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Chapter 1
Example 1.1 The Apex Television Company has to decide on the number of 27- and 20-inch sets to be produced at one of its factories. Market research indicates that at most 40 of the 27-inch sets and 10 of the 20-inch sets can be sold per month. The maximum number of work-hours available is 500 per month. A 27-inch set requires 20 work-hours and a 20-inch set requires 10 work-hours. Each 27-inch set sold produces a profit of $120 and each 20-inch set produces a profit of $80. A wholesaler has agreed to purchase all the television sets produced if the numbers do not exceed the maxima indicated by the market research. (a) Formulate a linear programming model for this problem. The decisions that need to be made are the number of 27-inch and 20-inch TV sets to be produced per month by the Apex Television Company. Therefore, the decision variables for the model are x1 = number of 27-inch TV sets to be produced per month, x2 = number of 20-inch TV sets to be produced per month. Also let Z = total profit per month. The model now can be formulated in terms of these variables as follows. The total profit per month is Z = 120 x1 + 80 x2. The resource constraints are: (1) Number of 27-inch sets sold per month: x1  40 (2) Number of 20-inch sets sold per month: x2  10 (3) Work-hours availability: 20 x1 + 10 x2  500. Nonnegativity constraints on TV sets produced: x1  x2  0, 0 With the objective of maximizing the total profit per month, the LP model for this problem is Maximize Z = 120 x1 + 80 x2 subject to x1  40 x2  10 20 x1 + 10 x2 500 and x1  x2  0, 0

(b) Use the graphical method to solve this model.

Example 1.2 Dwight is an elementary school teacher who also raises pigs for supplemental income. He is trying to decide what to feed his pigs. He is considering using a combination of pig feeds available from local suppliers. He would like to feed the pigs at minimum cost while also making sure each pig receives an adequate supply of calories and vitamins. The cost, calorie content, and vitamin content of each feed are given in the table below. Contents Calories (per pound) Vitamins (per pound) Cost (per pound) Feed Type A 800 140 units $0.40 Feed Type B 1,000 70 units $0.80

Each pig requires at least 8,000 calories per day and at least 700 units of vitamins. A further constraint is that no more than one-third of the diet (by weight) can consist of Feed Type A, since it contains an ingredient which is toxic if consumed in too large a quantity. (a) Formulate a linear programming model for this problem. Let A and B be the quantity (pounds) of Feed Type A and Feed Type B, respectively, used per day. Also let Z be the total daily cost of the feed per pig. Then, the daily cost is Z = $0.4 A + $0.8 B. The constraints on the minimum daily requirements of calories and vitamins are (1) Calories requirement: 800 A + 1000 B 8,000.

(2) Vitamins requirement: 140 A + 70 B  700. Also, Dwight needs to avoid using too much of Feed Type A because of the toxic ingredient in it. The toxic constraint is A  1/3 (A+B), which reduces to 2/3 A - 1 B≤ . / 3 0 Nonnegativity constraints: A  B  0, 0. The resulting linear programming model for this problem is Minimize Z = 0.4 A + $0.8 B, subject to 800 A + 1000 B  8000 140 A + 70 B  700 2/3 A - 1/3 B  0 and A  B  0, 0. (b) Use the graphical method to solve this model. What is the resulting daily cost per pig? As shown below, the optimal solution is (A, B) = (20/7, 40/7). The resulting daily cost per pig is Z = 40/7 = $5.71.

Example 1.3 A furniture store has set aside 800 square feet to display its sofas and chairs. Each sofa utilizes 50 sq. ft. and each chair utilizes 30 sq. ft. At least five sofas and at least five chairs are to be displayed.

a. Write a mathematical model representing the store's constraints. b. Suppose the profit on sofas is $200 and on chairs is $100. On a given day, the probability that a displayed sofa will be sold is .03 and that a displayed chair will be sold is .05. Mathematically model each of the following objectives: 1. Maximize the total pieces of furniture displayed. 2. Maximize the total expected number of daily sales. 3. Maximize the total expected daily profit. Solution: Let s and c be the number of sofa and chair respectively. a. 50s + 30c < 800 s> 5 c> 5 b. (1) Max s + c (2) Max 0.03s + 0.05c (3) Max 6s + 5c Example 1.4 Maxwell Manufacturing makes two models of felt tip marking pens. Requirements for each lot of pens are given below. Plastic Ink Assembly Molding Time Fliptop Model Tiptop Model 3 4 5 4 5 2 Available 36 40 30

The profit for either model is $1000 per lot. a. What is the linear programming model for this problem? b. Find the optimal solution. c. Will there be excess capacity in any resource? Solution: a. Let F = the number of lots of Fliptip pens to produce Let T = the number of lots of Tiptop pens to produce Max 1000F + 1000T s.t. 3F + 4T < 36 5F + 4T < 40 5F + 2T < 30 F,T > 0

b.
15

T

10

5

0 0 5 10

F
15

The complete optimal solution is:F = 2, T = 7.5, Z = 9500 c.There is an excess of 5 units of molding time available.

Chapter 2
Example 2.1 Consider the following linear programming model. Maximize Z = 3x1 + 2 x2, subject to x1 ≤4 x1 + 3x2 ≤5 1 2x1 + x2 ≤0 1 and x1 ≥ , 2 ≥ . 0x 0 (a) Use graphical analysis to identify all the corner-point solutions for this model. Label each as either feasible or infeasible. The graph showing all the constraint boundary lines and the corner-point solutions at their intersections is shown below.

The exact value of (x1, x2) for each of these corner-point solutions (A, B, ..., I) can be identified by obtaining the simultaneous solution of the corresponding two constraint boundary equations. The results are summarized as... Corner-point solutions A B C D E F G H I (x1, x2) (0, 5) (0,10) (3, 4) (4, 11/3) (4, 2) (4, 0) (5, 0) (15, 0) (0, 0) Feasibility Feasible Infeasible Feasible Infeasible Feasible Feasible Infeasible Infeasible Feasible

(b) Calculate the value of the objective function for each of the CPF solutions. Use this information to identify an optimal solution. The objective value of each corner-point feasible solution is calculated in the following table: Corner-point feasible solutions A C E F I (x1, x2) (0, 5) (3, 4) (4, 2) (4, 0) (0, 0) Objective Value Z 3*0+2*5 = 10 3*3+2*4 = 17 3*4+2*2 = 16 3*4+2*0 = 12 3*0+0*0 = 0

Since point C has the largest value of Z, (x1, x2) = (3, 4) must be an optimal solution. (c) Use the simplex method to get the solution and to identify which sequence of CPF solutions would be examined by the simplex method to reach an optimal solution. Therefore, the sequence of CPF solutions examined by the simplex method would be I → F→ E→ C. Example 2.2 Consider the following linear programming model. Maximize Z = x1 + 2x2, subject to 6x1 - 2x2 ≤ 3 2x1 + 3x2 ≤ 6 x1 + x2 ≤ 3 and x1 ≥ , 2 ≥ . 0x 0 (a) Use graphical analysis to identify all the corner-point solutions for this model. Label each as either feasible or infeasible. (b) Calculate the value of the objective function for each of the CPF solutions. Use this information to identify an optimal solution. (c) Use the simplex method to get the solution and to identify which sequence of CPF solutions would be examined by the simplex method to reach an optimal solution.

Solution: (a)
3

x1 + x2 = 3 6x1- 2x2 = 3

2

1

Feasible Region
2x1 + 3x2 = 6

x1 + 2x2 =0

0 0 1 2 3

x1 (b) obj. value 0 0.5 3.68 4
2 x2 -2 3 1 0 2 2 x2 0 1 0 2 0 0 0 0 sl1 sl2 sl3 bi  i - 1 0 0 3 0 1 0 6 2 0 0 1 3 3 0 0 0 ∑cBibi= 0 i 0 0 0 0 0 0 sl1 sl2 sl3 1 2/3 0

(a) corner-point feasibility (0, 0) feasible (0.5, 0) feasible (21/22, 15/11) feasible (0, 2) feasible (c) 1st tableau i 1 2 3 Max Cj cB xB cB1= 0 xB1= sl1 cB2= 0 xB2= sl2 cB3= 0 xB3= sl3 fj Cj-fj Max cB Cj xB 1 x1 6 2 1 0 1 1 x1 2/3 1/3 4/3 -1/3

2nd tableau (final) i bi 7

 i

1 cB1= 0 xB1= sl1 22/3 2 cB2= 2 xB2= x2 3 cB3= 0 xB3= sl3 fj Cj-fj

0 1/3 0 2 0 -1/3 1 1 0 2/3 0 ∑cBibi= 4 i 0 -2/3 0

Corner point (0, 0) → Corner point (0, 2)

Chapter 3
Example 3.1 The Fagersta Steelworks currently is working two mines to obtain its iron ore. This iron ore is shipped to either of two storage facilities. When needed, it then is shipped o t t cm ays t l l tT e i r bl dp tt s ir u o ntok n o h o pn’s ep n h d ga e w eish d tbt n e r, e e a. a m o c i si i w where M1 and M2 are the two mines, S1 and S2 are the two storage facilities, and P is the steel plant. The diagram also shows the monthly amounts produced at the mines and needed at the plant, as well as the shipping cost and the maximum amount that can be shipped per month through each shipping lane.
40 tons produced M1 $2,000/ton 30 tons max. $1,700/ton 30 tons max. P $1,600/ton 50 tons max. 60 tons M2 produced $1,100/ton 50 tons max. S2 $800/ton 70 tons max. 100 tons needed S1 $400/ton 70 tons max.

Management now wants to determine the most economic plan for shipping the iron ore from the mines through the distribution network to the steel plant. (a) Formulate a linear programming model for this problem. The decision variables are defined as follows: xm1-s1 : number of units (tons) shipped from Mine 1 to Storage Facility 1, xm1-s2 : number of units (tons) shipped from Mine 1 to Storage Facility 2, xm2-s1 : number of units (tons) shipped from Mine 2 to Storage Facility 1, xm2-s2 : number of units (tons) shipped from Mine 2 to Storage Facility 2, xs1 -p : number of units (tons) shipped from Storage Facility 1 to the Plant, xs2 -p : number of units (tons) shipped from Storage Facility 2 to the Plant. The total shipping cost is:
Z = 2000 xm1-s1 + 1700 xm1-s2 + 1600 xm2-s1 + 1100 xm2-s2 + 400 xs1-p + 800 xs2-p

The constraints we need to consider are: (1) Supply constraint on M1 and M2: xm1-s1 + xm1-s2 = 40 xm2-s1 + xm2-s2 = 60 (2) Conservation-of-flow constraint on S1 and S2: xm1-s1 + xm2-s1 - xs1-p = 0 xm1-s2 + xm2-s2 - xs2-p = 0 (3) Demand constraint on P: xs1-p + xs2-p = 100

(4) Capacity constraints: xm1-s1  30, xm1-s2  30 xm2-s1  50, xm2-s2  50 xs1-p  70, xs2-p  70 (5) Nonnegativity constraints: xm1-s1  xm1-s2  0, 0 xm2-s1  xm2-s2  0, 0 xs1-p  xs2-p  0, 0 The resulting linear programming model for this problem is: Maximize Z = 2000 xm1-s1 + 1700 xm1-s2 + 1600 xm2-s1 + 1100 xm2-s2 + 400 xs1-p + 800 xs2-p, subject to xm1-s1 + xm1-s2 = 40 xm2-s1 + xm2-s2 = 60 xm1-s1 + xm2-s1 - xs1-p = 0 xm1-s2 + xm2-s2 - xs2-p = 0 xs1-p + xs2-p = 100 xm1-s1  30, xm1-s2  30 xm2-s1  50, xm2-s2  50 xs1-p  70, xs2-p  70 and xm1-s1  xm1-s2  xm2-s1  xm2-s2  xs1-p  xs2-p  0, 0, 0, 0, 0, 0 Example 3.2 Minimize f = -1x1 + 2x2 - 3x3 subject to x1 + x2 + x3 = 6 -1x1 + x2 + 2x3 = 4 2x2 + 3x3 = 10 x3  2 and x1, x2, x3  0 Find the optimal solution by big M and two phase methods. Solution: The first tableau of big M method: i 1 2 3 Cj -1 cB xB x1 cB1= M xB1=a1 1 cB2= M xB2=a2 -1 cB3= M xB3=a3 0 0 0 fj Cj-fj Min 2 x2 1 1 2 0 4M -3 x3 1 2 3 1 6M M a1 1 0 0 0 M 0 M M a2 a3 0 0 1 0 0 1 0 0 0 0 M M 0 s1 0 0 0 1 0 0 bi 6 4 10 2 ∑cBibi= 20M+2 i  i

4 cB4= 0 xB3=s1

-1 2-4M -3-6M

The first tableau of Phase I in two phase method:
Min i cB Cj ’ xB 0 x1 1 0 0 0 0 0 x2 1 1 2 0 4 -4 0 x3 1 2 3 1 6 -6 1 a1 1 0 0 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 s1 0 0 0 1 0 0 bi 6 4 10 2  i 6 2 10/3 2 a2 a3

1 cB1= 1 xB1=a1 3 cB3= 1 xB3=a3 4 cB4= 0 xB3=s1 f ’ j Cj-f ’’ j

2 cB2= 1 xB2=a2 -1

∑cBibi= 20 i

The final tableau of Phase I in two phase method:
Min i cB Cj ’ xB 0 x1 1 0 0 0 0 0 0 x2 0 1 0 0 0 0 0 x3 0 0 0 1 0 0 1 a1 1 a2 1 a3 0 s1 bi 2 2 0 2 ∑cBibi= 0 i  i

1 cB1= 0 xB1=x1 2 cB2= 0 xB2=x2 3 cB3= 1 xB3=a3 4 cB4= 0 xB3=x3 f ’ j Cj-f ’’ j

1/2 -1/2 0 1/2 1/2 1/2 -1 0 -1 2 -1 0 -1 2 0 -3/2 1 0 1 0 0 1 0 0

**extra constraint The first & also final tableau of Phase II in two phase method:
Min i cB Cj xB -1 x1 1 0 0 -1 0 2 x2 0 1 0 2 0 -3 x3 0 0 1 -3 0 0 s1 1/2 -3/2 1 -7/2 7/2 bi 2 2 2 ∑cBibi= -4 i  i

1 cB1= -1 xB1=x1 2 cB2= 2 xB2=x2 4 cB4= -3 xB3=x3 fj Cj-fj

Example 3.3 A&C Distributors is a company that represents many outdoor products companies and schedules deliveries to discount stores, garden centers, and hardware stores. Currently, scheduling needs to be done for two lawn sprinklers, the Water Wave and Spring Shower models. Requirements for shipment to a warehouse for a national chain of garden centers are shown below.
Month March April May Shipping Capacity 8000 7000 6000 Product Water Wave Spring Shower Water Wave Spring Shower Water Wave Spring Shower Requirement 3000 1800 4000 4000 5000 2000 Unit Cost to Ship .30 .25 .40 .30 .50 .35 Per Unit Inventory Cost .06 .05 .09 .06 .12 .07

Let Sij be the number of units of sprinkler i shipped in month j, where i = 1 or 2, and j = 1, 2, or 3. Let Wij be the number of sprinklers that are at the warehouse at the end of a month, in excess of the requirement. (a.) Write the portion of the objective function that minimizes shipping costs. (b.) An inventory cost is assessed against this ending inventory. Give the portion of the objective function that represents inventory cost. (c.) There will be three constraints that guarantee, for each month, that the total number of sprinklers shipped will not exceed the shipping capacity. Write these three constraints. (d.) There are six constraints that work with inventory and the number of units shipped, making sure that enough sprinklers are shipped to meet the requirements. Write these six constraints.

Solutions (a.) Min 0.3S11 + 0.25S21 + 0.40S12 + 0.30S22 + 0.50S13 + 0.35S23 (b.) Min 0.06W11 +0.05W21 +0.09W12 +0.06W22 +0.12W13 +0.07W23 (c.) S11 + S21  8000 S12 + S22  7000 S13 + S23  6000 (d.) S11 - W11 = 3000 S21 - W21 = 1800 W11 + S12 - W12 = 4000 W21 + S22 - W22 = 4000 W12 + S13 - W13 = 5000 W22 + S23 - W23 = 2000 Example 3.4 (a.) Find the optimal solution of (a.) in Example 3.3 by the big-M method. (b.) Find the optimal solution of (b.) in Example 3.3 by the two phases method. (寫到 Phase II 的 1st tableau 就可以。) Solution for (a.) Min 0.3S11 + 0.25S21 + 0.40S12 + 0.30S22 + 0.50S13 + 0.35S23 S11 + S21  8000 S12 + S22  7000 S13 + S23  6000 S11 - W11 = 3000 S21 - W21 = 1800 W11 + S12 - W12 = 4000 W21 + S22 - W22 = 4000 W12 + S13 - W13 = 5000 W22 + S23 - W23 = 2000 S11,~, S23, W11,~, W23  0 S11 + S21 +s1 = 8000 S12 + S22 + s2 = 7000 S13 + S23 + s3 = 6000 S11 - W11 + a1 = 3000 S21 - W21 + a2= 1800 W11 + S12 - W12 + a3= 4000 W21 + S22 - W22 + a4 = 4000 W12 + S13 - W13 + a5 = 5000 W22 + S23 - W23 + a6 = 2000 S11,~, S23, W11,~, W23, s1,~, s3, a1,~, a6  0

(a) Simplex 1st tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 s2 0 1 0 0 0 0 0 0 0 0 0 0 s3 0 0 1 0 0 0 0 0 0 0 0 M M M M M M a1 a2 a3 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 a4 a5 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 a6 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=M xB3=a1 5 cB5=M xB3=a2 6 cB6=M xB3=a3 7 cB7=M xB3=a4 8 cB8=M xB3=a5 9 cB9=M xB3=a6 fj Cj-fj

0 8000 8000 0 7000 0 6000 0 3000 0 4000 0 4000 0 5000 1 2000

0 1800 1800
-

M M M M M M

0 -M 0 0 M 0

0 -M 0 0 M 0

M M M M M M 0 0 0 0 0 0

0.3 0.4 0.5 0.25 0.3 0.35 0 -M -M -M -M -M -M

Simplex 2nd tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 1 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 -1 0 1 0 0
-0.25

0 0 0 0 0 0 0 -1 0 1

0 0 0 0 0 0 0 0 0 -1

0 1 0 0 0 0 0 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 0 0 1 0 0 0 0 0 -1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 a4 a5 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 a6 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=M xB3=a1 6 cB6=M xB3=a3 7 cB7=M xB3=a4 8 cB8=M xB3=a5 9 cB9=M xB3=a6 fj Cj-fj

0 6200 6200 0 7000 0 6000 0 3000 0 1800 0 4000 0 5000 1 2000

5 cB5=0.25 xB3=S21 0

0 4000 4000
-

M M M 0.25 M M

0.3 0.4 0.5 0.3 0.35 0 0 -M -M -M -M -M

0 -M +M 0 -M 0 0 M -M 0
0.25

M 0.25 M M M M 0 +M 0
-0.25

M

0

0

0

0

Simplex 3rd tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 -1 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 -1 0 -1 0 1
-0.25

0 0 0 0 0 0 0 0 0 -1

0 1 0 0 0 0 0 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 0 0 1 0 0 0 0 0 -1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 a4 a5 -1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 a6 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=M xB3=a1 6 cB6=M xB3=a3 8 cB8=M xB3=a5 9 cB9=M xB3=a6 fj Cj-fj

0 2200 2200 0 7000 0 6000 0 3000 0 5800 0 4000 0 4000 0 5000

5 cB5=0.25 xB3=S21 0 7 cB7=0 xB3=W21 0

1 2000 2000

M M M 0.25 0.25 M

0.35 0.3 0.4 0.5 0 0.05 0 -M -M -M -M

0 -M 0 +M -M 0 0 M 0
0.25

M 0.25 M 0.25 M M 0 +M 0 +M 0
-0.25 -0.25

-M

M

0

0

Simplex 4th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 -1 -1 1 0 0 1 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 -1 0 -1 0 -1
-0.25

0 1 0 0 0 0 0 0 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 0 0 1 0 0 0 0 0 -1 0 0 0 1 0 0 0 0
-0.25

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1

a4 a5 -1 0 0 0 1 0 1 0 0
-0.25

a6

bi

 i
-

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=M xB3=a1 6 cB6=M xB3=a3 8 cB8=M xB3=a5 fj Cj-fj

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0

-1 200 200 0 7000 0 6000 1 7800 0 4000 1 6000 0 5000 1 2000
-0.25

0 3000 3000
-

5 cB5=0.25 xB3=S21 0 7 cB7=0 xB3=W21 0 9 cB9=0 xB3=W22 0

M M M 0.25 0.25 0.25 0 0.3 0.4 0.5 0 0.05 0.05 0 -M -M -M

0 -M 0 0 M 0

M 0.25 M 0.25 M 0.25 0 +M 0 +M 0 +M

0 0.25 0

Simplex 5th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 -1 -1 1 0 1 1 0 1 0 0 0 1 1 1 0 1 0 1 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 -1 0 -1 0 -1 0 1 0 0 0 0 0 0 0 0 s2 0 1 0 0 0 0 0 0 0 0 s3 0 0 1 0 0 0 0 0 0 0 0 M M M M M M a1 a2 a3 0 0 0 1 0 0 0 0 0 -1 0 0 1 1 0 0 0 0
-0.05

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 0 0 0

a4 a5 -1 0 0 1 1 0 1 0 0
-0.05

a6

bi

 i
-

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=M xB3=a1 6 cB6=M xB3=a3 8 cB8=M xB3=a5

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0

-1 200

0 7000 7000 0 6000
-

-1 -1

1 2800 2800 1 7800 7800 0 4000 0 5000 1 2000
-0.05

5 cB5=0.25 xB3=S21 0 7 cB7=0 xB3=W21 0 9 cB9=0 xB3=W22 0 fj Cj-fj

-

1 6000 6000
-

0.3 M M 0.25 +M +M 0 0.35 0.4 0.4 0.5 0 0 0 -M -M -M -M

-0.05 -0.05

0 -M 0 0 M 0

-0.05 0.3 0 -M 0 -M 0.05 -0.3 0 0 +M +M

M +M M +M M +M

0 0.05 0 0.05 0 0.05

Simplex 6th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 -1 1 1 0 0 0 0 1 -1 1 0 -1 1 1 1 0 0
-0.35

0 0 0 0 0 0 -1 0 1 0

0 0 0 0 0 0 0 0 -1 0

0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 0 1

0 0 1 0 0 0 0 0 -1

0 0 1 0 1 0 1 0 0

0 s2 0 1 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0

M M M M M M a1 a2 a3 1 0 0 0 0 0 0 1 0 0 0 -1 -1 0 1 -1 0 0 0 0 1 0 0 0 0
-0.3

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0

a4 a5 0 0 0 1 0 0 0 0 0
-0.3

a6

bi

 i
-

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3

0 0 0 0 0 0 0 1 0

0 3000

-1 4200 4200 0 6000 1 2800
-

4 cB4=0.3 xB3=S22 0 5 cB5=0.25 xB3=S21 0 6 cB6=M xB3=a3 8 cB8=M xB3=a5 0 0 7 cB7=0 xB3=W21 0 9 cB9=0 xB3=W22 0 fj Cj-fj

-1 -1

0 5000 5000 0 4000 4000 0 3200 3200 0 5000 1 2000
-0.3 -

-1 -1

0.3 M M 0.25 0.3 0.3 +M 0 -M 0 0.35 0.4 0.5 0 0 0 0.05 0 M 0 -M -M -M

0 -0.3 -0.05 0 0 0.3 0.05 0

0 0.35 0.3 M 0.3 M 0.3 0 0 +M 0 +M 0 +M

Simplex 7th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 -1 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 -1 0 0 1 -1 0 1 -1 -1 1 0 0
0.35
-0.35

0 0 0 0 0 0 0 0 0 1

0 0 1 0 -1 0 0 0 0 -1

0 1 0 0 0 0 -1 1 0 0
0.3

0 s2 0 1 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 0 0 0 0 1 0 0 -1 0 0 0 1 1 0 0
-0.05

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0

a4 a5 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0

a6

bi

 i
-

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3

0 0 0 0 0 1 0 0 0

0 6200

-1 1000 1000 0 6000 1 6000 0 1800 0 0 3200 0 5000 1 2000
-

4 cB4=0.3 xB3=S22 0 5 cB5=0.25 xB3=S21 0 6 cB6=M xB3=a3 8 cB8=M xB3=a5 0 0 7 cB7=0 xB3=W11 0 9 cB9=0 xB3=W22 0 fj Cj-fj

800 800

-1 -1

-

0.3 M M 0.25 0.3 0.3 0 0 0.4 0.5 0 -M -M 0 0.05 0

0 -M -M 0 -0.3 0 -M 0 M +M 0 0.3 0 +M
-0.3

M +M M 0.3 M 0.3 0 0.05 0
-0.3

+M

0

-0.3

+M

Simplex 8th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 -1 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 -1 0 1 0
-0.4

0 0 0 0 0 0 0 0 -1 0

0 1 0 0 1 -1 -1 1 0 0

0 0 0 0 0 0 0 0 0 1

0 0 1 0 -1 0 0 0 0 -1

0 1 1 0 0 0 -1 1 0 0

0 s2 0 1 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 -1 0 -1 -1 -1 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 a4 a5 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 a6 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3

0 6200

-1 200 200 0 6000 1 6000 0 1800 0
800

-

4 cB4=0.3 xB3=S22 0 5 cB5=0.25 xB3=S21 0 6 cB6=0.4 xB3=S12 0 7 cB7=0 xB3=W11 0 8 cB8=M xB3=a5 0 9 cB9=0 xB3=W22 0 fj Cj-fj

-1 -1

0 3200 1 2000

0 5000 5000
-

0.3 0.4 M 0.25 0.3 0.3 0 0 0 0.5 0 -M

-M -0.05 0 -0.3 -0.1 0 +M 0.4 0 0.05 0 M 0.05 0 0.3 0.1 0 -M

0 0.4 0.35 0.4 0.3 M 0.3 0
-0.4 -0.35 -0.4 -0.3

0 +M +M +M +M +M

-0.3

Simplex 9th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 -1 1 1 0 -1 0 1 1
-0.1

0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 -1 0

0 1 0 0 1 -1 -1 1 0 0
-0.05

0 0 0 0 0 0 0 0 0 1 0

0 0 1 0 -1 0 1 0 -1

0 1 1 0 0 0 0 1 0

0 s2 0 1 0 0 0 1 0 0

0 s3 0 0 1 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 -1 0 -1 -1 -1 0 0 0 0 1 0 0 0 1 0 1 0
-0.05

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 0 0 0 0 1 0 0 0 0

a4 a5 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0

a6

bi

 i
-

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=W12 0 3 cB3=0 xB3=s3 0 4 cB4=0.3 xB3=S22 0 5 cB5=0.25 xB3=S21 0 6 cB6=0.4 xB3=S12 0 7 cB7=0 xB3=W11 0 8 cB8=M xB3=a5 0 9 cB9=0 xB3=W22 0 fj Cj-fj

0 6200 -1 200

0 0 0 0 0 1 0

0 6000 6000 1 6000 6000 0 1800 -1 1000 0 3200
-

-1 -1

-1 -1 -1

1 4800 4800 1 2000 2000
-0.1

0.3 0.4 M 0.25 0.3 0 0 0.5 0 -M 0

+M
0.45

0 -M 0

0.1 0.3 0.4 0 -M -M -M
-0.1 -0.3 -0.4

M +M M 0.3 M +M 0
0.05

-M

M 0.05 0

+M +M +M

0

0

-0.3

+M

0 0.1

Simplex 10th tableau
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 1 0 0 1 -1 -1 1 0 0
-0.05

0 0 1 -1 -1 0 1 0 -1 1
0.45

0 0 0 1 0 0 0 0 0 -1
-0.35

0 1 1 0 0 0 0 1 0

0 s2 0 1 0 0 0 1 0 0

0 s3 0 0 1 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 -1 0 -1 -1 -1 0 0 0 0 1 0 0 0 1 0 1 0
-0.05

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1

a4 a5 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0

a6

bi

 i
-

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=W12 0 3 cB3=0 xB3=s3 0 4 cB4=0.3 xB3=S22 0 5 cB5=0.25 xB3=S21 0 6 cB6=0.4 xB3=S12 0 7 cB7=0 xB3=W11 0 8 cB8=M xB3=a5 0 9 cB9=0.35 xB3=S23 0 fj Cj-fj

0 6200 0 2200 0 4000 0 1800 0 3000 0 3200 1 2000

0 0 0 0 0 1 0

-1 4000 4000
-

-1 -1

-1 -1

0 2800 2800

0.3 0.4 M 0.25 0.3 0.35 0 0 0 0.5 0 -M 0 0 0

0 -M 0

-M

0.3 0.4 0 -M -M
-0.3 -0.4

M +M M 0.3 M 0.35 0
0.05

M 0.05 +M 0.35 0 +M +M

-0.45

0

-0.3

+M

0 +M

-0.35

Simplex 11th tableau - Final
Min i cB Cj xB 0.3 0.4 0.5 0.25 0.3 0.35 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 1 0 0 1 -1 -1 1 0 0
-0.05

0 0 1 0 -1 0 1 0 -1 1
0.45

0 0 0 1 0 0 0 0 0 -1
-0.35

0 1 1 1 0 0 0 1 0

0 s2 0 1 1 0 0 1 0 0

0 s3 0 0 1 0 0 0 0 0 0

M M M M M M a1 a2 a3 0 -1 0 -1 -1 -1 -1 -1 -1 0 0 0 1 0 0 1 0 1 0
-0.05

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1

a4 a5 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0

a6

bi

 i
-

1 cB1=0.3 xB1=S11 1 2 cB2=0 xB2=W12 0 3 cB3=0 xB3=s3 0 4 cB4=0.3 xB3=S22 0 5 cB5=0.25 xB3=S21 0 6 cB6=0.4 xB3=S12 0 7 cB7=0 xB3=W11 0 8 cB8=0.5 xB3=S13 0 9 cB9=0.35 xB3=S23 0 fj Cj-fj

0 6200 0 2200 0 4000 0 1800 0 3000 0 3200 0 2800 1 2000

-1 -1 1200
-

0 0 0 0 1 0

-1 -1

-1 -1

0.3 0.4 0.5 0.25 0.3 0.35 0 0 0 0 0 0 0 0

0 -M 0

-M

M 0.05 +M 0.35 0 +M +M

-0.45

-0.3 -0.4

0.3 0.4 0 -M -M

M +M M 0.3 M 0.35 0
0.05

0

-0.3

+M

0 +M

-0.35

Optimal cost: 0.3*6200 + 0.25*1800 + 0.40*3000 + 0.30*4000 + 0.50*2800 + 0.35*2000 = 6810

Solution for (b) Phase I : 1st tableau (由於考慮 Wij 所以 pivoting 以 Wij 優先)
Min i cB Cj xB 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1
0

0 0 0 0 -1 0 1 0 0 0 0 0

0 0 0 0 0 0 -1 0 1 0 0 0

0 0 0 0 0 0 0 0 -1 0 -1 1

0 0 0 0 0 -1 0 1 0 0 0 0

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 -1 1

0 1 0 0 0 0 0 0 0 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 0 0 0 1 0 0 0 0 0 1 0

1 0 0 0 0 1 0 0 0 0 1 0

1 0 0 0 0 0 1 0 0 0 1 0

1 0 0 0 0 0 0 1 0 0 1 0

1 0 0 0 0 0 0 0 1 0 1 0

1 a6 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1

a1 a2 a3

a4 a5

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=1 xB3=a1 5 cB5=1 xB3=a2 6 cB6=1 xB3=a3 7 cB7=1 xB3=a4 8 cB8=1 xB3=a5 9 cB9=1 xB3=a6 fj Cj-fj

0 8000 8000 0 7000 0 6000 0 1800 0 4000 0 4000 0 5000 1 2000 1 0

0 3000 3000
-

-1 -1 -1 -1 -1 -1

Phase I : 2nd tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1
0

0 1 0 0 -1 0 1 0 0 0 1

0 0 0 0 0 0 -1 0 1 0 0 0

0 0 0 0 0 0 0 0 -1 0 -1 1

0 0 0 0 0 -1 0 1 0 0 0 0

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 -1 1

0 1 0 0 0 0 0 0 0 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 -1 0 0 1 0 0 0 0 0 0 1

1 0 0 0 0 1 0 0 0 0 1 0

1 0 0 0 0 0 1 0 0 0 1 0

1 0 0 0 0 0 0 1 0 0 1 0

1 0 0 0 0 0 0 0 1 0 1 0

1 a6 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1

a1 a2 a3

a4 a5

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 5 cB5=1 xB3=a2 6 cB6=1 xB3=a3 7 cB7=1 xB3=a4 8 cB8=1 xB3=a5 9 cB9=1 xB3=a6 fj Cj-fj

0 5000 5000 0 7000 0 6000 0 3000 0 1800 0 4000 0 5000 1 2000 1 0

4 cB4=0 xB3=S11 1

0 4000 4000
-

-1 -1 -1 -1 -1 -1

Phase I : 3rd tableau
Min i cB Cj xB 0 0 0 0 1 0 0 0 0 0 0 0 -1 1 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1
0

0 0 0 0 0 0 1 0 0 0 0 0

0 1 0 0 -1 0 -1 0 1 0 1 -1

0 0 0 0 0 0 0 0 -1 0 -1 1

0 0 0 0 0 -1 0 1 0 0 0 0

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 -1 1

0 1 0 0 0 0 0 0 0 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 -1 0 0 1 0 0 0 0 0 0 1

1 0 0 0 0 1 0 0 0 0 1 0

1 -1 0 0 1 0 1 0 0 0 0 1

1 0 0 0 0 0 0 1 0 0 1 0

1 0 0 0 0 0 0 0 1 0 1 0

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1

a1 a2

a3 a4

a5 a6

1 cB1=0 xB1=s1 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=1 xB3=a2 7 cB7=1 xB3=a4 8 cB8=1 xB3=a5 9 cB9=1 xB3=a6 fj Cj-fj

0 1000 1000 0 7000 0 6000 0 7000 0 1800 0 4000 0 4000 1 2000 1 0

6 cB6=0 xB3=W11 0

0 5000 5000
-

-1 -1 -1 -1

Phase I : 4th tableau
Min i cB Cj xB 0 0 -1 1 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 1 0 1 0 1 0 0 1 1 1 0 -1 0 0
0

0 0 1 0 0 0 0 1 0 0 1

0

0 0 0 0 0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 -1 0 -1 1

0 0 0 0 0 -1 0 1 0 0 0 0

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 1

0 1 0 0 1 0 1 0 -1 0 1

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 -1 0 0 0 0 -1 0 1 0 1 0

1 0 0 0 0 1 0 0 0 0 1 0

1 -1 0 0 0 0 0 0 1 0 1 0

1 0 0 0 0 0 0 1 0 0 1 0

1 0 0 0 0 0 0 0 1 0 1 0

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0

a1 a2

a3 a4

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=1 xB3=a2 7 cB7=1 xB3=a4 8 cB8=1 xB3=a5 9 cB9=1 xB3=a6 fj Cj-fj

0 1000

0 7000 7000 0 6000 0 8000 0 1800 0 5000 0 4000 1 2000 1 0
-

6 cB6=0 xB3=W11 0

0 4000 4000
-

-1 -1

-1 -1

-1 -1

Phase I : 5th tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 -1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1 0 -1 0 1
-1

0 0 1 0 0 0 0 1 0 0 1

0

0 0 0 0 0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 -1 1 0 0 0 0 0 -1 0 0 0

0 0 0 0 0 -1 0 1 0 0 0 0

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 -1 1

0 0 1 0 1 0 1 0 -1 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 0 -1 0 0 0 -1 0 1 0 0 1

1 0 0 0 0 1 0 0 0 0 1 0

1 0 -1 0 0 0 0 0 1 0 0 1

1 0 0 0 0 0 0 1 0 0 1 0

1 1 -1 0 0 0 0 0 1 0 0 1

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0

a1 a2

a3 a4

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=1 xB3=a2 7 cB7=1 xB3=a4 8 cB8=0 xB3=S12 9 cB9=1 xB3=a6 fj Cj-fj

0 5000

0 3000 3000 0 6000
-

0 8000 8000 0 1800 1800 0 5000 5000 0 4000 0 4000 1 2000 1 0
-

6 cB6=0 xB3=W11 0

-1 -1

Phase I : 6th tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 -1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0
0

0 0 1 0 0 0 0 1 0 0 1

0

0 0 0 0 0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 -1 1 0 0 0 0 0 0 0 0

0 0 1 0 1 -1 1 1 0 1 -1

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 -1 1

0 0 1 0 1 0 1 0 -1 0 0 0

0 s2 0 1 0 0 0 0 0 0 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 0

1 0

1 0

1 0 0 0 0 0 0 1 0 0 1 0

1 1 -1 0 0 0 0 0 1 0 0 1

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0

a1 a2

a3 a4

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=s2 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=0 xB3=S21 7 cB7=1 xB3=a4 8 cB8=0 xB3=S12 9 cB9=1 xB3=a6 fj Cj-fj

0 5000

-1 -1 -1 0 0 0 0 1 0 0 1 0 -1 1 0 1 0 0 1 0 0 0 0 0 1 0 0 1

0 1200 1200 0 6000 0 1800
-

0 6200 6200
-

6 cB6=0 xB3=W11 0

-1 -1

0 3200 3200 0 4000 4000 0 5800 1 2000 1 0
-

-1 -1

-1 -1

Phase I : 7th tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 -1 1 1 -1 1 1 0 0 1 -1 0 0 0 0 0 1 0 0 0 0 0
0

0 0 1 0 -1 1 -1 0 1 0 0 0

0

0 0 0 0 0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 -1 1 0 -1 1 -1 -1 0 0 -1 1

0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 -1 0 1 0 0

0 0 0 0 0 0 0 0 0 -1 1

0 0 1 0 0 1 0 0 0 1

0 s2 0 1 0 -1 1 -1 1 0 1

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 0

1 0

1 0

1 0 0 0 0 0 0 1 0 0 1 0

1 1 -1 0 1 -1 1 1 0 0 1 0

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 1 0 0 0 0 0 1 1 -1

a1 a2

a3 a4

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=W21 0 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=0 xB3=S21 7 cB7=1 xB3=a4 8 cB8=0 xB3=S12 9 cB9=1 xB3=a6 fj Cj-fj 0 1 0 0 0 0 0 0

0 5000 0 1200 0 5000 0 3000 0 2000 0 2800 0 7000 1 0

-1 -1 -1 0 1 -1 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 -1 1 1 0 0 1 0

0 6000 6000
-

6 cB6=0 xB3=W11 0

-1 -1

1 2000 2000

-1 -1 -1

Phase I : 8th tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 -1 1 1 -1 1 1 0 0 1 -1 0 0 0 0 0 1 0 0 0 0 0
0

0 0 1 0 -1 1 -1 0 1 0 0 0

0

0 0 0 0 0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 -1 1 0 -1 1 -1 -1 0 0 -1 1

0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 -1 0 0 0 -1 0 1 1

0 0 0 1 0 0 0 0 0 -1 1

0 0 1 0 0 1 0 0 0 1

0 s2 0 1 0 -1 1 -1 1 0 1

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 0

1 0

1 0

1 0 0 0 0 0 0 1 0 0 1 0

1 1 -1 0 1 -1 1 1 0 0 1 0

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 0 0 0 0 0 0 1 0 0

a1 a2

a3 a4

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=W21 0 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=0 xB3=S21 7 cB7=1 xB3=a4 8 cB8=0 xB3=S12 9 cB9=0 xB3=S23 fj Cj-fj 0 1 0 0 0 0 0 0

0 5000 5000 0 1200

-1 -1 -1 0 1 -1 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 -1 1 1 0 0 1 0

-1 4000 4000 0 5000 5000 0 3000
-

6 cB6=0 xB3=W11 0

0 2000 2000 0 2800 2800 0 7000 1 2000 0 1
-

-1 -1

-1 -1 -1 -1

Phase I : 9th tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
0

0 1 0 1 0 0 -1 1 1 0 1 -1

0

0 -1 1 -1 -1 1 1 -1 0 0 -1 1

0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 -1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 -1 0 0 0 -1 0 1 -1 1

0 0 0 1 0 0 0 0 0 -1 0 0

0 0 1 0 0 1 0 -1 0 0 -1 1

0 s2 1 0 1 0 0 -1 0 1 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1 0

1 0

1 -1 0 -1 0 0 1 0 0 0 0 1

1 0 0 0 0 0 0 1 0 0 1 0

1 0 0 0 0 1 0 0 0 0 1

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 0 0 0 0 0 0 1 0 0

a1 a2 -1 -1 0 1 -1 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0

a3 a4

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=W21 0 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=0 xB3=S21 6 cB6=0 xB3=S13 7 cB7=1 xB3=a4 8 cB8=0 xB3=S12 9 cB9=0 xB3=S23 fj Cj-fj 0 1 0 0 0 0 0 0 0

0 3000 3000 0 3200 0 3000 0 5000 0 2000 0

-1 -1 2000 2000
-

800 800

0 7000 7000 1 2000 0 1
-

Phase I : 10th tableau
Min i cB Cj xB 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
0

0 0 0 0 0 0 0 1 0 0 0 0

0

0 0 1 0 -1 1 0 -1 1 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 -1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 -1 -1 1 1 0 0

0 0 0 1 0 0 0 0 0 -1 0 0

0 1 1 -1 0 1 -1 1 0 0 0

0 s2 1 0 1 0 0 0 1 0 0 0

0 s3 0 0 1 0 0 0 0 0 0 0 0

1

1

1

1

1 0 0 0 0 1 0 0 0 0 1

1 bi  i
-

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1 0 0 0 0 0 0 0 0 1 0 0

a1 a2 -1 -1 1 1 -1 1 1 0 0 1 1 0 0 1 1 0 0 1

a3 a4 0 -1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 -1 0 0 1

a5 a6

1 cB1=0 xB1=W12 0 2 cB2=0 xB2=W21 0 3 cB3=0 xB3=s3 4 cB4=0 xB3=S11 5 cB5=0 xB3=S21 6 cB6=0 xB3=S13 7 cB7=0 xB3=S22 8 cB8=0 xB3=S12 9 cB9=0 xB3=S23 fj Cj-fj 0 1 0 0 0 0 0 0 0

-1 -1 -1 -1

0 2200 0 3200 0 3000 0 5000 0 2800 0
800

-1 -1 2800

-1 -1

-1 -1

0 6200 1 2000 0 1

(b.) Min 0.06W11 +0.05W21 +0.09W12 +0.06W22 +0.12W13 +0.07W23 Phase II : 1st tableau
Min i cB Cj xB 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
0

0 0 0 0 0 0 0 1 0 0 0 0

0

0.06 0.09 0.12 0.05 0.06 0.07 0 1 0 -1 1 0 -1 1 0 0.01 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -1 0 0 0 0 0.12 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 -1 1 1 0 0 1 0 0 0 0 0 -1 0

0 s1 1 1 -1 0 1 -1 -1 1 0

0 s2 1 0 1 0 0 -1 0 1 0

0 s3 0 0 1 0 0 0 0 0 0 0 0 bi
2200 3200 2800 3000 5000 2800 800 6200 2000

S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 0 0 0 0 0 0 0 0 1 0 0

 i

1 cB1=0.09 xB1=W12 2 cB2=0.05 xB2=W21 3 4 5 6 7 8 9 cB3=0 cB4=0 cB5=0 cB6=0 cB7=0 cB8=0 cB9=0 xB3=s3 xB3=S11 xB3=S21 xB3=S13 xB3=S22 xB3=S12 xB3=S23 fj Cj-fj

0.05 0.09

0.05 0.09

0.14 0.09

-0.03 0.07 -0.14 -0.09

Chapter 4
Example 4.1 Consider the following primal linear programming model. Minimize f = 6x1 + 8x2 subject to 3x1 + x2 ≤ 4 5x1 + 2x2 ≤ 0 1 x1 + 2x2 = 3 and x1, x2 ≥ 0 Find its dual linear programming model. Solution: Change into canonical form: -3x1 - x2 -4 -5x1 - 2x2 -10 x1 + 2x2  3 -x1 - 2x2 -3 and x1, x2 ≥ 0 Thus,
  4  10  6    c   and A = b , , 3   8      3  1 3  5    2   i.e., AT = ,  1 2    2 1    5 1 3  1   1  2    2 2 

So, the dual form is: Maximize g = -4y1-10y2+3y3-3y4 -3y1-5y2+ y3- y4 ≤ 6 - y1-2y2+2y3-2y4 ≤ 8 y1, y2, y3, y4 ≥ 0

Example 4.2 Consider the the following linear programming model: Min f = 2x1 + 4x2 Subject to x1 + 5x2 80 4x1 + 2x2 6 x1 + x2 = 3 x1 , x2  0 (a) Rewrite the linear programming model into its canonical form. (b) Change into its dual form.

Solution for (a) Min f = 2x1 + 4x2 Subject to - x1 - 5x2  -80 4x1 + 2x2 6 -x1 - x2 -3 x1 + x2 3 x1, x2  0 Solution for (b) Max f= 80y1 -6y2+ 3y3 -3y4 ’ y1 -4y2 + y3 -y4  2 5y1 -2y2 + y3 -y4  4 y1, y2, y3, y4  0

Chapter 5
Example 5.1 Consider the following primal linear programming model. Maximize f = 4x1 - x2 + x3 subject to 2x1 + x2 + x3 ≤ 6 -2x1 + 2x2 ≤4 and x1, x2, x3 ≥ 0 The following is its simplex final tableau. Cj 4 i cB xB x1 1 cB1=4 xB1=x1 1 2 cB2=0 xB2=s2 0 fj 4 Cj-fj 0 -1 1 0 0 x2 x3 s1 s2 bi 0.5 0.5 0.5 0 3 3 1 1 1 10 2 2 2 0 ∑cBibi=12 i -3 -1 -2 0

Find the adjustable range of the follows such that the basic variables in the optimal solution can keep invariant. (1) The coefficient’range of non-basic variable x2 in objective function. s (2) The coefficient’range of basic variable x1 in objective function. s (3) The range of b1. Solutions: 1 0 0 (1) Cj 4 c2 i cB xB x1 x2 x3 s1 s2 bi 1 cB1=4 xB1=x1 1 0.5 0.5 0.5 0 3 2 cB2=0 xB2=s2 0 3 1 1 1 10 fj 4 2 2 2 0 ∑cBibi=12 i Cj-fj 0 c2 - 2 -1 -2 0 Let c2 be the coefficient of x2 in the objective function, then the following inequality c2 –2 ≤0  c2 ≤2 must be hold to keep x2 staying outside of the basic variables. This also implies that if the unit profit of x2 is greater than $2, the optimality should be changed. (2) Cj i cB xB 1 cB1=c1 xB1=x1 2 cB2=0 xB2=s2 fj Cj-fj -1 1 0 x1 x2 x3 s1 1 0.5 0.5 0.5 0 3 1 1 c1 0.5c1 0.5c1 0.5c1 0 -1-0.5c1 1-0.5c1 -0.5c1

c1

0 s2 bi 0 3 1 10 0 ∑cBibi=12 i 0

-1-0.5c1 ≤  1+0.5c1   c1 -2 0 0 1-0.5c1 ≤  c1 2 0 -0.5c1 ≤  c1 0 The concluding result is c1  . 0 2

Let c1 be the coefficient of x1 in the objective function, then the following inequalities must be hold to keep the basic variables unchanged.

If the unit profit of x1 is smaller than $2, x1 should leave from basic variables, and x3 will get into as a basic variable. (3) i cB Cj xB 4 x1 1 0 4 0 -1 x2 3 2 -3 1 x3 1 2 -1 0 s1 0 s2 bi 3 10

1 cB1=4 xB1=x1 2 cB2=0 xB2=s2 fj Cj-fj

0.5 0.5 0.5 0

1
2 -2

1
0

0 ∑cBibi=12 i

Observe that NxN + BxB = b  B-1NxN + B-1BxB = B-1b  xB = B-1b - B-1NxN -1 Since xN = 0 and xB  so we conclude that xB = B b  , thus the range of b1 is, 0, 0 0 0 .5 0b1    0 0 -4,   4   0.5b1 and b1 + 4   b1  the result is: b10.  0  1 1     Example 5.2 Maximize f = 9 x1+12 x2 subject to 3x1+2x2  40 2x1+3x2  30 2x1+ x2  250 and x1, x2  0 The following table is the final table of the above model. Max Cj 9 12 i cB x B x 1 x 2 1 9 x1 2 12 x2 3 0 s3 fj Cj-fj 1 0 0 0 0 1 0 0 0 s1 3/5 -2/5 -4/5 3/5 0 s2 -2/5 3/5 5/9 18/5 0 s3 0 0 1 0 bi 12 2 224 132  i

9 12

-3/5 -18/5 0

Under the concern of keeping the basic variables invariant, do the following exercises (a) Find the allowable range of coefficient’values (say, c1 and c2) for x1 and x2. s (b) Find the allowable range of resource amount (say, b1 and b2) for 1st and 2nd resources. (c) Find the marginal values of 1st and 2nd resources. Solution: (a) -0.6c1+4.8  0;0.4c1-7.2   c1  0 8;c1  18  8 c1  18 -5.4+0.4c2  0;0.4-0.6c2   c2  0 1.35;c2  ⅔  ⅔ c2 1.35 (b)
 .6  .4 0 b1   0.6b1  0 0 12          0 18  0   0.4 0.6 0 30    .4b1     0.8 5    1250  0.8b1  .67  266   9    

 0.6b1-12  0;-0.4b1+18  0;-0.8b1+266.67  0  b1  20;b1  45;b1  333.34  20 b1  45
 .6  .4 0 40    .4b2  0 0 24 0         .4 0.6 0 b2  16  .6b2  0 0 0       0.8 5   1250  5b2  218  9     9     24-0.4b2  0;16+0.6b2  0;218+0.8b2  0  b2  60;b2 -26.67;b2 -272.5  -26.67 b2  60

(c) Let y1, y2 and y3 be the marginal values of 1st, 2nd and 3rd resources, then the dual form of the original model is as follows: Min Min g = 40 y1+30 y2+250 y3 g = 40 y1+30 y2+250 y3+Ma1+Ma2 subject to subject to 3y1+2y2+2y3-s1+a1 = 9 3y1+2y2+2y3  9 2y1+3y2+ y3  12 2y1+3y2+ y3-s2+a2 = 12 y1, y2, y2 0 y1, y2, y2 s1, s2, a1, a2  0

Min Cj i cB x B 1 M a1 2 M a2 fj Cj-fj Min Cj i cB x B 1 40 y1 2 M a2 fj Cj-fj Min Cj i cB x B 1 40 y1 2 30 y2 fj Cj-fj

40 y1 3 2

30 y2 2 3

250 y3 2 1 3M

0 s1 -1 0 -M M 0 s1 -1/3 2/3
-40/3 +2M/3 40/32M/3

0 s2 0 -1 -M M 0 s2 0 -1 -M M 0 s2

M a1 1 0 M 0 M a1 1/3 -2/3
40/3 2M/3 -40/3 +2M/3

M a2 0 1 M 0 M a2 0 1 M 0 M a2

bi 9 12

 i 3 4

5M 5M

40-5M 30-5M 250-3M

40 y1 1 0 40 0 40 y1 1 0 40 0

30 y2 2/3 5/3
80/3+ 5M/3 10/35M/3

250 y3 2/3 -1/3
80/3 M/3 670/3 +M/3

bi 3 6

 i 4.5 3.6

30 y2 0 1 30 0

250 y3 0.4 -0.2 10 240

0 s1

M a1

bi

 i

-0.6 0.4 7/15 -0.4 0.6 0.4 -0.6 -0.2 0.6 3.6 -12 -18 12 18 略 略 略 略

Marginal value of Resource 1 is 0.6. Marginal value of Resource 2 is 3.6.

Chapter 6
Example 6.1 某電子公司,主要訂單來自 A、B、C 三市,而該公司在這三市都設有倉庫。公司 的三座工廠與倉庫不在同一地方,分別設於 I、J、K 三地。由於市場不景氣,業務 嚴重萎縮,如下表所示:
工廠 每年最大產能(千個) 倉庫 每倉庫預估來年之需求量(千個)

I J K 倉庫 工廠 I J K

210 140 290

A B C

80 200 200

各廠與倉庫間的運輸成本及其生產成本如下表: A 2 4 3 B 4 3 6 C 4 4 4 生產成本 11 14 12

公司目前的流通策略如下:所有 A 市倉庫的需求由 I 廠供應,所有 J 廠的產品都運 到 B 市倉庫,而 B 市倉庫不足的需求量都由 I 廠供應,C 市的需求量則完全由 K 廠 供應。試評估針對來年需求,採用現行策略的成本。以 VAM 法改善公司此一策 略。 Solution: (1)、試評估針對來年需求,採用現行策略的成本。解釋公司為何採用此一策略? 令成本=運輸成本+生產成本
倉庫 工廠

A
13 80 18 15

B
15 60 17 140 18

C
15

D(slack) Supply
0 70 18 0 0 90

I J K Request

210 140 290 640

16 200

80

200

200

160

Total cost=13×80+15×60+17×140+16×200=7420(千元)。

Apply company strategy: penalty penalty 13 2 17 1 15 2 1 倉庫 工廠 I J K 1 4 15 A 13 10 18 15 70 233 B 15 200 17 18 200 1 165 C 15 18 16 200 200 0 D(slack) 0 0 140 0 20 140 20 Supply 210 10 140 290 270 70 640

Request 80 70

Total cost=13×10+15×200+15×70+16×200=7380(千元)。 Example 6.2 下表所示為三處倉庫與四位顧客間的供需關係表。陰影部份是各相關之運輸成本。 顧客 倉庫 戒 定 慧 需求量 阿 7 10 6 20 彌 8 12 10 28 陀 11 5 11 17 佛 10 4 9 33 98 庫存量 30 45 35 110

以 VAM 法求出其可行解,並計算該解之總運算成本。 Solution: penalty 顧客 penalty 倉庫

1 阿
7 10 6

2 彌
8 12 10

6 陀
11 5 11

5

0

佛 D(slack) 庫存量
10 4 9 0 0 0

7 4 6

戒 定 慧 Request

30 45 35 110

20

28

17

33

12

penalty 顧客 penalty 倉庫

1 阿
7 10 6

2 彌
8 12 10

6 陀
11 5 11

5

0

佛 D(slack) 庫存量 12
10 4 9 0 0 0

17 14 36

戒 定 慧 Request penalty 30 18 45 35 110

20 1 阿
7 10 6

28 2 彌
8 12 10

17 6 陀
11 17 5 11

33 5

12 0 0

顧客 penalty 倉庫

佛 D(slack) 庫存量 12
10 4 9 0 0 0

1 61 3

戒 定 慧 Request penalty 18 45 28 35 110

20 1 阿
7 10 6

28 2 彌
8 12 10

17 0 6 陀
11 17 5 11

33 51

0 0

顧客 penalty 倉庫

佛 D(slack) 庫存量 12
10 28 4 9 0 0 0

1 6 3

戒 定 慧 Request

18 28 0 35 110

20

28

0

33 5

0

penalty 顧客 penalty 倉庫

1 阿
7 10 20 6

2 彌
8 12 10

6 陀
11 17 5 11

1

0

佛 D(slack) 庫存量 12
10 28 4 9 0 0 0

12 6 31

戒 定 慧

18 0 35 15 110

Request 20 0 penalty 顧客 penalty 倉庫

28 2 彌
18 8 12 10

0 6 陀
11 17 5 11

5 1

0 0

1 阿
7 10 20 6

佛 D(slack) 庫存量 12
10 28 4 9 0 0 0

2 6 1

戒 定 慧 Request penalty 18 0 0 15 110

0 1 阿
7 10 20 6

28 10 彌
18 8 12 10 10

0 6 陀
11 17 5 11

5 -

0 0

顧客 penalty 倉庫

佛 D(slack) 庫存量 12
10 28 4 5 9 0 0 0

2 6 1

戒 定 慧 Request

0 0 15 0 110

0

10 0

0

50

0

Final

penalty

顧客 penalty 倉庫


7 10 20 6


18 8 12 10 10


11 17 5 11

佛 D(slack) 庫存量 12
10 28 4 5 9 0 0 0

戒 定 慧 Request

30 45 35 110

20

28

17

33

12

總運送成本: 20×6+18×8+10×10+17×5+28×4+5×9 = 606。

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...Research Article Research is important to every business because of the information it provides. There is a basic process to researching information and that process begins by deciding what information needs to be researched. The next step is to develop a hypothesis, which describes what the research paper is about and what the researcher’s opinion is regarding the topic. The research article chosen for this paper is titled, “The Anchor Contraction Effect in International Marketing Research.” The hypothesis for this paper is, “This raises the issue of whether providing responses on rating scales in a person’s native versus second language exerts a systematic influence on the responses obtained.” Simply explained, the hypothesis of this paper is to determine whether research questions should be in a person’s native language rather than expecting them to respond to questions in a language in which they might not be fluent. The hypothesis of this paper was accepted based on the research data gathered by the research team. This hypothesis was supported by nine studies using a variety of research methods. The research methods provided data that demonstrated the level of inaccuracy based on questions being asked in a language that was not the respondent’s native language. The research data provided insight into the probability of more accurate results when the respondent was asked questions in a manner that related well with their culture. There are several implications...

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...ACE8001: What do we mean by Research? & Can we hope to do genuine Social Science Research (David Harvey)  What do we mean by research? What might characterise good research practice? There is no point in us trying to re-invent the wheel - other and probably more capable people than us have wrestled with this problem before us, and it makes good sense and is good practice to learn what they have discovered.  In other words - we need to explore more reliable and effective methods and systems for the pursuit of research than we have been doing so far. What is research? Dictionary Definitions of Research: * "The act of searching closely or carefully for or after a specified thing or person" * "An investigation directed to discovery of some fact by careful study of a subject" * "A course of scientific enquiry" (where scientific = "producing demonstrative knowledge") Howard and Sharp (HS) define research as:  "seeking through methodical processes to add to bodies of knowledge by the discovery or elucidation of non-trivial facts, insights and improved understanding of situations, processes and mechanisms".  [Howard, K. and Sharp, J.A. The Management of a student research project, Gower, 1983 - a useful and practical “how to do it” guide] Two other, more recent guides are: Denscombe, Martyn, 2002, Ground rules for good research: a 10 point guide for social research,  Open University Press. Robinson Library Shelf Mark: 300.72 DEN, Level 3 (several copies)...

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...solve analytic models or whatever, but they often fail to demonstrate that they have thoroughly thought about their papers—in other words, when you push them about the implicit and explicit assumptions and implications of their research models, it appears that they haven’t really given these matters much thought at all.[1] Too often they fall back on saying that they are doing what they are doing because that is the way it is done in the prior literature, which is more of an excuse than a answer. (Of course, once a researcher reaches a certain age, they all feel that youngsters aren’t as good as they were in the good old days!) Therefore, in this class we shall go beyond simply studying research in managerial accounting. For many of you, this is your first introduction to accounting research and to PhD level class. Hence, in these classes we shall also learn how to solve business problems systematically and to understand what it means to have thoroughly “thought through” a paper. We begin not with academic research, but with some real world cases, because we should never forget that ours is an applied research field: accounting research is a means towards the end of understanding business and is not an end in itself, in the way pure science research is. Developing a systematic procedure for solving a real world business problem is the starting point for developing a...

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...manger know about research when the job entails managing people, products, events, environments, and the like? Answer: Research simply means a search for facts – answers to questions and solutions to problems. It is a purposive investigation. It is an organized inquiry. It seeks to find explanations to unexplained phenomenon to clarify the doubtful facts and to correct the misconceived facts. Research is the organized and systematic inquiry or investigation which provides information for solving a problem or finding answers to a complex issue. Research in business: Often, organization members want to know everything about their products, services, programs, etc. Your research plans depend on what information you need to collect in order to make major decisions about a product, service, program, etc. Research provides the needed information that guides managers to make informed decisions to successfully deal with problems. The more focused you are about your resources, products, events and environments what you want to gain by your research, the more effective and efficient you can be in your research, the shorter the time it will take you and ultimately the less it will cost you. Manager’s role in research programs of a company: Managing people is only a fraction of a manager's responsibility - they have to manage the operations of the department, and often have responsibilities towards the profitability of the organization. Knowledge of research can be very helpful...

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...Contents TITLE 2 INTRODUCTION 3 BACKGROUND OF THE STUDY 3 AIM 4 OBJECTIVES 4 RESEARCH QUESTIONS 4 LITERATURE REVIEW 5 METHODOLOGY AND DATACOLLECTION 5 POPULATION AND SAMPLING 6 DATA ANALYSIS METHODS 6 PARTICIPANTS IN THE STUDY 7 STUDY PERIOD (GANTT CHART) 8 STUDY RESOURCES 9 REFERENCES 9 BIBLIOGRAPHY 9 APPENDICES: 10 * The Impact of Motivation through Incentives for a better Performance - Adaaran Select Meedhupparu Ahmed Anwar Athifa Ibrahim (Academic Supervisor) Applied Research Project to the Faculty of Hospitality and Tourism Studies The Maldives National University * * Introduction As it is clear, staff motivation is important in all the sectors especially in the tourism sector where we require highly skilled employees to get the best of their output to reach the organizational goals. Therefore, organizations spend a lot on their staff motivation in terms of different incentive approaches, such as financial benefits, training and development, appreciations, rewards and promotions. As mentioned in the title, the outline of the findings will be focused on the motivation of the staffs on improving their performances by the different incentive packages that they get at the resort. This study will be executed at Adaaran Meedhupparu by giving questionnaire to the staff working in different departments to fill up and return to the scholar to examine the current situation of staff satisfaction on motivation to do...

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...goal of the research process is to produce new knowledge or deepen understanding of a topic or issue. This process takes three main forms (although, as previously discussed, the boundaries between them may be obscure): * Exploratory research, which helps identify and define a problem or question. * Constructive research, which tests theories and proposes solutions to a problem or question. * Empirical research, which tests the feasibility of a solution using empirical evidence. There are two ways to conduct research: Primary research Using primary sources, i.e., original documents and data. Secondary research Using secondary sources, i.e., a synthesis of, interpretation of, or discussions about primary sources. There are two major research designs: qualitative research and quantitative research. Researchers choose one of these two tracks according to the nature of the research problem they want to observe and the research questions they aim to answer: Qualitative research Understanding of human behavior and the reasons that govern such behavior. Asking a broad question and collecting word-type data that is analyzed searching for themes. This type of research looks to describe a population without attempting to quantifiably measure variables or look to potential relationships between variables. It is viewed as more restrictive in testing hypotheses because it can be expensive and time consuming, and typically limited to a single set of research subjects. Qualitative...

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...Volume 3, number 2 What is critical appraisal? Sponsored by an educational grant from AVENTIS Pharma Alison Hill BSC FFPHM FRCP Director, and Claire Spittlehouse BSc Business Manager, Critical Appraisal Skills Programme, Institute of Health Sciences, Oxford q Critical appraisal is the process of systematically examining research evidence to assess its validity, results and relevance before using it to inform a decision. q Critical appraisal is an essential part of evidence-based clinical practice that includes the process of systematically finding, appraising and acting on evidence of effectiveness. q Critical appraisal allows us to make sense of research evidence and thus begins to close the gap between research and practice. q Randomised controlled trials can minimise bias and use the most appropriate design for studying the effectiveness of a specific intervention or treatment. q Systematic reviews are particularly useful because they usually contain an explicit statement of the objectives, materials and methods, and should be conducted according to explicit and reproducible methodology. q Randomised controlled trials and systematic reviews are not automatically of good quality and should be appraised critically. www.evidence-based-medicine.co.uk Prescribing information is on page 8 1 What is critical appraisal What is critical appraisal? Critical appraisal is one step in the process of evidence-based clinical practice. Evidencebased clinical practice...

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...How To Formulate Research Problem? Posted in Research Methodology | Email This Post Email This Post Formulating the research problem and hypothesis acts as a major step or phase in the research methodology. In research, the foremost step that comes into play is that of defining the research problem and it becomes almost a necessity to have the basic knowledge and understanding of most of its elements as this would help a lot in making a correct decision. The research problem can be said to be complete only if it is able to specify about the unit of analysis, time and space boundaries, features that are under study, specific environmental conditions that are present in addition to prerequisite of the research process. Research Process Research process is very commonly referred to as the planning process. One important point to be kept in mind here is to understand that the main aim of the research process is that of improving the knowledge of the human beings. The research process consists of the following stages – 1. The Primary stage :– This stage includes – a. Observation – The first step in the research process is that of the observation, research work starts with the observation which can be either unaided visual observation or guided and controlled observation.It can be said that an observation leads to research, the results obtained from research result in final observations which can play a crucial part in carrying out further research. Deliberate and guided...

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...activities for the quarter 4 which include weekly class discussion, class participation, midterm and final exam * Learned about what Research is and what Research is not. * Eight characteristics of research. * Sub problem – that is a question or problem that must be address before the main problem is resolved. * Hypothesis- that is a reasonable quests that needs to be proving. * I learned about assumption –that is a statement that is presume to be fact. * Learned about theory * Learned about methodology- that is a process a researchers use to collect data and information is research work. * Learned about internet – A researchers use internet to access information online. * Learned about two types of research report which is Juried or refereed – a reviewed report * Nonjuried or nonrefereed – none reviewed report. E.g. Journal report. * Learned about checklist evaluating research- that a report juried that is judge. * Learned that a research that is not screen or viewed by expert is not valid * Guidelines in reviewing research by going to library to sort for information needed for case study. * I learned as a researcher, you must read more than articles. * I learned about research paper / APA Style – that first thing is to choose the research topic. * Learned about what research paper entails, like cover page, table of content, abstract, introduction, summary, conclusion and references. * I learned about APA...

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