Question

QUESTION TWO A. Fiifi Auto mobile company manufactures cars and trucks. Each vehicle must be processed in the paint shop and body assembly shop. If the paint shop were only painting trucks, then 80 per day could be painted. If the paint shop were only painting cars, then 120 per day could be painted. If the body shop were only producing cars, then it could process 100 per day. If the body shop were only producing trucks, then it could process 100 per day. Each truck contributes GHS60,000 to profit, and each car contributes GHS40,000 to profit. Use linear programming to determine a daily production schedule that will maximize the company's profits. (10 marks) B. Macro farms Ltd can purchase two types of fertilizer which contain the following percentage Nitrate Phosphates Potash Type X 18 5 2 Type Y3 2 5 For certain type of crop, the following minimum quantities (kg) are required Nitrate Phosphate Potash 50 40 Type X cost GHS10 per kg and type Y cost GHS5 per kg. The company currently buys 1,000 kg of each type and wishes to minimize it expenditure on fertilizers. Required a. Construct a linear programming model for the above problem (5 marks) b. Solve the model graphically (5 marks) 100 Page 2 of 3 c. What is the minimum expenditure that can be incurred on fertilizer? (5 marks) d. Find the savings the company can make by switching from it current policy to your recommendation (5 marks) (Total: 30 marks)

Answer 1

A.

Linear Programming model is formulated as under:

Let T be the number of Trucks and C be the number of Cars to be produced daily

Total_Profit) Max 60000T+40000C

s.t.

Paint_Shop) (1/80)T+(1/120)C <= 1 or, 3T+2C<=240

Body_Shop) (1/100)T+(1/100)C <= 1 or, 1T+1C<=100

T, C >= 0

Solution by LINDO is as follows:

Lingo 18.0 - Solution Report - Lingo1 File Edit Solver Window Help 2 Lindo Model - Lingo1 Solution Report - Lingo1 Global optimal solution found. Objective value: Infeasibilities: Total solver iterations: Elapsed runtime seconds: Total_Profit) Max 60000T+40000C 4800000. 0.000000 1 0.04 s.t. Paint_Shop) 3T+2C<=240 Model Class : LP Body_Shop) 1T+10<=100 Total variables: Nonlinear variables: Integer variables: 2 0 0 Total constraints: Nonlinear constraints: 3 0 Total nonzeros: Nonlinear nonzeros: 6 0 Variable Value 80.00000 0.000000 Reduced Cost 0.000000 0.000000 с Row TOTAL PROFIT PAINT SHOP BODY SHOP Slack or Surplus 4800000. 0.000000 20.00000 Dual Price 1.000000 20000.00 0.000000

.

Optimal daily production mix:

Trucks = 80

Cars = 0

Daily profit = 4,800,000

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B)

a)

Linear Programming model is following:

Min 10X+5Y (total cost)

s.t.

0.18X+0.03Y >= 100   (nitrate)

0.05X+0.02Y >= 50 (phosphate)

0.02X+0.05Y >= 40   (potash)

X, Y >= 0

.

b)

Solution by graphical method

3500 + 10x + 5y = 10476.19 Х 10 (0,3333.333) Х .18x +.03y > 100 3000 3 05x +.02y > 50 Х 4 Х 02x +.05y > 40 5 2500 -2000 (238.095, 1904.762) -1500 1000 1 -500 (809.524, 476.19) (2000, 0) 500 1000 1500 2000 2500

Feasible region is the shaded unbounded region with extreme points at : (2000, ), (809.524, 476.19), (238.095, 1904.762), (0, 3333.333)

Value of objective function at each of the extreme points is:

x	y	z
2000	0	20000
809.524	476.19	10476.19
238.095	1904.762	11904.76
0	3333.333	16666.67

Minimum value of objective function is 10476.19 at extreme point (809.524, 476.19)

Optimal solution is:

X = 809.524 kg

Y = 476.19 kg

Total cost = GHS 10,476.19

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c)

Minimum expenditure that can be incurred on fertilizer = GHS 10,476.19

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d)

Total expenditure incurred in current policy = 10*1000+5*1000 = Ghs 15000

Savings by switching from current policy to recommended policy = 15000 - 10476.19

= Ghs 4523.81