Question

Problem 4-19 (Algorithmic)

The Silver Star Bicycle Company will be manufacturing both men’s and women’s models of its Easy-Pedal 10-speed bicycles during the next two months. Management wants to develop a production schedule indicating how many bicycles of each model should be produced in each month. Current demand forecasts call for 150 men’s and 125 women’s models to be shipped during the first month and 220 men’s and 170 women’s models to be shipped during the second month. Additional data are shown:

	Production	Labor Requirements (hours)		Current
Model	Costs	Manufacturing	Assembly	Inventory
Men's	$120	2.2	1.3	20
Women's	$90	1.5	1.1	30

Last month the company used a total of 1000 hours of labor. The company’s labor relations policy will not allow the combined total hours of labor (manufacturing plus assembly) to increase or decrease by more than 100 hours from month to month. In addition, the company charges monthly inventory at the rate of 2% of the production cost based on the inventory levels at the end of the month. The company would like to have at least 25 units of each model in inventory at the end of the two months.

1. Establish a production schedule that minimizes production and inventory costs and satisfies the labor-smoothing, demand, and inventory requirements. What inventories will be maintained, and what are the monthly labor requirements? If required, round your answers to the nearest whole number.
   
   Production schedule:
   
   The optimal solution is to produce  of the men's model in month 1,  of the men's model in month 2,  units of the women's model in month 1, and of the women's model in month 2.
   
   Total Cost = $  
   

   Inventory Schedule
   Month 1	Men's	Women's
   Month 2	Men's	Women's

   Labor Levels
   Previous month	hours
   Month 1	hours
   Month 2	hours

2. If the company changed the constraints so that monthly labor increases and decreases could not exceed 50 hours, what would happen to the production schedule? If required, round your answers to the nearest whole number.
   
   Production schedule:
   
   The optimal solution is to produce  of the men's model in month 1,  of the men's model in month 2,  units of the women's model in month 1, and of the women's model in month 2.
   
   Total Cost = $  
   
   How much will the cost increase?
   
   $ ___________

Answer 1

Decision variables:

x11 = number of men's model produced in month 1

x21 = number of women's model produced in month 1

x12 = number of men's model produced in month 2

x22 = number of women's model produced in month 2

y11 = inventory of men's model at the end of month 1

y12 = inventory of women's model at the end of month 1

y12 = inventory of men's model at the end of month 2

y22 = inventory of women's model at the end of month 2

Labour hours per unit of men's model = manufacturing hours per unit + assembly hours per unit

= 2.2 + 1.3 = 3.5 hrs per unit

Labour hours per unit of women's model = manufacturing hours per unit + assembly hours per unit

= 1.5 + 1.1 = 2.6 hrs per unit

Minimize: Total production costs + Total inventory costs

= Production cost per unit*(Men's model produced in months 1 & 2) + 2%*production costs*(Men's model inventory in months 1 & 2) + Production cost per unit*(Women's model produced in months 1 & 2) + 2%*production costs*(Women's model inventory in months 1 & 2)

= 120*(x11 + x12) + 2.40*(y11 + y12) + 90*(x21 + x22) + 1.80*(y21 + y22)

Constraints:

1). 20 + x11 - y11 = 150 or x11 - y11 = 130 (Demand for men's model in month 1)

2). 30 + x21 - y21 = 125 or x 21 - y11 = 95 (Demand for women's model in month 1)

3). y11 + x12 - y12 = 220 (Demand for men's model in month 2)

4). y21 + x22 -y22 = 170 (Demand for women's model in month 2)

5). y12 >= 25 (inventory in month 2 has to be at least 25 units)

6). y22 >= 25 (inventory in month 2 has to be at least 25 units)

7). 3.50*x11 + 2.60*x21 <= 1100 (labour hours constraint for month 1)

8). 3.50*x11 + 2.60*x21 >= 900 (labour hours constraint for month 2)

9). 3.50*x11 + 2.60*x21 - 3.50*x12 - 2.60*x22 <= 100 (difference in labor hours has to be at most 100)

10). 3.50*x11 + 2.60*x21 - 3.50*x12 - 2.60*x22 >= -100 (difference in labor hours has to be at most 100)

11). x11, x12, x21, x22, y11, y12, y21, y22 >= 0

Solving this LP, we get:

Number of models produced
	Month 1	Month 2
Men	210	165
Women	95	195
Inventory
	Month 1	Month 2
Men	80	25
Women	-	25

Labor hours:

	Month 1	Month 2
Men hours	736.25	576.25
Women hours	247.00	507.00
Total labor hours	983.25	1,083.25

Total cost = 71,398

b). Change the hours constraints to 50:

7). 3.50*x11 + 2.60*x21 <= 1050 (labour hours constraint for month 1)

8). 3.50*x11 + 2.60*x21 >= 950 (labour hours constraint for month 2)

9). 3.50*x11 + 2.60*x21 - 3.50*x12 - 2.60*x22 <= 50 (difference in labor hours has to be at most 50)

10). 3.50*x11 + 2.60*x21 - 3.50*x12 - 2.60*x22 >= -50 (difference in labor hours has to be at most 50)

Solving this, we get:

Number of models produced
	Month 1	Month 2
Men	218	158
Women	95	195
Inventory (Sij)
	Month 1	Month 2
Men	88	25
Women	-	25

Labor hours:

	Month 1	Month 2
Men hours	761.25	551.25
Women hours	247.00	507.00
Total labour hours	1,008.25	1,058.25

Total cost = 71,415

Cost increases by 71,415 - 71,398 = 17