + x25 + x35 + x45 + x56 + x57 = 5x16 + x26 + x36 + x46 + x56 + x67 = 14x17 + x27 + x37 + x47 + x57 + x67 = 9and xij ≥ 0, i = 1..6, j = 2..7, i < j.


xij = money invested at beginning of month i for a period of j months.

LP formulation:

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Maximize Z = 1.08x14 + 1.03x23 + 1.01x32 + 1.001x41subject to

x11 + x12 + x13 + x14 = 200 (month 1)x21 + x22 + x23 = 300 + 1.001x11 (month 2)x31 + x32 = −200 + 1.01x12 + 1.001x21 (month 3)x41 = 50 + 1.03x13 + 1.01x22 + 1.001x31 (month 4)and xij ≥ 0, i = 1..4, j = 1..5− i.


x1 = amount of germanium melted with method 1

x2 = amount of germanium melted with method 2

g1R = amount of grade 1 germanium refired



dR = amount of detective germanium refired

LP formulation:

Minimize Z = 50x1 + 70x2 + 25(g1R + g2R + g3R + dR)subject to

x1 + x2 + g1R + g2R + g3R + dR ≤ 20000 (capacity constraint)0.3x1 + 0.2x2 − (1− 0.3)g1R + 0.25dR ≥ 3000(monthly demand grade 1 transistors)0.2x1 + 0.25x2 + 0.3g1R − 0.6g2R + 0.15dR ≥ 3000(monthly demand grade 2 transistors)0.15x1 + 0.2x2 + 0.2g1R + 0.3g2R − 0.5g3R + 0.2dR ≥ 2000(monthly demand grade 3 transistors)0.05x1 + 0.15x2 + 0.2g1R + 0.3g2R + 0.5g3R + 0.1dR ≥ 1000(monthly demand grade 4 transistors)0.3x1 + 0.2x2 ≥ dR (amount of detective germanium available for refiring)0.3x1 + 0.2x2 ≥ g1R (amount of grade 1 germanium available for refiring)0.2x1 + 0.25x2 ≥ g2R (amount of grade 2 germanium available for refiring)0.15x1 + 0.20x2 ≥ g3R (amount of grade 3 germanium available for refiring)and x1, x2, g1R, g2R, g3R, g4R, dR ≥ 0.


Let shift i =

1: Midnight - 6 AM

2: 6 AM - Noon

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3: Noon 6 - PM

4: 6 PM - Midnight

x1 = Workers who work shifts 1 and 2




it = customers left after end of shift t

st = number of people served during shift t

LP formulation:

Minimize Z = 120(x1 + x2 + x3 + x4) + 5(i1 + i2 + i3 + i4)subject to

i1 = 100− s1i2 = i1 + 200− s2i3 = i2 + 300− s3i4 = i3 + 200− s4i4 = 0s1 ≤ 50x1 + 50x4s2 ≤ 50x1 + 50x2s3 ≤ 50x3 + 50x2s4 ≤ 50x4 + 50x3and all variables ≥ 0.


L = gallons bought at Los Angeles

H = gallons bought at Houston

N = gallons bought at New York

M = gallons bought at Miami

IL = gallons on hand when landing in LA

IH = gallons on hand when landing in Houston

IN = gallons on hand when landing in NY

IM = gallons on hand when landing in Miami

LP formulation:

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Minimize Z = 88L+ 15H + 105N + 95M ,subject to

IL = IM +M − 2700(1 + (IL+ IM +M)/2000

)IH = IL+ L− 1500

(1 + (IL+ L+ IH)/2000

)IN = IH +H − 1700

(1 + (IH +H + IN)/2000

)IM = IN +N − 1300

(1 + (IM + IN +N)/2000

)IL+ L ≤ 12, 000IH +H ≤ 12, 000IN +N ≤ 12, 000IM +M ≤ 12, 000IL, IH, IN, IM ≥ 600and L,H,N,M ≥ 0.


Let XY represent the amount of currency X converted to currency Y, where eachletter represents the first letter of the corresponding currency (P = pound, ...). Fur-thermore, FX represents the final amount of the currency, while X is the unconvertedamount.

LP formulation:

Maximize FD + 1.696993FP + 0.573721FM + 0.007230999FYsubject to

FD −D − 1.697PD − 0.57372MD = 0FP − P − 0.58928DP − 0.33808MP = 0FM −M − 1.743DM − 2.9579PM = 0

FY − 138.3DY − 234.7PY − 79.346MY = 0FD >= 6FP >= 3FM >= 1FY >= 10

D +DP +DM +DY = 8PD + P + PM + PY = 1

MD +MP +M +MY = 8and all variables ≥ 0

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Exercise session 2

1. Graphical solution:

The solution space is convex, so an optimal solution exists. We note that the linerepresenting the objective function (drawn here for z = 5) is parallel to the cornconstraint. All solutions in the line segment (x1 = 10, x2 = 5) − (x1 = 12, x2 = 0)are optimal solutions, with z = 60.

4. Second tableau:

Basic var Eq. z x1 x2 x3 s1 s2 RHS Ratioz (0) 1 0 0 5 0 1 15s1 (1) 0 0 2/5 9/5 1 -1/5 3 15/2x1 (2) 0 1 3/5 6/5 0 1/5 3 5

All coefficients ≥ 0, so this solution is optimal. The NBV x2 has coefficient 0, soalternative optima exist. We enter x2 into the basis (instead of x1).

Third tableau:

Basic var Eq. z x1 x2 x3 s1 s2 RHSz (0) 1 0 0 5 0 1 15s1 (1) 0 -2/3 0 1 1 -1/3 1x2 (2) 0 5/3 1 2 0 1/3 5

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The two optimal solutions are (x1 = 3, s1 = 3),(x2 = 5, s1 = 1) with objective valuez = 15.

5. Final tableau:

Basic var Eq. z x1 x2 s1 s2 RHSz (0) 1 0 -7/3 -4/3 0 -8x1 (1) 0 1 1/3 1/3 0 2s2 (2) 0 0 7/3 1/3 1 2

6. Final tableau:

Basic var Eq. z x1 x2 a1 s2 s3 RHSz (0) 1 0 7/2 M+5/2 0 0 15x1 (1) 0 1 1/2 1/2 0 0 3s2 (2) 0 0 1/2 -1/2 1 0 1s2 (3) 0 0 3/2 -1/2 0 1 2

7. Any solution on the line segment with endpoints (1

2, 0) and (

1

3,2

3) is an optimal

solution for the LP. The middle point of this segment has coordinates (x1, x2)=

(5

12,1

3), which is a third optimal solution.

8. Final tableau:

Basic var Eq. z x1 x2 e1 a1 e2 a2 RHSz (0) 1 0 -M+5 -M +3 -3 -M 0 5M -6x1 (1) 0 1 -2 -1 1 0 0 2a2 (2) 0 0 -1 -1 1 -1 1 5

Since all coefficients are negative, this is an ’optimal’ tableau. In fact,a2 is still inthe basis, which means that the LP is infeasible.

9. (a) The bfs are the corner points of the feasible region, shown in the graph:

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(b) Initial tableau:

Basic var Eq. z x1 x2 s1 s2 RHS Ratioz (0) 1 -10 -1 0 0 0s1 (1) 0 1 0 1 0 1 1s2 (2) 0 20 1 0 1 100 5

Second tableau:

Basic var Eq. z x1 x2 s1 s2 RHS Ratioz (0) 1 0 -1 10 0 10x1 (1) 0 1 0 1 0 1 -s2 (2) 0 0 1 -20 1 80 80

Third tableau:

Basic var Eq. z x1 x2 s1 s2 RHS Ratioz (0) 1 0 0 -10 1 90x1 (1) 0 1 0 1 0 1 1x2 (2) 0 0 1 -20 1 80 -

Final tableau:

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Basic var Eq. z x1 x2 s1 s2 RHSz (0) 1 10 0 0 1 100s1 (1) 0 1 0 1 0 1x2 (2) 0 20 1 0 1 100

10. (a) b ≥ 0 and −c ≥ 0.

(b) b ≥ 0 and −c = 0

- If a2 > 0, a3 ≤ 0 we can pivot in x4 to obtain an alternative optimum.- If a3 > 0, a2 ≤ 0 we can pivot in x5 and obtain an alternative optimum.- If a2, a3 > 0 Ratio Test has to be used to determine leaving variable

(c) −c, a2, a3 < 0 ensures that x1 can be made arbitrarily large causing z to becomearbitrarily large.

11. (a) b ≥ 0 is required since we assume that the LP is written in standard form.

- If c1 = 0 and c2 ≥ 0 we can pivot in x1 to obtain an alternative optimum.- If c1 ≥ 0, c2 ≥ 0 and a2 > 0 we can pivot in x5 and obtain an alternativeoptimum.- If c2 = 0, a1 > 0 and c1 ≥ 0 we can pivot in x2 and obtain an alternativeoptimum.

(b) b < 0

(c) b = 0

(d) b ≥ 0 makes the solution feasible. If c2 < 0 and a1 ≤ 0 we can make x2 as largeas desired and obtain an unbounded solution.

(e) b ≥ 0 makes the current basic solution feasible. For x6 to replace x1 we needc1 < 0 (this ensures that increasing x1 will increase z and we need Row 3 to winthe ratio test for x1. This requires 12 ≤ a3 b.

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Exercise session 3

1. A graphical representation of the problem:

(a)−2 ≤ −c1/50 ≤ −1/2→ 25 ≤ c1 ≤ 100

(b) 20 ≤ c2 ≤ 80

(c) Current basis remains optimal for 20 ≤ b1 ≤ 50. The shadow price S.P.1 = 20.

(d) The current basis remains optimal for b2 ≥ 80/3. S.P.2 = 0 since Hops is not abinding constraint.

(e) 20 ≤ b3 ≤ 50, S.P.3 = 10.

(f) On pound = 16 ounces, so each S.P. would be divided by 16.

(g) If we look at the graph we can see:0 ≤ c1 ≤ 25 : (0, 20) is optimal → z = c1(0) + 50(20) = 1000

25 ≤ c1 ≤ 100 : (40

3,40

3) is optimal → z = c1(

40

3) + 50(

40

3) =

2000

3+

40

3c1

c1 ≥ 100 : (20, 0) is optimal → z = 20c1

Plot:

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(h) Graph:

(i) Graph:

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(j) Graph:

2. (a) S.P. = 50 cents

(b) S.P. + $ 1 = $ 1.75

(c) $ 6 + R.C. for x1 = $ 6.25

(d) $ 97.5 + 10 $0.75 = $ 105

(e) $ 97.5 + 7.5 $ 2 = $ 112.5

3. (a) The dual of the LP is:

Minimize Y = 6y1 + 8y2 + 2y3,subject to

y1 + 6y2 ≥ 5y1 + y3 ≥ 1

y1 + y2 + y3 ≥ 2and y1, y2, y3 ≥ 0.

The optimal solution is given by the coefficients of the slack variables: y2 = 0,

y2 =5

6, y3 =

7

6, giving the optimal solution Y = Z = 9.

(b) 0 ≤ c1 ≤ 6.

(c) The basis remains optimal as long as:

c2 ≤ 7/6

4. (a) $32.540 + $10x 88 = $33.420.

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(b) The allowable increase for Type 1 machines is less than one, so we cannot answerthis question.

(c) At the moment we have 260 tons available, of which 1.2 is slack. As a conse-quence, the dual price is 0 and Carco should not be willing to pay anything foran extra ton of steel.

(d) Now, the requirement is on 88 cars. The allowable decrease is 3 so we now thatthe current basis will not change. The dual price is -20, so if we only need 86cars, our profit will increase to $32.540− $20 ∗ (−2) = $32.850.

(e) $600− 1, 2$400 = $120 > 0, so Carco should consider producing jeeps.

5. (a) The dual of this LP is:


y1 + 2y2 + y3 ≥ 3y1 − y2 + y3 ≥ 4y1 + y2 ≥ 1

and y1 ≥ 0, y2 ≤ 0, y3 u.r.s.

The optimal solution can be read from the optimal primal tableau:

y1 = 1, y2 = M −M = 0, y3 = 3 +M −M = 3, which gives Y = 80.

(b) c1 ≤ 4.

(c) To avoid that entering x1 into the basis improves the objective function, weneed c2 ≥ 3.

6. (a) $2, 50 + $10, 00 = $12, 50

(b) $75

(c) $4250− 5($75) = $3875

(d) Objective function coefficient for wheat is now 26(5) = $130. The allowabledecrease for wheat is $30 so the current basis remains optimal. The decisionvariables remain unchanged, but the new z value is given by 130(25)+200(20)−10(350) = $3750.

(e) $120− 1$75− 3$12, 5 = $7, 5 > 0, so Leary should consider producing barley.



2y1 + 3y2 + 4y3 ≥ 3y1 + 2y2 + 2y3 ≥ 1

and y1 ≥ 0, y2 ≤ 0, y3 u.r.s.

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The optimal dual solution is w = 9/2, y1 = 0, y2 = 1, y3 = 3/2.

(b) If we increase the RHS of the third constraint by ∆b3, we find that:

−1 ≤ ∆b3 ≤ 1

If ∆b3 = 1/2, the new optimal solution is:

9/2 + 1/2(3/2) = 21/4.



4y1 + y2 + 3y3 ≥ −43y1 + 2y2 + y3 ≥ −1

and y1 ≤ 0, y2 ≥ 0, y3 u.r.s.

The optimal dual solution is w = −18/5, y1 = 0, y2 = 1/5, y3 = −7/5.

(b) Our dual would have an additional constraint:

y1 + y2 + y3 ≥ −1

which is not satisfied for the current solution, so the basis would change.


Minimize Y = 18000y1 + 18000y2 + 9000y3 + 1000y4,subject to

15y1 + 15y2 + 3y3 + y4 ≥ 310y1 + 15y2 + 4y3 ≥ 58y1 + 4y2 + 2y3 ≥ 4

and y1, y2, y3 ≥ 0,y4 ≤ 0

The optimal dual solution is w = 4500, y1 = 0.5, y2 = y3 = 0, y4 = −4.5

(b) By pivoting in the non-basic variable s2 (with coefficient 0 in the optimaltableau!) we can obtain an alternative optimal solution z = 4, 500, HB =375, s2 = 1, 500, s3 = 5, 250, RS = 1, 000. This solution uses only 16,500 sewingminutes.

(c) The shadow price of sewing time is 0. Thus even though all 18,000 minutes ofavailable sewing time is used, additional sewing time will not increase revenue.This is because additional sewing time is worthless in the alternative optimalsolution which uses only 16,500 minutes of sewing time.

(d) Revenue will decrease by 100(−$4.5) = −$450.

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Exercise session 4

1. (a) Graphical representation:

LP formulation:

Minimize 400x11 + 420x12 + 440x13 + 425x21 + 420x22+440x23 + 420x31 + 415x32 + 410x33,

subject tox11 + x12 + x13 = 35x21 + x22 + x23 = 30x31 + x32 + x33 = 35x11 + x21 + x31 = 30x12 + x22 + x32 = 30x13 + x23 + x33 = 20x1D + x2D + x3D = 20xij ≥ 0, ∀i = 1, 2, 3, j = 1, 2, 3, D

(b) Vogel’s Method:

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(c) Stepping stone:

All indices positive, so solution is optimal.

Transportation simplex:

u1 = 0

u1 + v1 = c11 = 400→ v1 = 400

u1 + v2 = c12 = 420→ v2 = 420

u2 + v2 = c22 = 420→ u2 = 0

u2 + v4 = c24 = 0→ v4 = 0

u3 + v2 = c32 = 415→ u3 = −5

u3 + v3 = c33 = 410→ v3 = 415

For each unused cell, we now calculate the reduced cost:

c13 = c13 − u1 − v3 = 440− 0− 415 = 25

c14 = c14 − u1 − v4 = 0− 0− 0 = 0

c21 = c21 − u2 − v1 = 425− 0− 400 = 25

c23 = c23 − u2 − v3 = 440− 0− 415 = 25

c31 = c31 − u3 − v1 = 420− (−5)− 400 = 25

c34 = c34 − u3 − v4 = 0− (−5)− 0 = 5

2. 200 word processing files should be stored on the hard disk, 100 word processing filesin the computer memory, 100 packaged programs on tape, and 100 data files on tape.

3. Car 2/Car 4 for call 1, Car 5 for call 2, Car 4/Car 3 for call 3. 14 city blocks have tobe traveled in total.

7. Let i = flights that leave from NY, j = flights that leave from Chicago. Optimalassignment: x13 = x24 = x35 = x41 = x56 = x67 = x72 = 1. Total downtime = 25hours.

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8. LP:

Table:

10. (a) Supply point ij = engine j of company i and Demand point ij = j’th engine togo to fire i. All supplies and demands equal 1.

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This formulation picks up the correct costs because earlier engines arriving at afire have a larger effect on cost than later engines arriving at the same fire. Thisimplies, for example, that the transportation simplex or assignment method willnever assign an engine which takes 6 minutes to get to a fire to ’arrive’ earlierthan an engine taking 4 minutes to get to the same fire (because interchangingthese assignments would reduce cost and leave the same engines available forother assignments).

(b) it might be optimal for an assignment to have t12 < t11, and this would resultin incorrect costs being assigned.

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Exercise session 5

1. Shortest path = 1-2-5, cost = 14.

2. Dijkstra: 1-3-4, cost = 2. Shortest path = 1-2-3-4, cost =1. Since we cannot updatenode 3 from node 2 (since its distance is already fixed in the second iteration), wefail to obtain the shortest path with Dijkstra’s algorithm.

4. Max flow:

The total flow = 45. Minimum cut: so-2 with capacity 20 + 10 + 15 = 45.

5. Max flow:

Optimal flow = 9 units. The minimum cut is so-1-3 with capacity 3+1+3+2=9.

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6. Network:

If max flow= 21 then all packages can be loaded.

7. Network representation:

Clearly, 1200 cars can be sent (300 directly in first and second hour, 300 each via C2and C3).

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8. (a) The shortest path from NY to LA = {NY, St.-L.,Ph.,LA}, requiring 2450 gallonsof gas.

(b) Balanced transportation problem:

9. (a)

(b) Max flow = 750, min cut ={LA}

10. Prim:

Length of the MST = 600 + 600 + 800 + 400 + 900 + 400 + 600 = 4300

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Kruskal:

Length of MST = 400 + 400 + 600 + 600 + 600+ 800 + 900 = 4300

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Exercise session 6

1. Let{yi = 1 if we buy any computers from vendor i

yi = 0 otherwise.

xi = Number of computers purchased from vendor i.

Then our IP becomes:

Minimize Z = 500x1 + 350x2 + 250x3 + 5000y1 + 4000y2 + 6000y3,subject to

x1 + x2 + x3 = 1100x1 ≤ 500y1x2 ≤ 900y2x3 ≤ 400y3

yi ∈ {0, 1}, xi ≥ 0, ∀i = 1..3

2. Tree:

3. Solving the LP relaxation gives us the integer optimal solution x1 = x2 = z = 0.

4. After many iterations, we find out that the problem is infeasible:

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5. Let xi = 1 if surgeon i is used and xi = 0 otherwise. Then we can write:

Minimize Z = x1 + x2 + x3 + x4 + x5 + x6,subject to

x1 + x4 ≥ 1x2 + x3 ≥ 1x1 + x5 ≥ 1x1 + x6 ≥ 1x2 + x3 + x6 ≥ 1x1 + x4 ≥ 1x1 + x2 ≤ 1xi ∈ {0, 1}, ∀i = 1..6

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6. Let xij = Time that job i begins its processing on Machine j. For the objectivefunction, we took the average processing time.

Minimize Z = (x14 + 30 + x24 + 18 + x33 + 28)/3,subject to

x13 ≥ x11 + 20x14 ≥ x13 + 25x22 ≥ x21 + 15x24 ≥ x22 + 20x33 ≥ x32 + 35

x11 + 20− x21 ≤My1x21 + 15− x11 ≤M(1− y1)x22 + 20− x32 ≤My2x32 + 35− x22 ≤M(1− y2)x13 + 25− x33 ≤My3x33 + 28− x13 ≤M(1− y3)x24 + 18− x14 ≤My4x14 + 15− x24 ≤M(1− y4)

xij ≥ 0, ∀i = 1..3, j = 1..4yi ∈ {0, 1}, ∀i = 1..4

7. Let xit = 1 if plant i is built at the beginning of year t, and 0 otherwise. We find:

Minimize Z =∑5

t=1(20x1t + 16x2t + 18x3t + 14x4t)+7.5x11 + 6x12 + 4.5x13 + 3x14 + 1.5x15+4x21 + 3.2x22 + 2.4x23 + 1.6x24 + 0.8x25+6.5x31 + 5.2x32 + 3.9x33 + 2.6x34 + 1.3x35+3x41 + 2.4x42 + 1.8x43 + 1.2x44 + 0.6x45,

subject to ∑5t=1 xit ≤ 1, ∀i = 1..4∑tj=1(70x1j + 50x2j + 60x3j + 40x4j) ≥ dt, ∀t = 1..5

and xit ∈ {0, 1}, ∀i = 1..4, t = 1..5

, where dt is the capacity demand during year t.

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8. The IP:

Minimize Z =∑5

t=1(1.5x1t + 0.8x2t + 1.3x3t + 0.6x4t)

+∑4

t=1(1.7y1t + 1.2y2t + 1.3y3t + 0.8y4t)

+∑5

t=2(1.9z1t + 1.5z2t + 1.3z3t + 1.1z4t),subject to

70x1t + 50x2t + 60x3t + 40x4t ≥ dt, ∀t = 1..5

xit − xi,t+1 ≤ 1− wt, ∀t = 1..41− yit ≤ wt, ∀t = 1..4

xit − xi,t−1 ≤ 1− w′t, ∀t = 2..51− zit ≤ w′t, ∀t = 2..5

and all variables ∈ {0, 1}

10. Let yi = 1 if an auditor is based in city i (1 = NY, 2 = Chic, 3 = LA, 4 = Atl) andyi = 0 otherwise. Also let xij = the number of trips made by an auditor based incity i to region j(j = 1 is Northeast, etc.) Then an appropriate formulation is

Minimize Z = 100, 000(y1 + y2 + y3 + y4) + 1100x11 + 1400x12 + 1900x13 + 1400x14+1200x21 + 1000x22 + 1500x23 + 1200x24 + 1900x31 + 1700x32+1100x33 + 1400x34 + 1300x41 + 1400x42 + 1500x43 + 1050x44

subject tox11 + x21 + x31 + x41 ≥ 500x12 + x22 + x32 + x42 ≥ 400x13 + x23 + x33 + x43 ≥ 300x14 + x24 + x34 + x44 ≥ 400

x11 + x12 + x13 + x14 ≤ 1600y1x21 + x22 + x23 + x24 ≤ 1600y2x31 + x32 + x33 + x34 ≤ 1600y3x41 + x42 + x43 + x44 ≤ 1600y4

and yj ∈ {0, 1}, ∀j = 1..4, xij integer, ∀i = 1..4, j = 1..4

30

Exercise session 7

1. Let x be the location of the store. We wish to choose x to minimize

f(x) = (x− 3)2 + (x− 4)2 + (x− 5)2 + (x− 6)2 + (x− 17)2

Since we have no side constraints, we can easily find the minimum by:

f ′(x) = 2(x− 3 + x− 4 + x− 5 + x− 6 + x− 17) = 0

, giving x = 7 as a local minimum. However, f ′′(x) = 10 > 0 for all values of x,meaning our function is convex. The local minimum is therefore a global minimum,so the store should be located at x = 7. In general, we get:

f ′(x) = 2(x − x1 + x − x2 + x − x3 + ... + x − xn) = 0, giving x =∑n

i=1 xi/n. Thestore should be located at the arithmetic mean of the location of the n customers.

2. (a) Let R be the number of units of raw materials purchased. We wish to solve:

Maximize Z = (49− x1)x1 + (30− 2x2)x2 − 5R,subject to

x1 ≤ 2Rx2 ≤ R

x1, x2, R ≥ 0, ∀i = 1..3

We derive the Hessian of the objective function:

H(x1, x2, R) =

−2 0 00 −4 00 0 0

.

Since all elements in this matrix are ≤ 0, so the objective function is concave.Since the constraints are linear, the K-T conditions will yield an optimal solu-tion. The K-T conditions are:

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49− 2x1 − λ1 ≤ 0

30− 4x2 − λ2 ≤ 0

−5 + 2λ1 + λ2 ≤ 0

λ1(x1 − 2R) = 0

λ2(x2 −R) = 0

x1 − 2R ≤ 0

x2 −R ≤ 0

(49− 2x1 − λ1)x1 = 0

(30− 4x2 − λ2)x2 = 0

(−5 + 2λ1 + λ2)R = 0

x1, x2, R, λ1, λ2 ≥ 0

3. After several iterations, we find that the interval of uncertainty is [1.18, 1.63). Thisinterval has width less than 0.50, so we are finished. (actual maximum occurs forx = 1.5)

4. Iteration 1

∇f(x1, x2) = (e−(x1+x2)(1− x1 − x2)− 1, e−(x1+x2)(1− x1 − x2)).∇f(0, 1) = (−1, 0). Thus new point is (−t, 1) where t ≥ 0 maximizes f(−t, 1) =(1− t)e(t−1) + t.f ′(−t, 1) = (1 − t)e(t−1) − e(t−1) + 1 = 0 for 1 = te(t−1) or t = 1. Thus new point is(−1, 1)

Iteration 2

∇f(−1, 1) = (0, 1) so new point is (−1, 1 + t) and we choose t ≥ 0 to maximizef(−1, 1 + t) = te−t + 1.f ′(−1, 1 + t) = −te−t + e−t = 0 for t = 1. Thus new point is (−1, 2).

5. Using the lagrangian multiplier λ, we find x = 6, y = 4, w = 3, λ = 6, z = 72.

6. Using the K-T conditions yields y = 2 and x = 5, with z = 30. Since this is the onlypoint satisfying KT conditions, it must be the optimal solution.

7. We want to maximize Π = (10, 000 + 5√a− 100p)(p− 10)− a. Deriving to p and a,

we find p = 58 and a = 14, 400.

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8. Using the hint (with −x ≤ −a), we get the K-T conditions:

1− ln(x)

x2+ λ = 0

λ(a− x) = 0

x ≥ a

λ ≥ 0

We find the unique solution to the K- T conditions occurs for x = a, and λ = ln(a)−1a2

.

Thus for b > a ≥ e we know that ln(b)b

< ln(a)a

or a ln(b) < b ln(a). Taking e to bothsides we obtain ba < ab. To solve the problem now let a = e and b = π.

10. x = location of the store.We wish to choose x to minimize: f(x) = |x−3|+ |x−4|+ |x−5|+ |x−6|+ |x−17|.

Problem 4: For any two points on the curve y = |x| the line joining the two pointsis never below the curve. Thus |x| is a convex function of x.

From Problem 4 (and the fact that the sum of convex functions is convex) we knowthat f(x) is a convex function. Thus any local minimum of f(x) will minimize f(x).

We claim that choosing x = 5 (the median of the customer’s locations) yields a localminimum. To see this suppose that you move the store location λ units to the leftof 5 (λ is small). Then customers 3 and 4 travel λ less units to store but customers5, 6, and 17 travel λ units more to store, so total travel time increases. Suppose youmove the store location λ units to the right of 5. Then customers 3,4 and 5 travelλ more units to the store while customers 6 and 17 travel λ less units to the store,and again total travel time increases. Thus x = 5 is a local minimum, and hence aminimum over all possible store locations.

In general, if there are n customers, the store should be located at the median valueof the customer’s locations. If there are an even number (say 2n) of customers, locatethe store anywhere between the n’th and n + 1 customer.

11. (a) K-T conditions are (ignoring non-negativity) original constraints:

x1p′1(x1) + p1(x1)− λ1 = 0 (1)

x2p′2(x2) + p2(x2)− λ2 = 0 (2)

−c+ λ1k1 + λ2k2 = 0 (3)

λ1(x1 − k1z) = 0 (4)

λ2(x2 − k2z) = 0 (5)

λ1, λ2 ≥ 0, x1 ≤ k1z, x2 ≤ k2z (6)

33

(b) Now firm wishes to maximize f(x1, x2)) = x1p1(x1) + x2p2(x2) − λ̄1x1 − λ̄2x2.The optimum is found when:

x1p′1(x1) + p1(x1)− λ̄1 = 0

x2p′2(x2) + p2(x2)− λ̄2 = 0

Clearly any solution to the original problem will satisfy these equations.

Also the revenues in (a) are identical to the revenues in (b). Can we show thatthe costs in (b) are identical to the cost cz incurred in the original problem?

Case 1: λ̄1 > 0, λ̄2 > 0

Then x̄1 = k1z̄ and x̄2 = k2z̄ and from (3) total cost iscz̄ = λ̄1k1z̄+λ̄2k2z̄ = λ̄1x1+λ̄2x2, and both problems incur the same productioncost.

Case 2: λ̄1 = 0, λ̄2 = 0:

From (3) this cannot occur unless c = 0.

Case 3:λ̄1 > 0, λ̄2 = 0:

Then (3) and (4) yield x̄1 = k1z̄ and k1λ̄1 = c. Now total cost is cz̄ = k1λ̄1z̄ =x̄1λ̄1 + x̄2λ̄2 , as desired.

Case 4: λ̄1 = 0, λ̄2 > 0:

The reasoning here is identical to Case 3.

(c) Shadow prices.

34

Exercise session 8

1. The maximin decision is to mount a small advertising campaign, maximax + minimaxregret decision is to mount a large advertising campaign.

2. The reward matrix is:

Noble Greek PricePizza King Price $6 $8 $10

$5 $125 $175 $225$6 $200 $300 $400$7 $225 $375 $525$8 $200 $400 $600$9 $125 $375 $625

The Maximin action is to charge $7, the Maximax action is to charge $9 and theMinimax regret action is to charge $8. Pizza King maximizes their expected rewardby charging $8.

3. Let p = Alden’s bid. Then the reward table is as follows:

Forbes’ bidAlden’s bid = p $6000 $8000 $11000

p = $6000 0 0 0p = $8000 0 $2000 $2000p = $11000 0 0 $5000

Observe that p = $6,000 is dominated by p =$8,000.

Either $8,000 or $11,000 is the Maximin action, the Maximax and Minimax regretbid is $11,000. Bidding $11,000 also maximizes expected profit.

4. The optimal solution is to buy the forecast and then act optimally (don’t buy thepass if sunny forecast and buy the pass if rainy forecast).

If forecast were free it would be worth EVSI = 15.00 -13.56 = $1.44 > $1.00. TheEVPI = -9.6 - (-15) = $5.40.

5. The doctor should operate now.

35

6. The bank should run the survey. If the recommendation is favorable, grant the loan.If not, invest in bonds.

EVSI = 4280 + 500 - 3,760 = $1020 > $500.EVPI = 0.04*3000 + 0.96*6000 - 3,760 = $2,120.

7. We should hire geologist and if geologist predicts an earthquake, build at Roy Rogers.If the geologist predicts no quake build at Diablo.

EVSI = -13.90 +1 - (-14) = $1.1 million > $1 million, so it is worth it to pay $1million to hire the geologist!

EVPI = -12 - (-14) = $2 million.

8. Buy toss and if 1 comes up choose die in right hand and if 6 comes up choose die inleft hand.

EVSI = (312 + 15) - 310 = $17.

EVPI = 360 - 310 = $50.

10. (a) From the tree we see that university should not drug test (lower expected cost).Note: we assume that we must act on positive drug test and must not act onnegative drug test.

(b) From the decision tree we find that Expected Cost for Drug Testing> 0.14*(0.68)*c1> 0.14*(0.68)*c2 > Expected Cost for not Drug Testing = 0.05*c2, so it is op-timal not to test for drugs.

36

Exercise session 9

1. Shortest path from node 1 to node 10 is 1-4-8-10, shortest path from node 2 to node10 is 2-5-8-10.

2. Retracing the optimal path we find that the salesman should speak in Bloomingtonon Monday, Tuesday, and Wednesday and speak in Indianapolis on Thursday.

3. Define xt(i) to be a production level during period t with inventory level i at thebeginning of the month.

One optimal plan is to produce 0 units during month 1, x2(0) = 1 unit, and x3(0) = 4units during month 3. Other optimal production plans include:Plan three units produced during month 2 and two units produced during month 4.Plan one unit produced during month 2, two units produced during month 3, andtwo units produced during month 4.

All optimal plans incur a cost of $12.

4. It is optimal to produce x1(1) = 0 units during month 1; produce x2(0) = 0 unitsduring month 2; produce x3(−1) = 0 units during month 3; produce x4(−3) = 5 unitsduring month 4. Production is postponed because the backlogging cost is smallerthan the holding cost.

5. Define ft(i) be the minimum cost incurred during months t,t + 1,..3 if the inventoryat the beginning of month t is i.Let xt(i) be the quantity that should be producedduring month t in order to attain ft(i).

We find that f1(0) = $8,950 and x1(0) = 200 radios should be produced during month1. This yields a month 2 inventory of 200 - 200 = 0. Thus during month 2 we producex2(0) = 600 radios. At the beginning of month 3 the inventory will now be 0 + 600- 300 = 300. Hence during month 3 x3(300) = 0 radios should be produced.

Note that the total production cost of this plan is 250 + 200(10) + 250 + 600(10)= 8,500 and Total Holding Cost = 1.5(300) = 450 (assessed on inventory at end ofmonth 2). Thus total cost is 8500 + 450 = $8, 950 = f1(0).

6. Define gt to be minimum net cost incurred from time t to end of problem (time t =beginning of year t + 1) given that a new machine has just been purchased at timet. gt includes the cost of buying the machine purchased at time t. Then we find thatthe machine should always be kept for one year and then traded in.

7. (a) Let ft(w) = minimum cost incurred in meeting demands for years t,... 5 giventhat (before hiring and firing for year t) w workers are available. Further, let:

37

ht = Workers hired at beginning of year tdt = Workers fired (discharged) at beginning of year t.wt = Workers required during year t.We assume that newly hired workers do not quit during their first year, andthat salaries are paid to workers working any portion of a year. Then

ft(w) = min {10, 000ht + 20, 000dt + 30, 000(w+ht−dt) + ft+1(ht + .9(w−dt))}

, where ht and dt must satisfy 0 ≤ ht, 0 ≤ dt ≤ w, and ht + w − dt ≥ wt. Wework back to compute f1(20).

(b) Add the condition that if w < wt, ft(w) = ∞. This will ensure that eachmonth’s requirement is met. Also note that dt must satisfy w − dt ≥ wt.

8. Add 1 spare unit to each of the systems. The probability that all three systems willwork is .689.

9. We choose to begin by putting a Type 2 item in the knapsack. This leaves us withan 8 - 3 = 5 pound knapsack so we put another Type 2 item in the knapsack. Thisleaves us with a 5 - 3 = 2 pound knapsack so we next put in a Type 3 item to fill theknapsack. Thus the knapsack should be filled with two Type 2 items and one Type3 item. There are, of course, other optimal solutions such as using two Type 1 items.

38

Exercise session 10

1. Transition matrix:

P =

0 1 2 3 4

0 0 0 1/3 1/3 1/31 0 0 1/3 1/3 1/32 1/3 1/3 1/3 0 03 0 1/3 1/3 1/3 04 0 0 1/3 1/3 1/3

2. Transition matrix:

P =

0 1 2

0 0 0 11 1 1/3 2/32 1/9 4/9 4/9

3. (a) State 4 is transient.

(b) States 1, 2, 3, 5, 6

(c) {1,3,5} and{2,6} are closed sets.

(d) No, because state 4 is transient.

4. (a) π1 = 0.6, π2 = 0.4.

(b) π1 = 16/25, π2 = 1/5, π3 = 4/25.

(c) m11 = 1.56,m12 = 5,m13 = 6.25,m23 = 1.25,m21 = 2.81,m31 = 1.56,m22 =5,m32 = 5,m33 = 6.25.

5. If we do not replace a fair car we face the following transition matrix (note: weobserve state at beginning of year before a car is replaced):

P =

G F BD

G .85 .10 .05F 0 .70 .30BD .85 .10 .05

The Expected Cost per Year is given by 1000(.64) + 1500(.25) + 7000(.11) = $1785.

If we do replace a fair car, the Expected Net Cost Per Year is $1700, so the car shouldbe replaced as soon as it becomes fair.

39

6. Let the state be the age of the machine at the beginning of the month. For example,if state is 3, the machine is 3 months old at the beginning of the month and startsits fourth month of operation.

For Policy 1, the expected cost per month = $431.5, for Policy 2 the expected costper month = $410.5, for Policy 3 the Expected Cost Per Month= $600

Thus Policy 2(replace after 2 months) is the cheapest policy.

7. Expected Cost/Month for policy 1 = $30, for policy 2 this is $35. Thus Policy 1 hasa lower expected monthly cost.

8. (a) Element 1-1 of (I −Q)−1R = .705.

(b) Element 2-2 of (I −Q)−1R = .30.

(c) Sum of the elements in row 1 of (I −Q)−1 = 1 + .975 + .825 = 2.8.

9. (a) Sum of the elements in row 1 of (I −Q)−1 = 7.66 + 2.49 + .85 = 11.0 days.

(b) Good Prediction = 50 + 500(.65) + 300(.50) + 200(.51) = 627Fair Prediction = 40 + 500(.20) + 300(.30) + 200(.25) = 280Critical Prediction = 30 + 500(.05) + 300(.12) + 200(.20) = 131

(c) Let G = steady state number in good condition, F = steady state in fair condi-tion, C = steady state number in critical condition:20 + .5F + .51C = .35G10 + .2G + .25C = .70F10 + .05G + .12F = .80CSolving we find G = 289.6 F = 114 C = 47.7

(d) Entry 1-1 of (I −Q)−1R = .54

10. (a) Let A = steady state number of gallons of A sold per week, B = steady statenumber of gallons of B sold per week, C = steady state number of gallons of Csold per week.

Then

A = 1, 000, 000pB + pC − pApA + pB + pC

B = 1, 000, 000pA + pC − pBpA + pB + pC

40

C = 1, 000, 000pA + pB − pCpA + pB + pC

(b) Currently A’s market share is (.15+.20-.10)/(.10+.15+.20) = 55.6%. If we cutpercentage of defective juice cans in half pA = .05 and we find A’s market shareis now (.15+.20-.05)/(.05+.15+.20)= .75. Annually improved quality increasesprofit by $10, 088, 000 = 52($10, 000)(75− 55.6) > $1, 000, 000, so we should gofor the improved quality.

41

Exercise session 11

1. 6/7

4. (a) 5/6

(b) 25/6

(c) 1/2 minute

5. (a) 1.33 customers

(b) 3 minutes

(c) 16/81

6. (a) 2.45

(b) 14.8 cars/hour

(c) 0.23 h

7. (a) 8/13 - 1/13 - 4/13

(b) 6/13 lines

(c) 30/13 calls per hour will be lost

8. Let CNC = College ambulance is answering non-college callCC = College ambulance is answering college callCI = College ambulance is idleDNC = Downtown ambulance is answering non-college callDC = Downtown ambulance is answering college callDI = Downtown ambulance is idlePossible states (numbered 1-9) areState 1: (CNC, DNC) State 2: (CNC, DC) State 3: (CNC, DI)State 4: (CC, DNC) State 5: (CC, DC) State 6: (CC, DI)State 7: (CI, DNC) State 8: (CI, DC) State 9: (CI,DI)

(a) 0.24

(b) 0.228

(c) 8.1 % of all calls are considered lost to the system

(d) We assume that half of the ”service”time is the time required for an ambulanceto go from its base to a patient.A total of (1 - .081)3 = 2.76 college calls per hour are handled. 3(π3 + π6) callswait an average of 5/2 minutes and 3(π7+π8+π9) calls wait an average of 4/2 mi-

nutes. Thus a College call waits an average of3(π3 + π6)2, 5 + 3(π7 + π8 + π9)2

2.76=

42

2.08.In the same way we find that the average waiting time for a non-college call is2.25 min. Thus, on average, a non-college call waits longer than a college call.

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Introduction to Operations Research Answers to selected problems · 2014. 8. 14. · Introduction to Operations Research Answers to selected problems Prof. dr. E-H. Aghezzaf, assistant