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y (0) D = 12/4/2017 6:59 PM Q H o 11 6 A 9 wa A - G · · - Exam 3 - Word Lori Leh

ID: 3148358 • Letter: Y

Question

y (0) D = 12/4/2017 6:59 PM Q H o 11 6 A 9 wa A - G · · - Exam 3 - Word Lori Lehman 1 - 0 X File Home Insert Draw Design Layout References Mailings Review View Help V Tell me what you want to da Share XCut Calibri (Body- 11 - A A' Aa-A, =.=.=.9. P Find - " 2. "AD D bc ABbc:c: Aa Bbccc ADa A Be Copy Replace Paste *Format Painter er B I U - abre x,xA - ay -A- E=== IS--E 1 Norrmal 1 No Spac. Heading 1 Heading 2Title Subtitle Subtle Em. Emphasis Select - Styles to da - E Clipboard Font GEditing A company manufactures products A, B, and C. Each product is processed in three departments: I, II, and III. The total available labor hours per week for Departments I, II, and ill are 900, 1080, and 840, respectively. The time requirements (in hours per unit) and profit per unit for each product are as follows: Product A Product B Product C Department | Department || Department lli Profit 3 | 2 | $18 | $12 $15 Use the simplex method to find how many units of each product should the company produce to maximize its profit? What is the largest profit the company can realize? Are there any resources left over? Page 1 of 1 243 words IF E = - —+ 110%

Explanation / Answer

  

The problem is

Maximize Z = 18 x1 + 12x2 + 15x3

Subject to the constraints,

2x1 + x2+ x3 <= 900

3x1+ x2 + 2x3 <= 1080

2x1+2x2+x3 <= 840

x1,x2,x3>= 0.

After introducing slack variables, the problem is

Maximize Z = 18 x1 + 12x2 + 15x3 +os1+0s2+0s3

Subject to the constraints,

2x1 + x2+ x3 +s1    = 900

3x1+ x2 + 2x3    +s2    = 1080

2x1+2x2+x3    +s3    = 840

x1,x2,x3, s1,s2,s3 >= 0.

Maximum Cj-Zj is 18 and the corresponding column is x1. So the entering variable is x1

Minimum ratio is 360 and the corresponding row variable is s2. So s2 leave the basis.

Maximum Cj-Zj is 6 and it corresponds to column x2. So x2 enters the basis.

Minimum ratio is 90 , which corresponds to s3. So s3 leaves the basis. Max. Z = 6480.

Maximum Cj-Zj is 9/2 and it corresponds to column x3. So x3 enters the basis.

Minimum ratio is 440 , which corresponds to x1. So x1 leaves the basis. Max. Z =7020

Since all Cj- Zj <= 0, Solution is optimal.

Optimal solution is x1=0, x2 = 200; x3 = 440.

Maximum Z = 9000.

Hence,

Largest profit = 9000.

Resource that is left over is Product A.

Iteration1 cj 18 12 15 0 0 0 B Cb Xb x1 x2 x3 s1 s2 s3 Min.ratio = Xb/x1 s1 0 900 2 1 1 1 0 0 400 s2 0 1080 3 1 2 0 1 0 360 s3 0 840 2 2 1 0 0 1 420 Zj 0 0 0 0 0 0 Cj-Zj 18 12 15 0 0 0