For the practical use of supply chain management powerful planning tools are needed to support production and logistics decisions at various planning levels. Instead of using an off-the-shelf planning tool, the company decided to develop its own linear programming based optimization model.

Scenario 1:

Empire State Ski Products makes downhill and cross- country skis. A pair of downhill skis requires 4 man-hours for cutting, 1 man-hour for shaping and 3 man-hours for finishing while a pair of cross- country skis requires 3 man-hours for cutting, 2 man-hours for shaping and 1 man-hour for finishing. Each day the company has available 240 man-hours for cutting, 140 man-hours for shaping and 150 man-hours for finishing. How many pairs of each type of ski should the company manufacture each day in order to maximize profit if a pair of downhill skis yields a profit of $12 and a pair of cross-country skis yields a profit of $10?

An optimization problem is one where you have to make the best decision . In optimization models then, the words “minimize” and “maximize” come up a lot when articulating an objective

 The decision variables  x1 =pairs of Downhill    x2=pairs of Cross  Country

 Objective function  Max Profit = 12x1 + 10x2

 Constraints   4x1 + 3x2 ≤ 240,  1x1 + 2x≤  140,   3x1 + 1x2  ≤ 150, x1 ≥ 0,  x2 ≥ 0

screen-shot-2016-11-03-at-2-40-40-pm

How many pairs of each type of ski should the company manufacture each day in order to maximize profit if a pair of downhill skis yields a profit of $12 and a pair of cross-country skis yields a profit of $10?

x1 =pairs of Downhill =12    x2=pairs of Cross Country=64

x1* 12  +  x2 * 10 = Maximize profit=$784

 


Scenario 2 :

A company manufactures four types of the same product and in the final part of the manufacturing process there are assembly, polishing and packing operations. For each variant the time required for these operations is shown below (in minutes) as is the profit per unit sold.

Suppose that the company is free to decide how much time to devote to each of the three operations (assembly, polishing and packing) within the total allowable time of 205,000 minutes. How many of each variant should the company make per year and what is the associated profit?

Alternative scenario: Given the current state of the labour force the company estimate that, each year, they have 110,000 minutes of assembly time, 45,000 minutes of polishing time and 50,000 minutes of packing time available. How many of each type should the company make per year and what is the associated profit?

if you had an extra 100 hours to which operation would you assign it?

if you had to take 50 hours away from polishing or packing which one would you choose?

iii. what would the new objective function value be in these two cases?

What should the profit of types A and D should be for the optimal production to be higher than zero?

xi be the number of units of variant i (i=1,2,3,4) made per year

Tass be the number of minutes used in assembly per year
Tpol be the number of minutes used in polishing per year
Tpac be the number of minutes used in packing per year

where x>= 0 i=1,2,3,4    and Tass, Tpol, Tpac >= 0

Constraints

(a) operation time definition

Tass = 2x+ 4x+ 3x+ 7x4
Tpol = 3x1 + 2x2 + 3x3 + 4x4
Tpac = 2x+ 3x2 + 2x+ 5x4

(b) Operation time limits

In the first situation, where the maximum time that can be spent on each operation is specified, We simply have:

Tass <= 110000 (assembly)
Tpol <= 45000    (polish)
Tpac <= 50000    (pack)

Tass + Tpol + Tpac <= 205000 (total time)

Objective   Max profit 1.5x1 + 2.5x2 + 3.0x3 + 4.5x4

Solving via Solver the solution is:screen-shot-2016-11-03-at-2-41-04-pm


Scenario 3 :

A furniture company manufactures desks and chairs. Each desk uses 4 units of wood, and each chair uses 3 units of wood. A desk contributes $400 to profit, and a chair contributes $250. Marketing restrictions require that the number of chairs produced be at least twice the number of desks produced. There are 2000 units of wood available.

Use Solver to maximize the company profit : Max profit $ 180,000

Confirm graphically that the solution in part a maximizes the company’s profit.

Use SolverTable to see what happens to the decision variables and profit when the availability of wood varies from 1000 to 3000 in 100 unit increments. Based on your findings how much would the company be willing to pay for each extra unit of wood over its current 2000 units? How much profit would the company lose if it lost any of its current 2000 units?

screen-shot-2016-11-03-at-2-41-12-pm screen-shot-2016-11-03-at-2-41-23-pm

 

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